Lect 01 Separable Eq

7/31/2019 Lect 01 Separable Eq.

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TMS2033

Differential EquationsLecture 01:

Solutions of Some Differential Equations(1) Method of Calculus(2) Method of Separation of Variables


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Summary of Lect 00

ODE, PDE, system of ODE, order, solution

Mathematical models

Direction fieldsEquilibrium solutions

Without

solving theDE


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Objectives of Lect 01To understand the concept of solutions of DE

General solution, implicit, explicit solutionThe solution Integral curves

To identify an initial value problem

To employ(1) method of calculus(2) method of separation of variables


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Recall the free fall and owl/mice differential equations:

These equations have the general form y' = ay b We can use Methods of Calculus to solve differentialequations of this form.

4505.0,2.08.9 p pvv

Method of Calculus


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Revisit Model B: Mice and Owls (1 of 3)

We use methods of calculus , as follows.

4505.0 p p

C t

C t

C t

ek ke p

ee p

e p

C t p

dt p

dp pdt dp

,900

900

900

5.0900ln

5.0900

9005.0

5.0

5.0

5.0

t

ke p5.0

900

Ex. & Soln.

Solve the differential equationSolution:

[ Ans: where k is a constant]


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Revisit Model B: Integral Curves (2 of 3)

Thus we have infinitely many solutions to our equation,

since k is an arbitrary constant.

Graphs of solutions (integral curves ) for several values of k ,and direction field for differential equation, are given below.Choosing k = 0, we obtain the equilibrium solution, while fork 0, the solutions diverge from equilibrium solution.

,9004505.0 5.0 t ke p p p

Solution

General solution


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Revisit Model B: Initial Conditions (3 of 3)

A differential equation often has infinitely many solutions. If a point on the solution curve is known, such as an initialcondition , then this determines a unique solution.

In the mice/owl differential equation, suppose we know thatthe mice population starts out at 850. Then p(0) = 850, and

50

900850)0( 0

k

ke p

Solution

t et p 5.050900)(

:Solution

t ke p 5.0900


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Methods of Calculus

To solve the general equationwe use methods of calculus , as follows.

Thus the general solution is

where k is a constant.

bay y

C at

C at

C at

ek keab yeeab y

eab y

C t aab y

dt aab y

dy

a

b ya

dt

dy

, / /

/

/ ln /

,at keab

y

Proof


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Initial Value Problem

Next, we solve the initial value problem

From previous slide, the solution to differential equation is

Using the initial condition to solve for k , we obtain

and hence the solution to the initial value problem is

at ea

b y

a

b y 0

0)0(, y ybay y

at keab y

a

b yk ke

a

b y y 0

00)0(

Proof


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Equilibrium Solution & Solution Behaviour

Recall: To find equilibrium solution, set y' = 0 & solve for y:

From previous slide, our solution to initial value problem is:

Note the following solution behavior:If y0 = b / a, then y is constant, with y(t ) = b/aIf y0 > b / a and a > 0 , then y increases exponentially without boundIf y0 > b / a and a < 0 , then y decays exponentially to b / aIf y0 < b / a and a > 0 , then y decreases exponentially without bound

If y0 < b / a and a < 0 , then y increases asymptotically to b / a

ab

t ybay yset

)(0

at eab

yab

y 0

Note


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Ref. : Ch 2.2 Boyce & DiPrima

From method of calculus extended to Separable Equations

In this section, we examine a subclass of linear and nonlinearfirst order equations. Consider the first order equation

For example, let M ( x, y) = - f ( x, y) and N ( x, y) = 1. We canrewrite this in the form

(There may be other ways as well.)

If M is a function of x only and N is a function of y only, then

In this case, the equation is called separable .

0),(),(dx

dy y x N y x M

0)()( dy y N dx x M

),( y x f dx

dy

Definition


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Example 1.2 : Solving a Separable Equation

Solve the following first order nonlinear equation:

Solution : Separating variables, and using calculus, we obtain

The equation above defines the solution y implicit ly. Agraph showing the direction field and implicit plots of several

integral curves for the differential equation is given above.

