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Riccati Equation General Riccati Equation The Riccati equation is one of the most interesting nonlinear differential equations of first order. It's written in the form: where a(x), b(x), c(x) are continuous functions of x. The Riccati equation is used in different areas of mathematics (for example, in algebraic geometry and the theory of conformal mapping), and physics. It also appears in many applied problems. The differential equation given above is called the general Riccati equation. It can be solved with help of the following theorem: Theorem: If a particular solution y 1 of a Riccati equation is known, the general solution of the equation is given by Indeed, substituting the solution y = y 1 + u into Riccati equation, we have

Riccati Equation

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Page 1: Riccati Equation

Riccati Equation General Riccati Equation

The Riccati equation is one of the most interesting nonlinear differential equations of first order. It's written in the form:

where a(x), b(x), c(x) are continuous functions of x.

The Riccati equation is used in different areas of mathematics (for example, in algebraic geometry and the theory of conformal

mapping), and physics. It also appears in many applied problems.

The differential equation given above is called the general Riccati equation. It can be solved with help of the following theorem:

Theorem: If a particular solution y1 of a Riccati equation is known, the general solution of the equation is given by

Indeed, substituting the solution  y = y1 + u into Riccati equation, we have

The underlined terms in the left and in the right side can be canceled because y1 is a particular solution satisfying the equation. As a

result we obtain the differential equation for the function u(x):

Page 2: Riccati Equation

which is a Bernoulli equation. Substitution of  z = 1/u converts the given Bernoulli equation into a linear differential equation that

allows integration.

Besides the general Riccati equation, there is an infinite number of particular cases of Riccati equation at certain coefficients of a(x),

b(x), and c(x). Many of these particular cases have integrable solutions.

Returning to the general Riccati equation, we see that we can construct the general solution if a particular solution is known.

Unfortunately, there is no sctrict algorithm to find the particular solution, which depends on the types of the functions a(x), b(x), and

c(x).

Below we consider some well known particular cases of the Riccati equation. Special Case 1: Coefficients a, b, c are constants.

If the coefficients in the Riccati equation are constants, this equation can be reduced to a separable differential equation. The solution

is described by the integral of a rational function with a quadratic function in the denominator:

This integral can be easily calculated at any values of a, b and c (For more information see "Integration of Rational Functions"). Special Case 2: Equation of type y' = by 2 + cx n

Consider a Riccati equation of type y' = by 2 + cxn, where the function a(x) at the linear term is zero, the coefficient b at y2 is a constant,

and c(x) is a power function:

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This case of Riccati equation has nice solutions!

First of all, if n = 0, we get the Case 1 where the variables are separated and the differential equation can be integrated.

If n = −2, the Riccati equation is converted into a homogeneous equation with help of the substitution y = 1/z and then also can be

integrated.

This differential equation can be also solved at

Here the general solution is expressed through cylinder functions.

At all other values of the power n, the solution of the Riccati equation can be expressed through integrals of elementary functions. This

fact was discovered by the French mathematician Joseph Liouville (1809-1882) in 1841.

Many other special cases of the Riccati equation are presented on the EqWorld website.

   Example 1 Solve the differential equation  y' = y + y2 + 1.

Solution.

The given equation is a simple Riccati equation with constant coefficients. Here the variables x, y can be easily separated, so the

general solution of the equation is given by

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   Example 2

Solve the Riccati equation .

Solution.

We will seek for a particular solution in the form:

     

Substituting this into the equation, we obtain:

     

We get a quadratic equation for c:

     

We can take any value of c. For example, let c = 2. Now, when the particular solution is known, we make the replacement:

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Now substitute this into the original Riccati equation:

     

As it can be seen, we have a Bernoulli equation with the parameter m = 2. Make one more substitution:

     

Divide the Bernoulli equation by z2 (assuming that z ≠ 0) and rewrite it in terms of v:

     

The last equation is linear and can be easily solved using the integrating factor:

     

The general solution of the linear equation is given by

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From now on we will subsequently return back to the previous variables. Since z = 1/v, the general solution for z is written as follows:

     

Hence,

     

We can rename the constant: 3C = C1 and write the answer in the form:

     

where C1 is an arbitrary real number.

   Example 3 Find the general solution of the differential equation  x3y' + x2y − y2 = 2x4.

Solution.

Transform this equation to the standard form:

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As it can be seen, we have a Riccati equation. Try to find a particular solution in the form  y1 = cx2. Substituting this into the Riccati

equation, we can determine the coefficient c:

     

Solving this quadratic equation, we obtain the value of c::

     

Thus, there are even two particular solutions. However, we need only one of them. So we take, for example,  y1 = x2.

As a result, we can write the general solution of the Riccati equation in the form:

     

We get the following differential equation for the new function u(x):

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which is a Bernoulli equation. The substitution of converts it to a linear differential equation:

     

Then we calculate the integrating factor to solve the linear equation:

     

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One can take the function v(x) = x as the integrating factor. Indeed, we can check that the left side of the equation becomes equal to the

derivative of the product of z(x)v(x) after multiplying by v(x) = x:

     

Since z = 1/u, the function u(x) is defined by the formula

     

Hence, the general solution of the original Riccati equation is given by

     

where C is a constant.

   Example 4

Solve the equation .

Solution.

As it can be seen, this is a special Riccati equation of type  y' = by2 + cxn with the degree n = −2.

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By making the substitution y = 1/z we can convert the equation to a homogeneous one and then integrate.

Let . Then

     

To solve the homogeneous equation we make one more substitution:  z = tx,  z' = t'x + t. Hence,

     

The trinomial in the denominator of the left side can be factored as follows:

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so we may use partial decomposition to simplify the integrand:

     

As a result, we have:

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Rename the constant: , so the solution for the function t(x) will have the form:

     

Remember that . Therefore,

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Returning to the variable y which is related to z by the relationship , we get

     

The last expression is the general solution of the Riccati equation in the implicit form. Here the constant C is any real number. Indeed,

substituting C = 0, we see that this value also satisfies the differential equation:

     

Hence,

     

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Implicit Differential Equations Definition and Methods of Solution

An equation of type

where F is a continuous function, is called the first order implicit differental equation. If this equation can be solved for y', we get one

or several explicit differental equations of type

that can be solved by methods covered in other sections.

Further we suppose that the differential equation can not be solved in the explicit form, so we should use different methods. The main

techniques for solving an implicit differential equation is the method of introducing a parameter. Below we show how this method

works to find the general solution for some most important particular cases of implicit differential equations.

Here we note that the general solution may not cover all possible solutions of a differential equation. Besides the general solution, the

differential equation may also have so-called singular solutions. We consider this in more detail on the page Singular Solutions of

Differential Equations. Case 1. Implicit Differential Equation of Type x=f(y,y').

In this case, the variable x is expressed explicitly through y and the derivative y'. Introduce the parameter . Differentiate the

equation x = f(y,y') with respect to y. This produces:

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Since , the last expression can be written as follows:

We obtain an explicit differential equation such that its general solution is given by the function

where C is a constant.

Thus, the general solution of the original implicit differential equation is defined in the parametric form by the system of two algebraic

equations:

If the parameter p can be eliminated from the system, the general solution is given in the explicit form x = f(y,C). Case 2. Implicit Differential Equation of Type y=f(x,y').

Here we consider a similar case, when the variable y is an explicit function of x and y'. Introduce the parameter and

differentiate the equation y = f(x,y') with respect to x. As a result, we have:

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Solving the last differential equation, we get the algebraic equation g(x,p,C) = 0. Together with the original equation, they form the

following system of equations:

which is the general solution of the given differential equation in the parametric form. In some cases, when the parameter p can be

eliminated from the system, the general solution can be written in the explicit form y = f(x,C). Case 3. Implicit Differential Equation of Type x=f(y').

Here the differential equation does not contain the variable y. Using the parameter , it's easy to construct the general

solution of the equation. Since dy = pdx and

then

Integrating the last equation gives the general solution in the parametric form:

Case 4. Implicit Differential Equation of Type y=f(y').

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The equation of this kind does not contain the variable x and can be solved the similar way. Using the parameter , we can

write Then it follows from the equation that

Integrating the last expression gives the general solution of the original implicit equation in parametric form:

   Example 1 Find the general solution of the equation  g(y')2 − 4x = 0.

Solution.

This equation is of type x = f(y') (Case 3). Introduce the parameter p = y' and write the equation in the form:

     

By taking differentials of both sides, we obtain:

     

Since dy = pdx, the last expression can be presented as

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By integrating we find the dependence of the variable y on the parameter p:

     

where C is a constant.

