Introduction to Differential Equations

We begin by introducing differential equations, and the use of differential equationsfor modeling physical processes. We continue discussing different ways to investigatesolutions.

1. Differential Equations

2. Classifying Equations

3. Simple Modeling

4. Initial Value Problems (IVP)

5. Direction Fields and Autonomous equations

1 Differential Equations and Solutions

A differential equation is an equation which involves an unknown function (frequentlydenoted y), its independent variable (frequently denoted t), and any number of deriva-tives of the function (y′, y′′, y(n), etc...).

Example:

The followings are all examples of differential equations:

y′′ − 2y′ + y = 0

ty − y′ = sin(t)

y′′ = −y

tx′′ + 4x′ + 2 = et

The solution to a differential equation is a function (y(t)) which satisfies the equationfor all values of the independent variable (t).

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Example:

A solution to the differential equation y′ = y is the function y = et because y′ = et = y.It is easy to check and see if a given function is a solution to a differential equationby plugging in.

Example:

We can determine which (if either) of y = cos(t) and y = sin(t) are solutions toy′′ = −y by plugging in:

Here are a few more examples:

Example:

y = t2 is not a solution to 3t2y′′ + 11ty′ − 3y = 0, because y′ = 2t, y′′ = 2, and thus3t2y′′ + 11ty′ − 3y = 6t2 + 22t2 − 3t2 = 25t2, and this is zero only when t = 0.

Example:

The equation y′ + 3y = 2 has a solution y = 2/3 + Ce−3t where C is any constant.Verification:

In fact, it turns out every solution to y′ + 3y = 2 has the form 2/3 + Ce−3t for someconstant C, but we have not shown this yet. We will therefore say that 2/3 + Ce−3t

is the general solution to y′ + 3y = 2.

2 Classifying Equations

The order of a differential equation is the order of the highest derivative that appearsin the differential equation.

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Example:

y′′ − 2y′ + y = 0

ty − y′ = sin(t)

y(4) + 4y′′ − t2y′ = 0

A differential equation is called ordinary if the equation only involves a single in-dependent variable (t). Partial differential equations, however, involve two or moreindependent variables, i.e. y = f(x, t, ...).

A linear differential equation can be written of the form:

gn(t)yn + gn−1(t)yn−1 + ... + g1(t)y

′ + g0(t)y = h(t)

When h(t) is zero, the equation is called homogeneous; if h(t) is non-zero then theequation is nonhomogeneous.

Example:

t3y3 + sin(t)y′ = 0

yy′ + t2 = y ln(t)

tyy′ + cos(y) = sin(t)

Differential equations are useful because they allow us to express relationships amongvarious rates of change and a function.

3 Simple Modeling

We are interested in using differential equations to model physical situations. Tocreate a model, we write down equations that describe the relationship between afunction and its derivatives and independent variable.

Example:

According to Newton’s second law of motion forces are equal to mass times accelera-tion. One simple example is a falling object:A ball is thrown upward from 6 feet with an initial velocity of 20 ft/sec. Create amodel for the height of the ball after t seconds, ignoring air resistance. Use this modelto find the maximum height and time of impact. We know the acceleration is from

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gravity, so if s is the position at time t, we have the differential equation s′′ = g = −32,indicating the force of gravity acts downward at 32 ft/s2. We also have two initialconditions: s(0) = 6 and s′(0) = 20 (initial position and velocity). In this case, we canintegrate both sides with respect to t and get s′(t) = −32t+C1. We can solve for C1,since we know s′(0) = 20. Therefore C1 = 20, and we get s′(t) = −32t + 20. We nowfind s by integrating again with respect to t, and so we get s(t) = −16t2 + 20t + C2.Since s(0) = 6, we get C2 = 6, and so s(t) = −16t2 + 20t + 6 is the solution to ourinitial value problem.

The maximum height is attained at the top, when velocity s′(t) = 0, so we have−32t + 20 = 0, or t = 5/8 s. The maximum height is then s(5/8) = 12.25.

It hits the ground when s(t) = 0, or −16t2 + 20t + 6 = 0, which is when t = −1/4and t = 3/2. The correct time is t = 3/2, since it must hit the ground after it starts.Note in the above example it was possible to solve the differential equation by simplyintegrating both sides with respect to t. Usually this will not work. For example,if we attempt to integrate both sides of y′ = y with respect to t, we are left withy =

∫y dt, where y is still an unknown function of t. (We will see later however that

this approach can be adapted to so-called separable equations, which include y′ = y.)

Example:

The instantaneous rate of decay of a radioactive substance is proportional to theamount of material present, that is:

4 Initial Value Problems

An initial value problem is a differential equation together with one value of the un-known function (and possibly values of its derivatives at the same point); often (butnot always) we are told the value of y(0) (and perhaps y′(0), y′′(0), etc.).

Example:

The following is an initial value problem:

y′ = 2y, y(0) = 1

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The solution to an initial value problem requires that the unknown function is a solu-tion to the differential equation and also satisfies the given initial value. In practice,we will usually find a general solution to a differential equation, which may involvesome unknown constants, and then solve for the constants to find a particular orspecific solution to the equation which is a solution to the initial value problem.

Example:

Solve y′ + 3y = 2, y(1) = 2.We already said that y = 2/3 + Ce−3t was the general solution to y′ + 3y = 2, so letus try to solve for C using the fact that y(1) = 2:

2 = 2/3 + Ce−3·1

4/3 = Ce−3

C = 4/3 e3

So our particular solution is y = 2/3 + 4/3 e3e−3t, or y = 2/3 + 4/3 e3−3t.

