Answer to Question #212449 in Differential Equations for Prof

Question

Solve the equations by the use of the operator D

D²y -6Dy +9y= e^3x + e^-3x

Answer 1

$$(D-3)^2(y)=e^{3x} + e^{-3x}$$

$$(D-3)^2=e^{3x}D^2e^{-3x}$$

Then

$$e^{3x}D^2e^{-3x}(y)=e^{3x} + e^{-3x}$$

$$D^2e^{-3x}(y)=e^{3x} + e^{-3x}$$

We have the general solution od homogeneous differential equation

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

$$L(D)\bigg[\dfrac{e^{kx}}{L(k)}\bigg]=e^{kx}$$

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

Since

$$L(3)=(3)^2-6(3)+9=0$$

we use

$$y_1=\dfrac{1}{1}(\dfrac{1}{2!}x^2)e^{3x}=\dfrac{1}{2}x^2e^3x$$

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

$$L(-3)=(-3)^2-6(-3)+9=36$$

The particular solution is

$$y_p=\dfrac{1}{2}x^2e^{3x}+\dfrac{1}{36}e^{-3x}$$

The genral solution of nonhomogeneous differential equation is

$$y=(A+Bx)e^{3x}+\dfrac{1}{2}x^2e^{3x}+\dfrac{1}{36}e^{-3x}$$