11

2

2

y

x

dx

dy

C x x y y

dx xdy y

dx xdy y

33

22

22

3

1

3

1

11

11

Ex. & Soln.

C x x y y 33 33


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Example 1.3 :Implicit and Explicit Solutions (1 of 4)

Solve the following first order nonlinear equation:

Solution : Separating variables and using calculus, we obtain

The equation above defines the solution y implicit ly. Anexplicit expression for the solution can be found in this case:

12243 2

y

x x

dx

dy

dx x xdy y

dx x xdy y

24312

243122

2

2

224420222

23232 C x x x yC x x x y y

Ex. & Soln.

C x x x y y 222 232

C x x x y 221 23


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Example 1.3 : (2 of 4)

Solution : Using the implicit expression of y, we obtain

Thus the implicit equation defining y is

Using explicit expression of y,

It follows that411

221 23

C C

C x x x y

3)1(2)1(

2222

232

C C

C x x x y y

3222 232 x x x y y

4221 23 x x x y

Solution

(a) Suppose we seek a solution satisfying y(0) = -1 .


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Solution : Then, we choose the positive sign, instead of negative sign,

on square root term:

[Work it out ]

]4221:Ans.[ 23 x x x y

Solution

(b) Find the explicit solution if initial condition is y(0) = 3 .


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Thus the solutions to the initial value problem

are given by

From explicit representation of y, it follows that

and hence domain of y is 2 < y < . Note x = -2 yields y =1, which makes denominator of dy / dx zero (vertical tangent).Conversely, domain of y can be estimated by locating verticaltangents on graph (useful for implicitly defined solutions).

(explicit) 4221

(implicit) 322223

232

x x x y

x x x y y

2212221 22 x x x x x y

1)0(,

12243 2

y y

x x

dx

dy

Solution


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Consider the following initial value problem:

Solution : Separating variables and using calculus, we obtain

Using the initial condition, it follows that

1)0(,31

cos3 y y

x y y

C x y y

dx xdy y y

dx xdy y

y

sinln

cos31

cos31

3

2

3

1sinln 3 x y y

Ex. & Soln.


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Thus

The graph of this solution (black), along with the graphs of the direction field and several integral curves (blue) for thisdifferential equation, is given below.

1sinln1)0(,31

cos 33 x y y y y

x y y

Solution


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Revisit Model A: Free Fall Eq. (1 of 3)

Recall equation modeling free fall descent of 10 kg object,assuming an air resistance coefficient = 2 kg/sec:

Suppose object is dropped from 300 m. above ground.(a) Find velocity at any time t .(b) How long until it hits ground and how fast will it be moving then?

Solution : For part (a), we need to solve the initial value

problem Using result from previous slide, we have

0)0(,2.08.9 vvv

vdt dv 2.08.9 /

t at eveab

yab

y 2.0 2.08.9

02.08.9

Ex. & Soln.

t ev 2.149


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Revisit Model A: Free Fall Eq. (2 of 3)

The graph of the solution found in part (a), along with thedirection field for the differential equation, is given below.

t evvvv

2.149

0)0(,2.08.9

Solution


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Revisit Model A: Free Fall Eq.Part (b): Time and Speed of Impact (3 of 3)Next, given that the object is dropped from 300 m. aboveground, how long will it take to hit ground, and how fast willit be moving at impact?

Solution: Let s(t ) = distance object has fallen at time t .It follows from our solution v(t ) that

Let T be the time of impact. Then

Using a solver, T 10.51 sec, hence

24524549)(2450)0(

24549)(4949)()(2.

2.2.

t

t t

et t sC s

C et t set vt s

30024524549)( 2. T eT T s

ft/sec01.43149)51.10( )51.10(2.0ev

Solution


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SummaryUnderstood the concept of:

General solutionThe solution (implicit or explicit) of a DE

Able to identify an initial value problemWhat is initial condition?

Able to solve DE by using

(1) Method of Calculus(2) Method of separation of variables

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Lect 01 Separable Eq