## Calculus: Early Transcendentals 8th Edition

$y=c_{1}e^{2x}+c_{2}xe^{2x}+\frac{1}{4}+\frac{x}{4}-\frac{3}{25}sinx-\frac{4}{25}cosx$
$y''-4y'+4y=0$ 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, $y_{p}=\frac{1}{4}+\frac{x}{4}-\frac{3}{25}sinx-\frac{4}{25}cosx$ $y=y_{c}+y_{p}$ $y=c_{1}e^{2x}+c_{2}xe^{2x}+\frac{1}{4}+\frac{x}{4}-\frac{3}{25}sinx-\frac{4}{25}cosx$