1550078481-Ordinary_Differential_Equations__Roberts_

214 Ordinary Differential Equations

EXAMPLE 3 Wrong Initial Guess of the Form of the Particular

Solution

Solve the nonhomogeneous linear differential equat ion

(5) y" - 3y' + 2y = 4e^2 x.

SOLUTION

Again, the auxiliary equation of the associated homogeneous linear equa-
tion, y" - 3y' + 2y = 0, has roots 1 and 2. So the complementary solution of
equation (5) is

Yc(x) = c1ex + c2e^2 x.

Since the nonhomogenei ty b( x) = 4e^2 x, we guess that the particular solution

has the form Yp(x) = Ae^2 x , where A is an unspecified constant. Differentiating

Yp(x) = Ae^2 x twice, we find y;(x) = 2Ae^2 x and y~(x) = 4Ae^2 x. Substituting

Yp(x) and its derivatives into equation (5), we find A must satisfy

4Ae^2 x - 3(2Ae^2 x) + 2(Ae^2 x) = 4e^2 x or 0 = 4e^2 x.

Since the last equation above is clearly false, there is no particular solution

of (5) of the form we have assumed- namely, yp(x) = Ae^2 x. We guessed the

wrong form for a particular solution of the DE (5), and the resulting false
equation is telling us we made the wrong guess. Examining the complemen-
tary solution, Ye ( x) = c 1 ex + c 2 e^2 x, we see that the term Ae^2 x is already
present in the complementary solution, appearing as c 2 e^2 x. So, of course,
Ae^2 x is a solution of the associated homogeneo us equation (5) and produces
the value zero when substituted into the left-hand side of the nonhomogeneous
equation (6).

We need to guess a different form for a particular solution to the DE (5).
But, what should we guess? Since 2 is a single root of the auxiliary equation

p(r) = r^2 - 3r + 2 = 0, the function y(x) = Ae^2 x is one solution of the

associated homogeneous linear differential equation y" - 3y' + 2y = 0. If 2
were a double root of the auxiliary equation, then two linearly independent

solutions of the associated homogeneous equation would be y 1 ( x) = Ae^2 x and

Y2 ( x) = Bxe^2 x. Since 2 is a single root and it is not a double root of the

auxiliary equation, our second guess for the form of a particular solution is

Yp ( x) = Bxe^2 x, where B is an unspecified constant which is to be determined.

Differentiating Yp(x) = Bxe^2 x twice, we find

y~ ( x) = B ( 2x + 1) e^2 x and y~ ( x) = B ( 4x + 4) e^2 x.