Thus, we get the general solution of the equation in parametric form:

     

We can eliminate the parameter p from this system. It follows from the second equation that

     

Substituting this in the first equation, we obtain the general solution as the explicit function y = f(x):

     

   Example 2 Find the general solution of the differential equation  y = ln(25 + (y')2) = 0.

Solution.

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This differential equation is related to Case 4 because it contains only the variable y and its derivative y'. Using the parameter p we

rewrite this equation in the following way:

     

Take the differentials of both sides:

     

Since dy = pdx, we get

     

Now we can integrate the last expression to obtain x as function of p.

     

So we have the following parametric representation of the solution of the differential equation:

     

where C is an arbitrary constant.

   Example 3 Solve the differential equation  2y = 2x2 + 4xy' + (y')2.

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Solution.

This equation is related to Case 2. Let  y' = p, so we can rewrite the equation as

     

Find the differentials of both sides taking into acount that dy = pdx. This yields:

     

We have two solutions that satisfy the last equation:

     

Hence,

     

By integrating this simple equation, we obtain:

     

where C is a constant. To determine the value of C, we substitute this answer in the original differential equation:

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We see that the constant C must be equal to zero to satisfy the equation. Thus, the first solution is

     

Now we consider the second solution, which is defined by the differential equation

     

Then

     

At the beginning of the solution we have written the differential equation in the form

     

We can substitute the known expression for x (as a function of the parameter p) to find the dependence of y on p:

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Thus, the second solution is given parametrically by the following system:

     

where C is a constant. So, the final answer is

     

Singular Solutions of Differential Equations Definition of Singular Solution

A function φ(x) is called the singular solution of the differential equation F(x,y,y') = 0, if uniqueness of solution is violated at each

point of the domain of the equation. Geometrically this means that more than one integral curve with the common tangent line passes

through each point (x0,y0).

Note: Sometimes the weaker definition of the singular solution is used, when the uniqueness of solution of differential equation may be

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violated only at some points.

A singular solution of a differential equation is not described by the general integral, that is it can not be derived from the general

solution for any particular value of the constant C. We illustrate this by the following example:

Suppose that the following equation is required to be solved:  (y')2 − 4y = 0. It is easy to see that the general solution of the equation is

given by the function  y = (x + C)2. Graphically, it's represented by the family of parabolas (Figure 1).

Fig.1

Besides this, the function  y = 0 also satisfies the differential equation. However, this function is not contained in the general solution!

Since more than one integral curve passes through each point of the straight line  y = 0, the uniqueness of solution is violated on this

line, and hence it is a singular solution of the differential equation. p-discriminant

One of the ways to find a singular solution is investigation of the so-called p-discriminant of the differential equation. If the function

F(x,y,y') and its partial derivatives and are continuous in the domain of the differential equation, the singular solution can be

found from the system of equations:

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The equation ψ(x,y) = 0 obtained by solving the given system of equations is called the p-discriminant of the differential equation. The

corresponding curve determined by this equation is called a p-discriminant curve. Upon finding the p-discriminant curve, one should

check the following:

1. Whether it is a solution of the differential equation?

2. Whether it is a singular solution, i.e. are there any other integral curves of the differential equation that touch the p-discriminant

curve at each point?

This can be done as follows:

Find the general solution of the differential equation (denote it by y1);

Write the conditions of touching the singular solution (denote it by y2) and the general solution y1 at an arbitrary point x0:

        

If the system has a solution at the arbitrary point x0, the function y2 is a singular solution. The singular solution usually corresponds to

the envelope of the family of integral curves of the general solution of the differential equation. Envelope of the Family of Integral Curves and C-discriminant

Another way to find a singular solution as the envelope of the family of integral curves is based on using C-discriminant.

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Let Φ(x,y,C) be the general solution of a differential equation F(x,y,y') = 0. Graphically the equation Φ(x,y,C) = 0 corresponds to the

family of integral curves in the xy-plane. If the function Φ(x,y,C) and its partial derivatives are continuous, the envelope of the family

of integral curves of the general solution is defined by the system of equations:

To make sure whether a solution of the system of equations is really the envelope, one can use the method mentioned in the previous

section. General Algorithm of Finding Singular Points

A more common way of finding singular points of a differential equation is based on the simultaneous using p-discriminant and C-

discriminant.

Here first we find the equations of the p-discriminant and C-discriminant:

ψp(x,y) = 0 is the equation of the p-discriminant;

ψC(x,y) = 0 is the equation of the C-discriminant;

It turns out that these equations have a certain structure. In general case, the equation of the p-discriminant can be factored into the

product of three functions:

where E means the equation of the Envelope, T is the equation of the Tac locus, and C is the equation of the Cusp locus.

Similarly, the equation of the C-discriminant can be also factored into the product of three functions:

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where E is the equation of the Envelope, N is the equation of the Node locus, and C is the equation of the Cusp locus.

Here we meet with new kinds of singular points: C - Cusp loci, T - Tac loci, and N - Node loci. Their view in the xy-plane is shown

schematically in Figures 2-4.

Fig.2 Fig.3

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Fig.4 Fig.5

Three of the four types of points, namely, the Tac loci, Cusp loci and Node loci are extraneous points, i.e. they do not satisfy the

differential equation and, therefore, they are not singular solutions of the differential equation. Only the envelope of the considered

points is the singular solution. Since the envelope is presented in the equations of the both discriminants as a first degree factor, this

allows to find the equation of the envelope.

   Example 1

Find the singular solutions of the equation .

Solution.

We will use p-discriminant for investigation of the singular points. Differentiating the equation with respect to y' gives:

     

Putting this into the differential equation yields the equation of the p-discriminant:

     

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It follows from here that the equation of the p-discriminant describes two horizontal lines: . It is easy to check that this solution

satisfies the given differential equation:

     

Now we find the general solution of the differential equation. We can write it in the following form:

     

Make the replacement:

     

As a result, we get:

     

After integration we obtain the general solution of the differential equation:

     

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where C is an arbitrary constant.

The last expression can be written as follows:

     

This equation describes the family of circles of radius 1, filling in the band  −1 ≤ y ≤ 1 (Figure 5). As it can be seen from the Figure,

the p-discriminant lines are the envelopes for the given circles. However, we must prove formally that the uniqueness of

solution is violated on these straight lines.

Take an arbitrary point x0. Write the condition of touching two integral curves at this point:

     

Here y1(x) denotes our general solution that has the form (for the upper semicircle):

     

The function y2(x) corresponds to the horizontal line  y = 1. Both the lines will touch at the point x0 if only the following relations are

satisfied:

     

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The given conditions are satisfied, if we set C = −x0.

Thus, we have proved that at each point x0 on the straight line  y = 1 there exists a touching circle with  C = −x0. Hence, the

uniqueness of solution is violated at each point of the straight line. Therefore, the line  y = 1 is a singular solution of the given

differential equation. Similarly, we can prove that the line  y = −1 is also a singular solution.

   Example 2 Find the singular solution of the differential equation  y = (y')2 − 3xy' + 3x2. The general solution of the equation is known and given by

the function  y = Cx + C2 +x2.

Solution.

We will use C-discriminant to determine the singular solution. Since the general solution of the differential equation is known, we can

write:

     

The partial derivatives in C is

     

We get the following system of equations:

     

It follows from the second equation that . Substituting this in the first equation, we find the C-discriminant curve, which is a

parabola:

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Make sure that this function is a solution of the original differential equation:

     

Now we check that uniqueness of solution is violated on this curve. Denote:

     

Write the conditions of touching the two curves at a certain arbitrary point x0 as

     

As a result, we have:

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The given system of equations is consistent when the constant C at each point x0 is equal to

     

Thus, we have proved that the C-discriminant curve is the envelope (i.e. the singular solution) for the family of parabolas  y =

Cx + C2 +x2 representing the general solution of the differential equation.

   Example 3 Investigate the singular solutions of the differential equation  (y')2(1 − y)2 = 2 − y.

Solution.

First we find p-discriminant of the given equation. Differentiating with respect to x gives:

     

Eliminate y' from the system of the equations:

     

Then we obtain:

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Now we determine C-discriminant. Unfortunately, to find it, we need to solve the differential equation and find its general solution :(.

