Answer to Question #213691 in Differential Equations for anuj

Question

Solve the equations by the use of the operator D

D²y -6Dy +9y= e^3x + e^-3x

Answer 1

The homogeneous equation

$$L(y)=(D^2-6D+9)y=0$$

$$(D-3)^2y=0$$

The general solution of the homogeneous equation is

$$y_h=(A+Bx)e^{3x}$$

Find the particular equation.

$$L(D)y=e^{3x}$$

$L(D)=(D^2-6D+9)=(D-3)^2$ has $3$ as a double root. Then

$$y_1=\dfrac{x^2e^{3x}}{L''(3)}=\dfrac{x^2e^{3x}}{2}$$

$$L(D)y=e^{-3x}$$

$$y_2=\dfrac{e^{-3x}}{L(-3)}=\dfrac{e^{-3x}}{(-3-3)^2}=\dfrac{e^{-3x}}{36}$$

$$y_p=y_1+y_2=\dfrac{x^2e^{3x}}{2}+\dfrac{e^{-3x}}{36}$$

The general solution of the nonhomogeneous equation is

$$y=(A+Bx)e^{3x}+\dfrac{x^2e^{3x}}{2}+\dfrac{e^{-3x}}{36}$$