Page 69 - Advanced engineering mathematics

2.1 The Linear Second-Order Equation 49

THEOREM 2.5

Let Y p be any solution of the homogeneous equation (2.1). Let y 1 and y 2 be linearly independent
solutions of equation (2.2). Then the expression

c 1 y 1 + c 2 y 2 + Y p
contains every solution of equation (2.1).

For this reason, we call c 1 y 1 + c 2 y 2 + Y p the general solution of equation (2.1).
Theorem 2.5 suggests a strategy for finding all solutions of the nonhomogeneous
equation (2.1).
1. Find two linearly independent solutions y 1 and y 2 of the associated homogeneous
equation y + p(x)y + q(x)y = 0.

2. Find any particular solution Y p of the nonhomogeneous equation y + p(x)y + q(x)y =
f (x).

3. The general solution of y + p(x)y + q(x)y = f (x) is
y(x) = c 1 y 1 (x) + c 2 y 2 (x) + Y p (x)
in which c 1 and c 2 can be any real numbers.
If there are initial conditions, use these to find the constants c 1 and c 2 to solve the initial
value problem.

EXAMPLE 2.5

We will find the general solution of

y + 4y = 8x.
It is routine to verify that sin(2x) and cos(2x) are linearly independent solutions of y + 4y = 0.

Observe also that Y p (x)=2x is a particular solution of the nonhomogeneous equation. Therefore,

the general solution of y + 4y = 8x is
y = c 1 sin(2x) + c 2 cos(2x) + 2x.
This expression contains every solution of the given nonhomogeneous equation by choosing
different values of the constants c 1 and c 2 .
Suppose we want to solve the initial value problem

y + 4y = 8x; y(π) = 1, y (π) =−6.

First we need
y(π) = c 2 cos(2π) + 2π = c 2 + 2π = 1,
so c 2 = 1 − 2π. Next, we need
y (π) = 2c 1 cos(2π) − 2c 2 sin(2π) + 2 = 2c 1 + 2 =−6,

so c 1 =−4. The unique solution of the initial value problem is
y(x) =−4sin(2x) + (1 − 2π)cos(2x) + 2x.

We now have strategies for solving equations (2.1) and (2.2) and the initial value problem.
We must be able to find two linearly independent solutions of the homogeneous equation and any
one particular solution of the nonhomogeneous equation. We now will develop important cases
in which we can carry out these steps.

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October 14, 2010 14:12 THM/NEIL Page-49 27410_02_ch02_p43-76