Rewrite the equation in the following form:

     

By integrating both sides we have:

     

We change the variable in the left integral:

     

This yields:

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Differentiate the general solution with respect to C:

     

Putting (x + C) = 0 back into the general solution, we obtain the equation of the C-discriminant:

     

Now we can write both discriminants together:

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As it follows from the structure of the discriminants, the equation 2 − y = 0 is the equation of the envelope because it's contained in the

both discriminants as the first degree factor. We can also find the equation of the Tac locus from the expression for p-discriminant:

     

And similarly, as it follows from the expression for C-discriminant, the equation the Node locus is given by

     

In the given example, only the envelope y = 2 is the singular solution of the differential equation. Lagrange and Clairaut Equations Lagrange Equation

A differential equation of type

where φ(y') and ψ(y') are known functions differentiable on a certain interval, is called the Lagrange equation.

By setting  y' = p and differentiating with respect to x, we get the general solution of the equation in parametric form:

under condition of

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where p is a parameter.

Lagrange equation may also have a singular solution if the condition φ(p) − p ≠ 0 is failed. The singular solution is given by the

expression:

where c is the root of the equation φ(p) − p = 0. Clairaut Equation

The Clairaut equation has the form:

where ψ(y') is a nonlinear differentiable function. The Clairaut equation is a particular case of the Lagrange equation when φ(y') = y'. It

is solved in the same way by introducing a parameter. The general solution is given by

where C is an arbitrary constant.

Similarly to the Lagrange equation, the Clairaut equation may have a singular solution that is expressed parametrically in the form:

where p is a parameter.

   Example 1 Find the general and singular solutions of the differential equation  y = 2xy' − 3(y')2.

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Solution.

Here we see that we deal with a Lagrange equation. We will solve it using the method of differentiation.

Denote  y' = p, so the equation is written in the form:

     

Differentiating both sides, we find:

     

We can replace dy with pdx:

     

By dividing by p, we can write the following equation (later we check if p = 0 is a solution of the original equation):

     

As it can be seen, we obtain a linear equation for the function x(p). The integrating factor is

     

The general solution of the linear equation is given by

     

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Substituting this expression for x into the Lagrange equation, we obtain:

     

Thus, the general solution in parametric form is defined by the system of equations:

     

Besides that, the Lagrange equation can have a singular solution. Solving the equation  φ(p) − p = 0, we find the root:

     

Hence, the singular solution is expressed by the linear function:

     

   Example 2 Find the general and singular solutions of the equation  2y − 4xy' − ln y' =0.

Solution.

Here we have a Lagrange equation. By setting  y' = p, we can write:

     

Differentiate both sides of the equation:

     

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Since dy = pdx, we get

     

When we divided by p, we lost the root p= 0, which corresponds to the solution y= 0.

Thus, we get a linear differential equation for the function x(p). We solve it using the integrating factor:

     

The function x(p) is defined by

     

Substituting this into the original equation, we find the parametric expression for y:

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Hence, the general solution in parametric form is written as follows:

     

To find the singular solution, we solve the equation:

     

It follows from this that y = C. We can make direct substitution to make sure that the constant C must be equal to zero.

Thus, the differential equation has the singular solution y = 0. We have already met with this solution above when we divided the

equation by p.

   Example 3 Find the general and singular solutions of the differential equation  y = xy' + (y')2.

Solution.

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This is a Clairaut equation. By setting  y' = p we write it in the form

     

Differentiating with respect to x, we have

     

Replace dy with pdx to obtain:

     

By equating the first factor to zero, we have

     

Now we substitute this into the differential equation:

     

Thus, we obtain the general solution of the Clairaut equation, which is an one-parameter family of straight lines.

By equating the second term to zero we find that

     

This gives us the singular solution of the differential equation in parametric form:

     

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By eliminating p from this system, we get the equation of the integral curve:

     

From geometric point of view, the curve is the envelope of the family of straight lines defined by the general solution (see

Figure 1).

Fig.1 Fig.2   Example 4

Find the general and singular solutions of the differential equation .

Solution.

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As it can be seen, this is a Clairaut equation. Introduce the parameter  y' = p:

     

Differentiating both sides with respect to x, we get:

     

Since dy = pdx, one can write:

     

Consider the case dp = 0. Then p = C. Substituting this in the equation, we find the general solution:

     

Graphically, this solution corresponds to the family of one-parameter straight lines.

The second case is described by the equation . Find the corresponding parametric expression for y:

     

The parameter p can be eliminated from the formulas for x and y. Squaring and adding these equations, we obtain:

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The last expression is the equation of the circle with radius 1 and centered at the origin. Thus, the singular solution is represented by

the unit circle on the xy-plane, which is the envelope of the family of the straight lines (Figure 2).

Differential Equations of Plane Curves As is known, the solution of a differential equation is displayed graphically as a family of integral curves. It turns out that one can also

solve the inverse problem: construct a differential equation of the family of plane curves defined by an algebraic equation!

Suppose that a family of plane curves is described by the implicit one-parameter equation:

We assume that the function F has continuous partial derivatives in x and y. To write the corresponding differential equation of first

order, it's necessary to perform the following steps:

1. Differentiate F with respect to x considering y as a function of x:

2. Solve the system of equations:

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by eliminating the parameter C from it.

If a family of plane curves is given by the two-parameter equation

we should differentiate the last formula twice by considering y as a function of x and then eliminating the parameters C1 and C2 from

the system of three equations.

The similar rule is applied to the case of n-parametric family of plane curves.

   Example 1 Determine the differential equation for the family of curves defined by the equation  y = e x+C.

Solution.

Differentiating the given equation with respect to x gives:

     

We can easily eliminate the parameter C from the system of equations:

     

As a result, we obtain the following simplest homogeneous equation:

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   Example 2 Derive the differential equation for the family of plane curves defined by the equation  y = x2 − Cx.

Solution.

Differentiate the given implicit equation with respect to x:

     

Write this equation jointly with the original algebraic equation and eliminate the parameter C:

     

As a result, we obtain the implicit differential equation corresponding to the given family of plane curves.

   Example 3 Write the corresponding differential equation for the family of plane curves defined by the equation  y = cot(x − C).

Solution.

By differentiating the given equation with respect to x, we obtain:

     

Notice that

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Therefore we can write:

     

Hence, we get the following differential equation describing the given family of curves:

     

   Example 4

A family of curves is given by the expression , where C is a parameter, α is an arbitrary angle. Determine the

differential equation for this family of plane curves.

Solution.

First, we differentiate the equation with respect to the variable x assuming that y is a function of x:

     

Eliminate C from the system of two equations:

     

To make this we square both sides of the equation and then add them together:

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Inserting the found expression for C into the differential equation, we have:

     

Thus, the family of plane curves is described by the following implicit differential equation:

     

   Example 5 Derive the differential equation for the family of two-parameter plane curves  y = C1x2 + C2x.

Solution.

We differentiate the given equation twice with respect to x and write the following system of three equations:

     

Express C1 from the last equation and substitute it into the first two equations:

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Now we can express C2 through the derivatives of y and put this into the first equation to get the final differential equation:

      Orthogonal Trajectories Definition and Examples

Let a family of curves be given by the equation

where C is a constant. For the given family of curves, we can draw the orthogonal trajectories, i.e. another family of curves f(x,y) = C

that cross the given curves at right angles.

For example, the orthogonal trajectory of the family of straight lines defined by the equation y = kx, where k is a parameter (the slope

of the straight line), is any circle having center at the origin (Figure 1):

where R is the radius of the circle.

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Fig.1 Fig.2

Similarly, the orthogonal trajectories of the family of ellipses

are confocal hyperbolas satisfying the equation:

Both families of curves are sketched in Figure 2. Here a and b play the role of parameters describing the family of ellipses and

hyperbolas, respectively. General Method of Finding Orthogonal Trajectories

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The common approach for determining orthogonal trajectories is based on solving the partial differential equation:

where the symbol ∇ means the gradient of the function f(x,y) or g(x,y) and the dot means the dot product of the two gradient vectors.

Using the definition of gradient, one can write:

Hence, the partial differential equation is written in the form:

Solving the last PDE, we can determine the equation of the orthogonal trajectories f(x,y) = C. A Practical Algorithm for Constructing Orthogonal Trajectories

Below we describe an easier algorithm for finding orthogonal trajectories f(x,y) = C of the given family of curves g(x,y) = C using only

ordinary differential equations. The algorithm includes the following steps:

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1. Construct the differential equation G(x,y,y') = 0 for the given family of curves g(x,y) = C. See the web page Differential

Equations of Plane Curves about how to do this.

2. Replace y' with in this differential equation. As a result, we obtain the differential equation of the orthogonal

trajectories.

