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4.1 Basics of differential equations  (Page 3/14)

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A differential equation together with one or more initial values is called an initial-value problem    . The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation y ′ = 2 x , then y ( 3 ) = 7 is an initial value, and when taken together, these equations form an initial-value problem. The differential equation y ″ − 3 y ′ + 2 y = 4 e x is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be y ( 0 ) = 2 and y ′ ( 0 ) = −1 . These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is t , which represents time. Thus, a value of t = 0 represents the beginning of the problem.

Verifying a solution to an initial-value problem

Verify that the function y = 2 e −2 t + e t is a solution to the initial-value problem

y ′ + 2 y = 3 e t , y ( 0 ) = 3 .

For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that y satisfies the differential equation, we start by calculating y ′ . This gives y ′ = −4 e −2 t + e t . Next we substitute both y and y ′ into the left-hand side of the differential equation and simplify:

y ′ + 2 y = ( −4 e −2 t + e t ) + 2 ( 2 e −2 t + e t ) = −4 e −2 t + e t + 4 e −2 t + 2 e t = 3 e t .

This is equal to the right-hand side of the differential equation, so y = 2 e −2 t + e t solves the differential equation. Next we calculate y ( 0 ) :

y ( 0 ) = 2 e −2 ( 0 ) + e 0 = 2 + 1 = 3 .

This result verifies the initial value. Therefore the given function satisfies the initial-value problem.

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Verify that y = 3 e 2 t + 4 sin t is a solution to the initial-value problem

y ′ − 2 y = 4 cos t − 8 sin t , y ( 0 ) = 3 .
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In [link] , the initial-value problem consisted of two parts. The first part was the differential equation y ′ + 2 y = 3 e x , and the second part was the initial value y ( 0 ) = 3 . These two equations together formed the initial-value problem.

The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of C . The family of solutions to the differential equation in [link] is given by y = 2 e −2 t + C e t . This family of solutions is shown in [link] , with the particular solution y = 2 e −2 t + e t labeled.

A graph of a family of solutions to the differential equation y’ + 2 y = 3 e ^ t, which are of the form y = 2 e ^ (-2 t) + C e ^ t. The versions with C = 1, 0.5, and -0.2 are shown, among others not labeled. For all values of C, the function increases rapidly for t < 0 as t goes to negative infinity. For C > 0, the function changes direction and increases in a gentle curve as t goes to infinity. Larger values of C have a tighter curve closer to the y axis and at a higher y value. For C = 0, the function goes to 0 as t goes to infinity. For C < 0, the function continues to decrease as t goes to infinity.
A family of solutions to the differential equation y ′ + 2 y = 3 e t . The particular solution y = 2 e −2 t + e t is labeled.

Solving an initial-value problem

Solve the following initial-value problem:

y ′ = 3 e x + x 2 − 4 , y ( 0 ) = 5 .

The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation

∫ y ′ d x = ∫ ( 3 e x + x 2 − 4 ) d x ,

namely,

y + C 1 = 3 e x + 1 3 x 3 − 4 x + C 2 .

We are able to integrate both sides because the y term appears by itself. Notice that there are two integration constants: C 1 and C 2 . Solving [link] for y gives

y = 3 e x + 1 3 x 3 − 4 x + C 2 − C 1 .

Because C 1 and C 2 are both constants, C 2 − C 1 is also a constant. We can therefore define C = C 2 − C 1 , which leads to the equation

y = 3 e x + 1 3 x 3 − 4 x + C .

Next we determine the value of C . To do this, we substitute x = 0 and y = 5 into [link] and solve for C :

5 = 3 e 0 + 1 3 0 3 − 4 ( 0 ) + C 5 = 3 + C C = 2 .

Now we substitute the value C = 2 into [link] . The solution to the initial-value problem is y = 3 e x + 1 3 x 3 − 4 x + 2 .

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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