4.1 Basics of differential equations (Page 3/14)

Calculus volume 2 Page 3 / 14

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:

\begin{matrix} 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} . \end{matrix}

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 $y (0) :$

\begin{matrix} y (0) & = 2 e^{−2 (0)} + e^{0} \\ = 2 + 1 \\ = 3 . \end{matrix}

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

\int y^{'} d x = \int (3 e^{x} + x^{2} - 4) d x,

namely,

y + C_{1} = 3 e^{x} + \frac{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} + \frac{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} + \frac{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 $C :$

\begin{matrix} 5 & = & 3 e^{0} + \frac{1}{3} 0^{3} - 4 (0) + C \\ 5 & = & 3 + C \\ C & = & 2 . \end{matrix}

Now we substitute the value $C = 2$ into [link] . The solution to the initial-value problem is $y = 3 e^{x} + \frac{1}{3} x^{3} - 4 x + 2 .$

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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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