3. Solve the new differential equation to determine the algebraic equation of the family of orthogonal trajectories f(x,y) = C.

   Example 1 Find the orthogonal trajectories of the family of straight lines  y = Cx, where C is a parameter.

Solution.

We apply the algorithm described above.

1) First, we construct the differential equation for the family of straight lines  y = Cx. By differentiating the last equation with respect

to x, we get:

     

Eiminate the constant C from the system of equations:

     

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We obtain the differential equation of the initial set of straight lines.

2) Replace y' with . This gives the differential equation of the orthogonal trajectories:

     

3) Now we solve the last differential equation to find the algebraic equation of the family of orthogonal trajectories:

     

By replacing 2C with R2 we see that the orthogonal trajectories for the family of straight lines are concentric circles (Figure 1):

     

   Example 2

A family of hyperbolic curves is given by the equation . Find the orthogonal trajectories for these curves.

Solution.

1) Determine the differential equation for the given family of hyperbolas. Differentiating the equation with respect to x gives:

     

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Now we eliminate the parameter C from the system of two equations:

     

It follows from the first equation that C = xy. Substituting into the second equation yields:

     

2) Replace y' with :

     

3) Now we integrate the differential equation of the orthogonal trajectories:

     

In the last equation we replaced 2C with just a constant C. Thus, we have obtained the equation of the family of orthogonal

trajectories. As it can be seen, these orthogonal trajectories are also hyperbolas. Both the families of hyperbolas are shown

schematically in Figure 3.

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Fig.3 Fig.4   Example 3 Find the orthogonal trajectories of the family of curves given by the power function  y =Cx4.

Solution.

1) Determine the differential equation for the given family of power curves:

     

By solving the system of two equations and eliminating C, we get:

     

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2) Replacing y' with gives:

     

The last expression is the differential equation of the orthogonal trajectories.

3) By integrating we can find the algebraic equation of the orthogonal trajectories:

     

Divide both sides by 2C:

     

We obtain the equation of the family of ellipses, which are the orthogonal trajectories for the given family of power curves  y =Cx4.

The ratio of the lengths of semiaxes for these ellipses is

     

Schematically the graphs of the families of curves are shown in Figure 4 above.

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   Example 4 Determine the orthogonal trajectories of the family of sinusoids  y =Csinx.

Solution.

1) Differentiating the given equation with respect to x gives:

     

By substituting we find the differential equation of the given sinusoidal curves:

     

2) Replace y' with to write the diferential equation of the orthogonal curves:

     

3) Now we can integrate this differential equation:

     

It follows from here that

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By denoting , we obtain the final implicit equation of the orthogonal trajectories:

     

Radioactive Decay In nature, there are a large number of atomic nuclei that can spontaneously emit elementary particles or nuclear fragments. Such a

phenomenon is called radioactive decay. This effect was studied at the turn of 19-20 centuries by Antoine Becquerel, Marie and Pierre

Curie, Frederick Soddy, Ernest Rutherford, and other scientists. As a result of the experiments, F.Soddy and E.Rutherford derived the

radioactive decay law, which is given by the differential equation:

where N is the amount of a radioactive material, λ is a positive constant depending on the radioactive substance. The minus sign in the

right side means that the amount of the radioactive material N(t) decreases over time (Figure 1).

The given equation is easy to solve, and the solution has the form:

To determine the constant C, it is necessary to indicate an initial value. If the amount of the material at the moment t = 0 was N0, then

the radioactive decay law is written as

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Fig.1 Fig.2

Further, we introduce two useful parameters that follow from the given law.

The half life or half life period T of a radioactive material is the time reguired to decay to one-half of the initial value of the material.

Hence, at the moment T:

The formula for the half life follows from here:

The average lifetime τ of a radioactive atom is given by

As it can be seen, the half life T and the average lifetime τ are related to each other by the formula:

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These 2 parameters vary widely for different substances. For example, the half life of Polonium-212 is less than 1 microseconds, but

the half life of Thorium-232 is more than 1 billion years. A wide range of isotopes with different half lives was thrown from the atomic

reactors and cooling pools in Chernobyl and Fukushima disasters (Figure 2).

   Example 1 Find the mass of a radioactive isotope if 3 half lives occured. The initial mass of the material was 80g.

Solution.

The mass of a radioactive material decreases as a result of decay twice after each half life. So, after 3 half lives the quantity of the

material will be of the initial amount. Hence, the mass after decay is .

   Example 2 The initial mass of an Iodine isotope was 200g. Determine the Iodine mass after 30 days if the half life of the isotope is 8 days.

Solution.

According to the radioactive decay law the mass of an isotope depends on time as follows:

     

Here the decay constant λ is equal to

     

Calculate the mass of the Iodine isotope in 30 days:

     

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   Example 3 The radioactive isotope Indium-111 is often used for diagnosis and imaging in nuclear medicine. Its half life is 2.8 days. What was the

initial mass of the isotope before decay, if the mass in 2 weeks was 5g?

Solution.

Using the radioactive decay law, we can write:

     

Solve this equation for N0:

     

Substituting the known values T = 2.8 days, t = 14 days, and N(t = 14) = 5g, we have:

     

   Example 4 Find the half life of a radioactive element, if its activity decreases for 1 month by 10%.

Solution.

Activity of an isotope is measured by the number of nuclei decaying for a time unit. Suppose that dNd nuclei decay for a short period of

time. Then the isotope activity A is expressed by the formula

     

It follows from the radioactive decay law that

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where N(t) is the quantity of the remaining nuclei. Therefore,

     

By differentiating with respect to t, we find the expression for activity:

     

The initial isotope activity is equal to

     

Hence,

     

As it can be seen, the activity decreases over time by the same law as the amount of undecayed material. Substituting the expression

for the half life in the last formula, we can write:

     

The value of T can be found from the last expression:

     

In our case, the half life period of the given isotope is

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Barometric Formula In this section we derive how the gas pressure P depends on the height over sea level h in the gravitational field of Earth.

If we take an arbitrary gas column with intersection area S and height h, then the weight of this column is given by

where ρ is the gas density. Then the gas pressure is expressed by the following formula:

Now imagine such a column in the atmosphere and separate a thin layer of air with the height dh (Figure 1). It's clear that such a layer

causes the pressure change by the value of

We have put the minus sign because the pressure must decrease as the altitude increases.

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Fig.1 Fig.2

Considering atmospheric air as an ideal gas, we can use the ideal gas law to express the density ρ through pressure P:

Here T is the absolute temperature, R is the universal gas constant equal to , M is the molar mass, which is for air equal

to . It follows from here that the density is given by the formula

Putting this into the differential relation for dP gives:

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We obtain a differential equation describing the gas pressure P as a function of the altitude h. Integrating gives the relation:

Getting rid of the logarithms, we obtain the so-called barometric formula

The constant of integration C can be determined from the initial condition P(h = 0) = P0, where P0 is the average sea level atmospheric

pressure.

Thus, dependency of the barometric pressure on the altitude is given by the formula

Substituting the known constant values (see Figure 2 above), we find that the dependence P(h) (in kilopascals) is expressed by the

formula:

where the height h above sea level is expressed in meters.

If the pressure is given in millimeters of mercury (mmHg), the barometric formula is written in the form:

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In case when the height h is given in feet, and pressure in inches of mercury (inHg), this formula is written as

The barometric formula is often used for estimating the air pressure under different conditions, although it gives slightly higher values

compared with the real ones.

   Example 1 Determine at what altitude under standard conditions the air pressure is twice less than on the sea level?

Solution.

To estimate the altitude, we use the barometric formula

     

When h = 0, the pressure P(h) is equal to the average atmospheric sea level pressure P0. At a certain altitude H, the pressure is twice

less:

     

It follows from here that

     

Taking logarithms of both sides, we find the altitude H:

     

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   Example 2 Find the air pressure in a mine at a depth of 1 km at the temperature of 40 degrees Celsius.

Solution.

The air pressure in the mine can be estimated using the general barometric formula:

     

We substitute the following values into the formula: h = −1000 m (the sign is minus because the mine is under sea level); T = 40 +

273.15 = 313.15 K. The remaining parameters are standard: M = 0.02896 kg/mol, R = 8.3143 (N*m)/(mol*K), g = 9.807 m/s2.

After simple calculations we find:

     

Since the atmospheric sea level pressure is P0 = 760 mmHg, the air pressure in the mine will be 848 mmHg, that is about 12 percent

greater than on the sea level. Rocket Motion In the given section we consider moving bodies with variable mass. Such kind of motion often occurs in nature and technology. We

can mention here for example:

Falling of an evaporating raindrop;

Movement of a melting ice block or an iceberg on the ocean surface;

Movement of a squid or jellyfish;

Rocket flight.