We also know it is easy to verify whether or not something is a solution to a differ-ential equation or an initial value problem. We also know how to find the particularsolution to an initial value problem given the general solution to the differential equa-tion, but we do not know how to find a general solution. We are also not yet surethere must be a solution to a given differential equation, or how many there mightbe.

5 Direction fields and Autonomous Equations

If we have a differential equation of the form y′ = f(t, y), then a slope field is aqualitative way of seeing what solutions to that differential equation might look likewhen graphed. Given a differential equation of the form y′ = f(t, y), you can findthe slope of y at each possible coordinate (t, y) in the plane. We can then draw in ashort tangent line segment at a grid of points in the plane, which will suggest whatthe curve will look like.

So we do the following:

• Pick a grid of points in the plane,

• Evaluate y′ at each point (t, y),

• Draw a short “tangent line” segment at that point.

We will get what we refer to as a direction field (sometimes called a slope field) forthe equation y′ = f(t, y).

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Example:

Let’s sketch a direction field for y′−t = y2. We will just use four points: (0, 0), (0, 1),(1, 0) and (1, 1). (This will make a crude sketch, but cut down on our calculations.)Our equation is y′ − t = y2, or y′ = t + y2. This means we have y′ = f(t, y) wheref(t, y) = t + y2. So now let us find slopes:At (0, 0): y′ = f(0, 0) = 0 + 02 = 0.At (0, 1): y′ = f(0, 1) = 0 + 12 = 1.At (1, 0): y′ = f(1, 0) = 1 + 02 = 1.At (1, 1): y′ = f(1, 1) = 1 + 12 = 2.We then draw in short tangent line segments at the four points listed above, and wehave a crude direction field:

1t

1

y

It is obvious that constructing a direction field by hand will be tedious. Creating ann × n grid of points will require n2 calculations. However, generating such a fieldusing a computer is not difficult. The following sketch of a slope field for y′ = t + y2

(which we used in example 5 above) was generated using Matlab:

−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

Slopefield: y’= t + y2

Although we would not want to generate this field by hand, we can recognize featuresof the equation in the sketch of the slope field:

• Notice that as t increases, f(t, y) should increase, and slope does increase withincreasing t.

• Similarly, the slopes increase with increasing |y| values (as points move away

from the t-axis.

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• Where will the slopes be negative? It would be where t + y2 < 0, or wheret < −y2. We can see where t < −y2 by first identifying the curve t = −y2 onthe t − y plane; we see that slopes indeed seem to be negative where t < −y2.

Let’s look at another example, using a fairly special differential equation.

Example: Let’s sketch a direction field for the differential equation y′ = y:

−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

Slopefield: y’= y

There is something unusually simple about the direction field in example above. Theslope depends only on y, and not t. This is because the differential equation hadthe form y′ = f(y), where f does not depend on t at all. We say such equationsare autonomous. It is clear that sketching a slope field by hand for an autonomousequation is fairly easy, since we need only figure out the slope for a given y value, andwe can repeat this slope at all t values with the same y coordinate.

Example:

Each of the following are autonomous:

y′ = y6 − y′ + 2y

y′ = 2y

y′ = 2 −√y

The following equations are not autonomous:

y′ = t

y′ = y − t

One nice thing about autonomous equations is that it is possible to make some qual-itative statements about the solutions to the equations, without having to find asolution. In particular, since we know what y′ is in terms of y, it is possible to de-termine in what regions of the plane the slope is positive, negative, or zero. (In fact,

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we saw that it is easy to make slope plots for autonomous equations, since the slopeis the same along all t values with the same y-value.) A slope of zero is particularlyinteresting: When we have y′ = f(y) and there is a point y0 where f(y0) = 0, thisis referred to as an equilibrium solution. We see that the constant function y(t) = y0

must then be a solution to y′ = f(y).

Let’s analyze the solutions to y′ = y again:

Consider the equation y′ = y, with y(0) = y0. (We will assume a solution alwaysexists to this initial value problem. Later, we will show that this is so.)First, note that y = 0 is an equilibrium solution. So if we start with y0 = 0, we getthe (equilibrium) solution y(t) = 0.Second, note that if we start with y(0) = y0 > 0, then as y′ = y, the function startsout with positive slope. Thus, y increases. Thus, the slope will continue to increase.For this reason, we know that if we start with y0 > 0, the solution y(t) → ∞ ast → ∞.Third, if we start with y0 < 0, we have y′ < 0, so the y value decreases. It continuesto decrease without bound, since y′ = y. Thus, y(t) → −∞ as t → ∞ if y0 < 0.We will later be able to confirm that all of these things are true, since we will showthat y = y0e

t is always the solution to the initial value problem.Let’s look at another example:

Example:

Let’s try to determine the behavior of the solutions of y′ = 4y−y2 for different initialconditions.First, we look for equilibrium solutions:

4y − y2 = 0 if

Thus, the equilibrium solutions are

Sketch a direction field for −2 ≤ y ≤ 6, using y-values −2, 0, 2, 4, and 6.

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−4 −3 −2 −1 0 1 2 3 4−2

−1

0

1

2

3

4

5

6

Now describe what happens to solutions starting from different points:

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