Question

Solve $(D^2-6D+9)y=x+e^{2x}$

Answer 1

Given, $(D^2-6D+9)y=x+e^{2x}$

The characteristic equation
$p^2-6p+9=0$
$\Rightarrow (p-3)(p-3)=0$
$\Rightarrow p=3,3$ ....(real and equal roots)
C.F. $(A+Bx)e^{3x}$
$P.I_1=C_0+C_{1}x$
$\Rightarrow (D^2-6D+9)(C_0+C_{1}x)=x$
$\Rightarrow -6C_1+9C_0+9C_{1}x=x$
$\Rightarrow -6C_1+9C_0=0$
$\Rightarrow 9C_1=1$; $9C_0=\frac{2}{3}$
$\Rightarrow C_1=\frac{1}{9};C_0=\frac{2}{27}$
$\Rightarrow P.I.=(\frac{x}{9}+\frac{2}{27})$
$\Rightarrow P.I_2=\frac{1}{D^2-6D+9}{e}^{2x}=\frac{1}{4--12+9}{e}^{2x}=e^{2x}$
The general solution is
$y=(A+Bx)e^{3x}+\frac{x}{9}+\frac{2}{27}+e^{2x}$