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Below we derive a simple differential equation for the motion of body with variable mass considering as an example rocket motion. Differential Equation of Rocket Motion

Rocket motion is based on Newton's third law, which states that "for every action there is an equal and opposite reaction". Hot gases

are exausted through a nozzle of the rocket and produce the action force. The reaction force acting in the opposite direction is called

the thrust force. The thrust force just causes the rocket acceleration.

Let the initial mass of the rocket be m and its initial velocity be v. In certain time dt, the mass of the rocket decreases by dm as a result

of the fuel combustion. This leads the rocket velocity to be increased by dv. We apply the law of conservation of momentum to the

system of the rocket and gas flow. At the initial moment the momentum of the system is equal to mv. In a small time dt the momentum

of the rocket becomes

and the momentum of the exhaust gases in the Earth's coordinate system is

where u is the exhaust gas velocity with respect to the Earth. Here we took into account that the exhaust velocity is in the opposite

direction to the rocket movement (Figure 1). Therefore we have put the minus sign in front of u.

By the law of conservation of the total momentum of the system, we can write:

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Fig.1

By transforming the given equation, we obtain:

We can neglect the term dmdv in the last expression considering small increments of these values. As a result, the equation is written

as

We divide both sides by dt to convert the equation into the form of Newton's second law:

The given equation is called the differential equation of rocket motion. The right side of the equation represents the thrust force T:

As it can be seen from the last formula, the thrust force is proportional to the exhaust velocity and the fuel burn rate. Of course, the

differential equation we derived describes an ideal case. It does not take into account the gravitational force or aerodynamic force.

Their inclusion leads to significant complication of the differential equation.

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Ideal Rocket Equation or Tsiolkovsky Rocket Equation

If we integrate the differential equation, we can get the dependence of the rocket velocity on the burned fuel mass. The resulting

formula is called the ideal rocket equation or Tsiolkovsky rocket equation who derived it in 1897.

To get this formula it's convenient to use the differential equation in the form:

Separating the variables and integrating gives:

Take into account that dm denotes mass decrease. Therefore we take the increment dm with the negative sign. As a result, the equation

is written as follows:

where v0 and v1 are the initial and final velocities of the rocket, m0 and m1 are the initial and final masses of the rocket, respectively.

By setting v0 = 0, we obtain the formula derived by Tsiolkovsky:

This formula determines the rocket velocity depending on its mass change while the fuel is burning. It allows rough estimation of the

fuel capacity necessary to accelerate the rocket to a given velocity.

   Example 1

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Estimate the fuel mass needed to launch a small "nanosatellite" with mass of 50 kg to a low orbit using a single-stage rocket. The

specific impulse of the rocket is 3,000 m/s.

Solution.

We can use the ideal rocket equation for rough estimates:

     

The exhaust velocity is approximately equal to the specific impulse, so we can set:

     

The final mass of the satellite is m = 50 kg. The initial mass m0 includes the mass m of the satellite itself and the fuel mass mp:

     

Suppose that the satellite velocity v on the low orbit is equal to the first space velocity 7.91 km/s = 7,910 m/s.

Express the necessary fuel mass mp through the remaining parameters and calculate its value:

     

Thus, the fuel mass needed to launch the given satellite is 700 kg (14 times more than the mass of the satellite itself).

Of course, this is an estimation of the lower mass boundary, because it's based on the ideal rocket equation.

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   Example 2 Estimate the acceleration of a rocket at the moment when the spacecraft reaches the orbit. Take the following parameters:

the specific impulse (exhaust velocity) is u = 3000 m/s; the mass of the spacecraft on the orbit is m = 5000 kg; the fuel burn rate is µ =

100 kg/s.

Solution.

We again apply the ideal rocket equation:

     

In this formula the rocket velocity v depends on the rocket mass m(t), which decreases as fuel is burning. We suppose for simplicity

that the fuel burn rate is constant, so that the dependence of the mass on time is described by the linear function:

     

The final mass m of the spacecraft is known in the given problem. Assuming that the exhaust velocity is constant, we can write the

ideal rocket equation in the form:

     

By differentiating with respect to t, we find the acceleration of the rocket:

     

Thus, we have obtained the rocket acceleration formula in the form:

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Notice that the acceleration a increases if any of the three parameters (the time t, the fuel burn rate µ and the exhaust velocity u) is

increased. This can be easily proved by calculating the partial derivatives with respect to each of the variables. For example, the partial

derivative in t is determined by the expression:

     

Similarly, the partial derivative in µ is given by

     

We can estimate the acceleration of the rocket when it reaches the orbit, assuming that the final mass is m = 5000 kg:

     

Thus, the rocket at the final stage of the trajectory can experience a big acceleration. Since it is a noninertial frame of reference, the

force of inertia of the same magnitude will affect the astronauts in the opposite direction (i.e. directed towards the Earth). As it can be

seen, the positive g-forces the astronauts can experience at take off can reach several g.

Newton's Law of Cooling

In the late of 17th century British scientist Isaac Newton studied cooling of bodies. Experiments showed that the

cooling rate approximately proportional to the difference of temperatures between the heated body and the

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environment. This fact can be written as the differential relation:

where A is the surface area of the body through which the heat is transferred, T is the temperature of the body, TS is

the temperature of the surrounding environment, α is the heat transfer coefficient depending on the geometry of the

body, state of the surface, heat transfer mode, and other factors.

Since Q = CT, where C is the heat capacity of the body, we can write:

The given differential equation has the solution in the form:

where T0 denotes the initial temperature of the body.

Thus, the temperature of any body approaches exponentially to the temperature of the surrounding environment in the

cooling process. The cooling rate depends on the parameter . With increase of the parameter k (for example,

due to increasing the surface area), the cooling occurs faster (Figure 1.)

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Fig.1 Fig.2

   Example 1

The temperature of a body dropped from 200o to 100o for the first hour. Determine how many degrees the body cooled

in one hour more if the environment temperature is 0o?

Solution.

First, we solve this problem for an arbitrary environment temperature and then determine the final body's temperature

when the surrounding environment temperature is 0o.

Let the initial temperature of the heated body be T0 = 200o. The further temperature dynamics is described by the

formula:

     

At the end of the first hour the body has cooled to 100o. Therefore, we can write the following relation:

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After 2nd hour the body's temperature becomes equal to X degrees:

     

Thus, we obtain the system of two equations with three unknowns: TS, k and X:

     

We cannot determine uniquely the body's temperature X after the 2nd hour from this system. However, we can derive

the dependence of X on the environment temperature TS. Express the function e−k from the first equation:

     

Hence,

     

Then the dependence X(TS) has the form:

     

If, for example, the surrounding environment temperature is zero degrees, the body's temperature X in 2 hours will be

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In the given example the value of X depends on TS as shown in Figure 2 above.

   Example 2

A body at the initial temperature T0 is put in a room at the temperature of TS0. The body cools according to the

Newton's law with the constant rate k. The temperature of the room slowly increases by the linear law:

where β is the known parameter. Determine the time τ when the body's temperature and the surrounding environment

temperature become equal.

Solution.

First of all, we should emphasize the difference with the case of the constant environment temperature. In the last

case, a body's temperature will be approaching the environment temperature an infinitely long time. In the given

problem, the environment temperature increases linearly, and therefore, sooner or later both the temperatures become

equal in some time. Thus, the problem has a solution.

The cooling process is described by the differential equation:

     

In the given case TS =TS0 + βt. Hence, the last equation can be written in the form:

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We have obtained a linear differential equation, which can be solved using, for example, the integrating factor:

     

The general solution of the equation is written as

     

The second integral in the numerator can be calculated by integrating by parts:

     

Thus, we get the cooling law of the body in the following form:

     

The constant C can be found from the initial condition T(t = 0) = T0. Then

     

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Thus, the body's cooling law is given by the formula

     

At a certain moment τ, the temperature of the body and temperature of the surrounding environment will become

equal:

     

Thus, the time τ is found from the equation:

     

We can estimate the time τ for some typical values of the parameters:

     

As a result, we have:

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Fluid Flow from a VesselTorricelli's Law

The Italian scientist Evangelista Torricelli investigating fluid flow experimentally found in 1643 that the velocity of

fluid flowing out through a small hole at the bottom of an open tank (Figure 1) is given by the formula:

where h is the height of fluid above the opening, g is the gravitational acceleration.

