Answer
$y=c_{1}e^{2x}+c_{2}xe^{2x}+\frac{1}{4}+\frac{x}{4}-\frac{3}{25}sinx-\frac{4}{25}cosx$
Work Step by Step
$y''-4y'+4y=0$
Use auxiliary equation
$r^{2}-4r+4=0$
$(r-2)^{2}=0$
$r=2$
$y_{c}=c_{1}e^{r_{1}x}+c_{2}xe^{r_{2}x}$
$y_{c}=c_{1}e^{2x}+c_{2}xe^{2x}$
The particular solution is of the form
$y_{p}=A+BxCsinx+Dcosx$
$y_{p}'=B+Ccosx-Dsinx$
$y_{p}''=-Csinx-Dcosx$
Plug back into the original equation
$(-Csinx-Dcosx)-4(B+Ccosx-Dsinx)+4(A+Bx+Csinx+Dcosx)=x-sinx$
$-Csinx-Dcosx-4B-4Ccosx+4Dsinx+4A+4Bx+4Csinx+4Dcosx=x-sinx$
$(4A-4B)+(4B)x+(4C+4D-C)sinx+(-D-4C+4D)cosx=x-sinx$
$(4A-4B)+(4B)x+(3C+4D)sinx+(-4C+3D)cosx=x-sinx$
$4B=1$
$B=\frac{1}{4}$ (1)
$4A-4B=0$
$4A-1=0$
$A=\frac{1}{4}$ (2)
$3C+4D=-1$ (3)
$-4C+3D=0$ (4)
Multiply both equations
$12C+16D-12C+9D=-4$
$25D=-4$
$D=-\frac{4}{25}$
Substitute $D=-\frac{4}{25}$ in equation(3)
$3C+4(-\frac{4}{25})=-1$
$3C=-1+\frac{16}{25}$
$C=-\frac{3}{25}$
Therefore,
$y_{p}=\frac{1}{4}+\frac{x}{4}-\frac{3}{25}sinx-\frac{4}{25}cosx$
$y=y_{c}+y_{p}$
$y=c_{1}e^{2x}+c_{2}xe^{2x}+\frac{1}{4}+\frac{x}{4}-\frac{3}{25}sinx-\frac{4}{25}cosx$
