Answer
$y=\frac{1}{6}e^{-2x}+\frac{17}{15}e^{x}-\frac{1}{4}-\frac{x}{2}-\frac{3}{20}sin2x-\frac{1}{20}cos2x$
Work Step by Step
$y''+y'-2y=0$
Use auxiliary equation
$r^{2}+r-2=0$
$(r+2)(r-1)=0$
$r_{1}=-2$
$r_{2}=1$
$y_{c}=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$
$y_{c}=c_{1}e^{-2x}+c_{2}e^{x}$
The particular solution is the form $y_{p}=A+Bx+Csin2x+Dcos2x$
$y_{p}'=B+2Ccos2x-2Dsin2x$
$y_{c}''=-4Csin2x-4Dcos2x$
Substitute into the main differential equation
$(-4Csin2x-4Dcos2x)+(B+2Ccos2x-2Dsin2x)-2(A+Bx+Csin2x+Dcos2x)=x+sin2x$
$(B-2A)-2Bx+sin2x(-4C-2D-2C)+cos2x(-4D+2C-2D)=x+sin2x$
$(B-2A)-2Bx+sin2x(-6C-2D)+cos2x(-6D+2C)=x+sin2x$
$-2B=1$
$B=-\frac{1}{2}$
$B-2A=0$
$A=\frac{B}{2}$
$A=-\frac{1}{4}$
$-6C-2D=1$
$2C-6D=0$
Multiply both equations
$6C-18D-6C-2D=1$
$D=-\frac{1}{20}$
$2C-6(-\frac{1}{20})=0$
$C=-\frac{3}{20}$
Therefore
$y_{p}=-\frac{1}{4}-\frac{x}{2}-\frac{3}{20}sin2x-\frac{1}{20}cos2x$
$y=y_{c}+y_{p}$
$y=c_{1}e^{-2x}+c_{2}e^{x}-\frac{1}{4}-\frac{x}{2}-\frac{3}{20}sin2x-\frac{1}{20}cos2x$
The first condition is $y(0)=1$
$1=c_{1}+c_{2}-\frac{1}{4}-\frac{1}{20}$
$\frac{13}{10}=c_{1}+c_{2}$ (1)
The second condition is $y'(0)=0$
$y'=-2c_{1}e^{-2x}+c_{2}e^{x}-\frac{1}{2}-\frac{6}{20}cos2x+\frac{2}{20}sin2x$
$0=-2c_{1}+c_{2}-\frac{1}{2}-\frac{6}{20}$
$\frac{2}{5}=-c_{1}+\frac{1}{2}c_{2}$ (2)
Add equation (1) and (2) together
$\frac{3}{2}c_{2}=\frac{13}{10}+\frac{4}{10}=\frac{17}{10}$
$c_{2}=\frac{17}{15}$
Substitute $c_{2}=\frac{17}{15}$ in equation (1)
$c_{1}=\frac{13}{10}-\frac{17}{15}=\frac{39}{30}-\frac{34}{30}=\frac{5}{30}=\frac{1}{6}$
The solution of the initial problem is
$y=\frac{1}{6}e^{-2x}+\frac{17}{15}e^{x}-\frac{1}{4}-\frac{x}{2}-\frac{3}{20}sin2x-\frac{1}{20}cos2x$