Fig.1 Fig.2

The same formula describes the velocity of a free solid particle falling from the height h in the Earth's gravitational

field in a vacuum. Actually this formula is not quite precise. In fact, the velocity of fluid depends on the shape and

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size of the opening, the fluid viscosity and flowing mode. Therefore, the Torricelli's formula is often written with an

additional factor φ:

where the coefficient φ is close to 1. The values of the coefficient φ for openings of different shape and size are given

in hydraulic handbooks.Fluid Flowing out of a Thin Pipe

Fluid flow through a thin long pipe (Figure 2) has a number of features. The various capillary effects caused by the

surface tension and wetting due to contact with the walls of the vessel play an important role.

The velocity of fluid flowing out from the capillary pipes is approximately proportional to the height of the fluid

above the opening, i.e.:

where k is a certain constant depending on the fluid viscosity, geometry and material of the pipe.

Further we describe the fluid flow using differential equations for both types of the vessels.Differential Equation of Fluid Flowing Out

We can derive the differential equation considering fluid balance in a vessel. Take as an example a cylindrical vessel

with a broad base of radius R. Suppose that fluid flows out through a small opening of radius a at the bottom of the

vessel (Figure 3).

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Fig.3 Fig.4

The fluid velocity is described by the Torricelli's formula:

where z is a the height of the fluid above the opening. Then the fluid flow is given by

Here πa2 corresponds to the area of the opening through which the fluid flows out, and the "minus"

sign means that the height of the fluid decreases when it flows out of the reservoir.

The fluid balance equation in the reservoir is written as follows:

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Since the volume change dV can be expressed as

we have the following differential equation:

Putting the function q(z) into this equation gives:

The cross section S(z) of the cylindrical vessel does not depend on the height z and is given by

where R is the base radius of the cylinder. Then

As a result, we obtain the separable equation:

Now we integrate this equation assuming that the initial height of the fluid is H and the fluid level

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decreases

to 0 for the time T.

It follows from here that the time T is defined by the expression:

Interestingly, that the resulting formula for the case a = R (when the cross sections of the opening and the cylinder are

equal) is transformed into the known formula that determines the fall time of a solid particle from the

height H.

The dependence of the time T on the height H is schematically shown in Figure 4 above.

Similarly, we can describe fluid leakage from the vessels of other shape.

   Example 1Derive the differential equation of fluid leakage from a conical vessel and determine the total flow time T. The upper

base radius of the conical vessel is R, the lower base radius is a. The initial height of the fluid is H (Figure 5).

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Fig.5 Fig.6

Solution.

The change of the fluid level at the height z is described by the differential equation

     

where S(z) is the cross-sectional area at the height z and q(z) is the fluid flow depending on its height z.

Based on the geometry of the system, we can assume that the Torricelli's law holds. Therefore, one can write:

     

where a is the radius of the opening at the bottom of the cone. Taking into account that this opening is small enough,

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we can consider the cross section of the cone as a triangle (Figure 6 above). Then it follows from the similarity of the

triangles that

     

Hence, the surface area of the fluid at the height z is given by

     

Substituting S(z) and q(z) into the differential equation gives:

     

After some transformations we get the following differential equation:

     

Integrate both sides taking into account that the level of the fluid decreases from its initial value H to zero for the time

T:

     

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Here we can again see the analogy with the fall of a solid particle from the height H in the Earth's gravitational field.

As it's known, the fall time is given by the formula:

     

If we compare this result with the case when the fluid flows out of a cylinder, we see that under the same values of H,

R and a the fluid flowing out time for the cone is exactly 5 times less than for the cylinder (though the volume of fluid

in the conical vessel is only 3 times less!) Such integer-valued ratios in nature seem surprising, don't they?

   Example 2Investigate the fluid flowing out of a thin pipe of the radius R and height H assuming that the pipe is completely filled.

Fig.7 Fig.8

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Solution.

Similar to the previous examples, we can write the fluid balance equation at an arbitrary height z in the form:

     

In the given case the cross-sectional area S(z) is constant:

     

and the flow of the fluid flowing out of the pipe is determined by the formula:

     

where k depends on the opening size, wettability and other parameters.

As a result we obtain the simple equation:

     

or after separating variables:

     

Now we can integrate this equation assuming that the fluid level decreases from H to h for the time from 0 to t:

     

The dependence of t on the ratio H/h is shown schematically in Figure 8. This curve is similar to the dependence of

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time T on the height H for a wide cylindric vessel for which the Torricelli's law is valid. Interestingly that the flowing

out time t formally approaches infinity as h → 0 in the given simple model.

Population GrowthPopulation growth is a dynamic process that can be effectively described using differential equations. We consider

here a few models of population growth proposed by economists and physicists.Malthusian Growth Model

The simplest model was proposed still in 1798 by British scientist Thomas Robert Malthus. This model reflects

exponential growth of population and can be described by the differential equation

where a is the growth rate (Malthusian Parameter). Solution of this equation is the exponential function

where N0 is the initial population.

The given simple model properly describes the initial phase of growth when population is far from its limits.

However, the accuracy of the exponential model drops at a later stage due to saturation or other nonlinear effects

(Figure 1).

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Fig.1 Fig.2

Fig.3 Fig.4Logistic Model

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This kind of population models was proposed by French mathematician Pierre Francois Verhulst in 1838. This model

is also called the logistic model and is written in the form of differential equation:

where M is the maximum size of the population. The right side of this equation can be presented as

where the first term is responsible for growth of population and the second term limits this growth due to lack of

available resources or other reasons (Figures 2,3). The logistic differential model has exact solution, which we derive

below.

The integrand in the left integral can be found using the partial fraction decomposition method:

Then the integral in the left side is

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Thus, the general solution of the logistic differential equation is given by

The last algebraic equation can be solved for N:

The constant C can be determined from the initial condition N(t = 0) = N0, so that

Substituting this value for C into the general solution, we obtain:

The graph of the logistic function has a nice view. Figure 2 shows a few logistic curves at different values of N0, and

Figure 3 shows how the shape of the curve changes depending on the growth rate a.

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We see that the family of logistic curves on the segment t > 0 can describe nonlinear population growth with

saturation, when the maximum allowed value has a limit.Hyperbolic Growth Models

The models we considered above are useful in the analysis of demographic processes on a scale of centuries. If

consider population growth for several thousand years (Figure 4), it can be seen that the main explosive growth from

2 to 7 billion people occured on the past 50 years. This type of dependancy is similar to the hyperbolic curve. A

simple hyperbolic growth model was suggested by several researchers (von Forster (1960), von Hoster (1975), and

Shklovskii (1980)) in the following form:

As it follows from this model the world population goes off to infinity as the year 2025 approaches.

However, the real growth dynamics demonstrates that the so-called demographic transition follows after the

explosive growth phase. This new state is characterized by declining fertility and mortality. Such a transition has

already occurred in many developed countries. As a result of the demographic transition, the population growth ceases

and may even fall. The global world population had just entered the phase of demographic transition in the beginning

of 21st century.

It turns out that such a complex population dynamics can be also well described using differential equations! A model

of this type was recently (in 1997) developed by Russian physicist Sergey Kapitsa. Kapitsa proposed to describe the

explosive growth using the following equation:

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where T0, C and τ are certain approximation parameters. This differential equation has the exact solution as a function:

The given function describes the explosive population growth remarkably well at the following values of the

parameters: C = 1.86*1011, T0 = 2007, τ = 42. Besides that, the model covers the demographic transition phase when

the population growth reaches saturation (see Figure 4 above). According to this model, the global world population

will reach about 12 billion in 2200-2300.

dvertising AwarenessDifferential equations are widely used to describe a variety of dynamic processes in Economic Sciences, Business and

Marketing. Below we consider how an advertising campaign can be simulated using differential equations.

Imagine that a company has developed a new product or service. The marketing strategy of the company involves

aggressive advertising. To construct the simple model we introduce two variables:

q(t) represents advertising activity that is described by spending rate, for example, by the amount of dollars

(euros, pounds) the company spends for advertising per week;

A(t) represents awareness of the target group of potential consumers about the new goods or services.

Thus, we will consider the market niche as a black box (Figure 1). The advertising activity q(t) is the input variable,

and awareness of consumers A(t) is the output variable that measures response of the system to the advertising

campaign.

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Fig.1 Fig.2

A simple model of such type was proposed in 1962 and is called the advertising model of Nerlove and Arrow (the N-

A model). This model relates advertising activity q(t) and awareness of consumers A(t) and is given by the differential

equation:

where b is a constant describing advertising effectiveness, k is a constant corresponding to decay (or forgetting) rate.

The given equation contains two terms in the right side. The first term bq(t) provides the linear growth of awareness

of consumers as a result of advertising. The second term  −kA reflects the opposite process, i.e. forgetting about the

product.

In first approximation, we can assume that the forgetting rate is proportional to the current level of awareness A.

This equation is a Linear Differential Equation of First Order. It's convenient to rewrite it in the standard form:

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The integrating factor is the exponential function:

Therefore the general solution of the given differential equation is given by

The constant of integration C can be found as usual from the initial condition A(t0) = A0.

In the examples below we consider how awareness of customers A(t) varies at different advertising modes.

   Example 1Management of a company decided to advertise a new product permanently during the year. The advertising budget is

$12,000. The coefficients k and b are k = 1/4, b = 25. Construct and solve the differential equation describing the

number of people A(t) aware of this product.

Solution.

The equation of dynamics of A(t) is written in the form

     

We assume that the time t is measured in months. By the condition of the problem, the advertising costs are constant

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during the year, so every month they are equal to:

     

Substituting the known data leads to the following differential equation:

     

The integrating factor for this equation has the form:

     

The general solution is written as

     

We determine the constant of integration C from the initial condition A(t = 0) = 0. Hence, C = −100,000. Then the

particular solution is expressed by the formula

     

The graph of this function is shown in Figure 2 above. Thus in case of permanent advertising, the number of potential

buyers aware of the product grows non-linearly approaching the maximum value of

     

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   Example 2Under the conditions of the previous problem 1, figure out how will the number of potential buyers vary by the end of

the year if the advertising budget is spent evenly for the first 6 months?

Solution.

In this problem the advertising mode has a stepwise shape. The advertising costs are shown schematically in Figure 3

below. We will investigate how awareness of consumers A(t) is changed compared with the case of constant

advertising over all year considered in Example 1.

The problem is divided into two stages. The value of A is easily calculated by the end of sixth month using the

formula

     

derived in the previous Example. The coefficients have the values: k = 1/4, b = 25, q0 = 2000. Then

     

At the moment t = 6, we find that the number of consumers aware of the new product is

     

In the second phase from the 7th to the 12th month, the advertising is completely absent. As a result, the level of

awareness A(t) decreases according to the equation

     

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The solution of the homogeneous equation is the exponential function:

     

where t > 6 months, and the constant C can be found from the initial condition for the 2nd phase:

     

Thus, the law of changing A(t) in the second half-year is given by the formula

     

Hence, the complete solution of the problem is written in the form:

     

The graph of the function A(t) is given in Figure 4.

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Fig.3 Fig.4

For comparison, the curve A(t) from the previous Example 1 is also shown on this chart. As it can be seen, awareness

of consumers by the end of the year in the second case will be lower than in the case of homogeneous advertising. The

exact values of A for both these cases are

     

Interestingly, that the average value of A over the year is greater for the second mode:

     

One can roughly assume that the volume of sales is proportional to the awareness of consumers about the product, so

the mode of stepwise advertising (with the same total budget!) may be more profitable from this point of view.

   Example 3Investigate the dynamics of awareness A(t) for the case of Linear Advertising Function q(t). Use the same data as in

Examples 1,2.

Solution.

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In this problem we make our marketing model a bit more complicated. We will assume that the advertising budget is

spent over year according to the linear law:

     

The function q(t) may be as increasing or decreasing (Figure 5).

Fig.5 Fig.6

In any case, the total yearly advertising budget remains unchanged (let it be equal to U). Graphically, this means that

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the area of all trapezoids (or triangles in the limiting case) shown in Figure 5 is the same.

The parameters q0 and α are related by the equation:

     

The left side of this formula is the area of the trapezoid. Then the coefficient α is expressed through q0 as follows:

     

The time dependence of advertising costs will be given by the formula:

     

Substitute this expression into the general solution A(t) and then integrate it:

     

The integral in the numerator can be found using integrating by parts. We set:

     

Hence,

     

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As a result, we obtain the following expression for A(t):

     

We determine the constant C from the initial condition  A(t = 0) = 0. This gives:

     

By substituting C into the formula for A(t), we get:

     

Finally we substitute the known values: k = 1/4, b = 25, U = 12,000:

     

If we set q0 = 1000 (the mode of homogeneous advertising considered in Example 1), we get the formula found above:

     

Using the general solution A(t), we compare the dynamics of awareness for the following limiting cases (see Figure 6

above).

1. Scenario 1: Advertising costs grow linearly from 0 to 2000;

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2. Scenario 2: Advertising costs decrease linearly from 2000 to 0;

3. Scenario 3: Advertising costs are constant during the year.

The results of the computations are presented in Figure 7.

Fig.7 Fig.8

The mean values of A(t) for the year for the indicated scenarios are also given in the table (Figure 8). As it can be

seen, the most agressive scenario 2 can lead to sales growth, though scenario 1 has advantages in terms of long-lasting

effect.

Of course, these conclusions are limited by the accuracy of the model. The dynamics of A(t) in a real business

environment can be more complex.

Reduction of Order A second order differential equation is written in general form as

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where F is a function of the given arguments.

If the differential equation can be resolved for the second derivative y'', it can be represented in the following explicit form:

In special cases the function f in the right side may contain only one or two variables. Such incomplete equations include 5 different

types:

With the help of certain substitutions, these equations can be transformed into first order equations.

In the general case of a second order differential equation, its order can be reduced if this equation has a certain symmetry. Below we

discuss two types of such equations (cases 6 and 7):

The function F(x, y, y', y'') is a homogeneous function of the arguments y, y', y'';

The function F(x, y, y', y'') is an exact derivative of the first order function Ô(x, y, y').

Thus, consider these cases of reduction of order in more detail. Case 1. Equation of type y''= f (x)

For an equation of type y'' = f(x), its order can be reduced by introducing a new function p(x), such that y' = p(x). As a result, we

obtain the first order differential equation

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Solving it, we find the function p(x). Then we solve the second equation

and obtain the general solution of the original equation. Case 2. Equation of type y''= f (y)

The right-hand side of the equation depends only on the variable y. We introduce a new function p(y), setting y' = p(y). Then we can

write:

so the equation becomes:

Solving it, we find the function p(y). Then we find the solution of the equation y' = p(y), that is, the function y(x). Case 3. Equation of type y''= f (y' )

In this case, to reduce the order we introduce the function y' = p(x) and obtain the equation

which is a first order equation with separable variables p and x. Integrating, we find the function p(x), then the function y(x). Case 4. Equation of type y''= f (x,y' )

Here we use the substitution y' = p(x), where p(x) is a new unknown function, and obtain the first order equation

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Integrating, we find the function p(x). Next, we solve one more equation of the 1st order

and find the general solution y(x). Case 5. Equation of type y''= f (y,y' )

To solve this equation, we introduce a new function p(y), setting y' = p(y), similar to case 2. Differentiating this expression with

respect to x leads to the equation

As a result, our original equation is written as an equation of the 1st order

Solving it, we find the function p(y). Then we solve another first order equation

and determine the general solution y(x).

The above 5 cases of reduction of order are not independent. Based on the structure of the equations, it is clear that case 2 follows from

the case 5 and case 3 follows from the more general case of 4. Case 6. Function F(x, y, y', y'') is homogeneous with respect to the arguments y, y', y''

If the left side of the differential equation

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satisfies the condition of homogeneity, ie for any k, the relation

is valid, the order of the equation can be reduced by substituting

After finding the function z(x) the original function y(x) is determined by the integration formula

where C2 is the constant of integration. Case 7. Function F(x, y, y', y'') is an exact derivative

If one can find a function Ô(x, y, y'), which does not contain the second derivative y'' and satisfies the equation

then the solution of the original equation is given by the integral

Using this way the second order equation can be reduced to first order equation.

In some cases, the left part of the original equation can be transformed into an exact derivative, using an integrating factor.

   Example 1 Solve the equation  y'' = sin x + cos x.

Solution.

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This example is related to the Case 1. Consider the function y' = p(x). Then  y'' = p'. Consequently,

     

Integrating, we find the function p(x):

     

Given that y' = p(x), integrate one more equation of the 1st order:

     

The latter formula gives the general solution of the original differential equation.

   Example 2

Solve the equation .

Solution.

This is an equation of type 2, where the right-hand side depends only on the variable y. We introduce the parameter p = y'. Then the

equation can be written as

     

We obtain the equation of the 1st order for the function p(y) with separable variables. Integrating gives:

     

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where C1 is a constant of integration.

Taking the square root of both sides, we find the function p(y):

     

Now recall that y' = p and solve another equation of the 1st order:

     

Separate the variables and integrate:

     

To calculate the integral on the left-hand side, make the change:

     

Then the left-hand integral is equal to

     

As a result, we obtain the following algebraic equation:

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where C1, C2 are constants of integration.

The last expression is the general solution of the differential equation in implicit form.

   Example 3

Solve the equation .

Solution.

This equation does not contain the function y and the independent variable x (Case 3). Therefore, we set y' = p(x). Then this equation

takes the form

     

The resulting first-order equation for the function p(x) is a separable equation and can be easily integrated:

     

Replacing p by y', we obtain

     

Integrating again, we find the general solution of the original differential equation:

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   Example 4

Solve the equation .

Solution.

This equation does not explicitly include the variable y, i.e. it corresponds to the type 4 in our classification. We introduce a new

variable y' = p(x). The original equation is transformed into the first order equation:

     

which is solved by separation of variables:

     

Integrating the resulting equation once more yields the function y(x):

     

To compute the last integral we make the substitution: x = t2, dx = 2tdt. As a result, we have

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Returning to the variable x, we finally obtain

     

   Example 5 Solve the equation  y'' = (2y + 3)(y' )2.

Solution.

This equation does not explicitly contain the independent variable x, i.e. refers to the Case 5. Let y' = p(y). Then the equation can be

written as

     

Separate variables and integrate:

     

Integrating again, we obtain the final solution in implicit form:

     

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where C1, C2 are constants of integration.

   Example 6

Solve the equation .

Solution.

The equation satisfies the condition of homogeneity. Therefore, we make the following change of variable: . The

derivatives will be equal

     

Then the differential equation becomes:

     

The function z(x) is easily found:

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The original function y(x) is defined by the formula

     

The calculations give the following answer:

     

Note that in addition to the general solution, the differential equation also contains a singular solution y = 0.

   Example 7 Solve the equation  yy'' + (y' )2 = 2x + 1.

Solution.

You may notice that the left side of the equation is the derivative of yy'. Therefore, denoting z = yy', we obtain the following

differential equation:

     

The last equation is easily solved by separation of variables:

     

Now we integrate one more equation for y(x):

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where C1, C2 are arbitrary constants.

Bessel Differential Equation The linear second order ordinary differential equation of type

is called the Bessel equation. The number v is called the order of the Bessel equation.

The given differential equation is named after the German mathematician and astronomer Friedrich Wilhelm Bessel who studied this

equation in detail and showed (in 1824) that its solutions are expressed through a special class of functions called cylinder functions or

Bessel functions.

Concrete representation of the general solution depends on the number v. Further we consider separately two cases:

The order v is non-integer;

The order v is an integer.

Case 1. The Order v is Non-Integer

Assuming that the number v is non-integer and positive, the general solution of the Bessel equation can be written as

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where C1, C2 are arbitrary constants and Jv(x), J−v(x) are Bessel functions of the first kind.

The Bessel function can be represented by a series, the terms of which are expressed through the so-called Gamma function:

The Gamma function is the generalization of the factorial function from integers to all real numbers. It has, in particular, the following

properties:

The Bessel functions of the negative order (−v) (it's assumed that v > 0) are written in similar way:

The Bessel functions can be calculated in most mathematical software packages. For example, the Bessel functions of the 1st kind of

orders v = 0 to v = 4 are shown in Figure 1. The corresponding functions are also available in MS Excel.

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Fig.1 Fig.2Case 2. The Order v is an Integer

If the order v of the Bessel differential equation is an integer, the Bessel functions Jv(x) and J−v(x) can become dependent from each

other. In this case the general solution is described by another formula:

where Yv(x) is the Bessel function of the second kind. Sometimes this family of functions is also called Neumann functions or Weber

functions.

The Bessel function of the second kind Yv(x) can be expressed through the Bessel functions of the first kind Jv(x) and J−v(x):

The graphs of the functions Yv(x) for several first orders v are shown above in Figure 2.

Note: Actually the general solution of the differential equation expressed through Bessel functions of the first and second kind is valid

for non-integer orders as well. Some Differential Equations Reducible to Bessel's Equation

1. One of the well-known equations tied with the Bessel's differential equation is the modified Bessel's equation that is obtained by

replacing x to −ix. This equation has the form:

The solution of this equation can be expressed through the so-called modified Bessel functions of the first and second kind:

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where Iv(x) and Kv(x) are modified Bessel functions of the 1st and 2nd kind, respectively.

2. The Airy differential equation known in astronomy and physics has the form:

It can be also reduced to the Bessel equation. Its solution is given by the Bessel functions of the fractional order :

3. The differential equation of type

differs from the Bessel equation only by a factor a2 before x2 and has the general solution in the form:

4. The similar differential equation

is reduced to the Bessel equation

by using the substitution

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Here the parameter n2 denotes:

As a result, the general solution of the differential equation is given by

The special Bessel functions are widely used in solving problems of theoretical physics, for example in investigating

wave propagation;

heat conduction;

vibrations of membranes

in the systems with cylindrical or spherical symmetry.

   Example 1 Solve the differential equation  x2y'' + xy' + (3x2 − 2)y = 0.

Solution.

This equation has order √2 and differs from the standard Bessel equation only by factor 3 before x2. Therefore, the general solution of

the equation is expressed by the formula

     

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where C1, C2 are constants, J√2(√3x) and Y√2(√3x) are Bessel functions of the 1st and 2nd kind, respectively.

   Example 2

Solve the equation .

Solution.

This equation differs from the modified Bessel equation by factor 4 in front of x2. The order of the equation is v = 1/√2. Then the

general solution is written through the modified Bessel functions in the following way:

     

where C1 and C2 are arbitrary constants.

   Example 3 Find the general solution of the differential equation  x2y'' + 2xy' + (x2 − 1)y = 0.

Solution.

We make the substitution:

     

Put these expressions back into the equation:

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Indeed, we see that

     

Thus, the general solution for the function z(x) can be written in the form

     

Then the solution for the original function y(x) is given by

     

where C1 and C2 are arbitrary constants. Chebyshev Differential Equation

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Definition and General Solution

The differential equation of type

where |x| < 1 and n is a real number, is called the Chebyshev equation after the famous Russian mathematician Pafnuty Chebyshev.

This equation can be converted to a simpler form using the substitution x = cos t. Indeed, in this case we have

Hence,

Substituting the expressions of derivatives into the differential equation gives:

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As a result we can write the equation in the compact form:

The general solution of the last equation is given by the formula

which can be also written as

Here C1, C2, and C, α are arbitrary real numbers. For simplicity, we can set α = 0. Then the general solution of the original Chebyshev

equation will be given by the formula:

In this expression, n may be any real number. But if n is an integer, the given function is the Chebyshev polynomial of the first kind. Chebyshev Polynomials of the First Kind

The Chebyshev polynomial of the first kind is called the function

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where |x| ≤ 1 and n = 0,1,2,3,.... Next, we show that this function is really an algebraic polynomial. For n = 0 and n = 1 we have

By setting x = cos t, we can write:

Since

we obtain the following recursive relationship for the Chebyshev polynomials of the first kind:

Now we can easily calculate the Chebushev polynomials of higher orders:

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and so on... Chebyshev Polynomials of the Second Kind

The Chebyshev polynomials of the second kind can be also defined by the recursive relationship:

The polynomials of the second kind are solutions to the Chebyshev differential equation of the type

The graphs of the Chebyshev polynomials of the 1st and 2nd kind are shown in Figures 1 and 2, respectively.

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Fig.1 Fig.2   Example 1 Find the general solution of the equation  (1 − x2)y'' − xy' + 2y = 0 at |x| < 1.

Solution.

The given equation is the Chebyshev differential equation with the fractional parameter n = √2. Its general solution can be written in

the trigonometric form:

     

where C1, C2 are constants. Note that the solution in this case is not expressed via the Chebyshev polynomials because of the fractional

parameter √2.

   Example 2 Find the general solution of the differential equation  (1 − x2)y'' − xy' + 4y = 0 at |x| < 1.

Solution.

Here we deal with the Chebyshev equation with the parameter n =2. Therefore, we can directly write the general solution in the form:

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where C1, C2 are arbitrary constants.

We can express this solution via the Chebyshev polynomials of the first kind. Since

     

we obtain the final answer: