Answer
$y=\frac{3}{40}e^{-2x}+\frac{9}{8}e^{2x}+\frac{1}{10}e^{x}sinx-\frac{1}{5}e^{x}cosx$
Work Step by Step
$y''-4y=0$
Use auxiliary equation
$r^{2}-4=0$
$r^{2}=4$
$r=±2$
$r_{1}=-2$
$r_{2}=2$
$y_{c}=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$
$y_{c}=c_{1}e^{-2x}+c_{2}e^{2x}$
The particular solution is of the form $y_{p}=Ae^{x}sinx+Be^{x}cosx$
$y_{p}'=Ae^{x}sinx+Ae^{x}cosx+Be^{x}cosx-Be^{x}sinx$
$y_{p}'=(A-B)e^{x}sinx+(A+B)e^{x}cosx$
$y_{p}''=((A-B)-(A+B))e^{x}sinx+((A-B)+(A+B))e^{x}cosx$
$y_{p}''=-2Be^{x}sinx+2Ae^{x}cosx$
Plug back into the main differential equation
$(-2Be^{x}sinx+2Ae^{x}cosx)-4(Ae^{x}sinx+Be^{x}cosx)=e^{x}cosx$
$(-2B-4A)e^{x}sinx+(2A-4B)e^{x}cosx=e^{x}cosx$
$-2B-4A=0$
$2A-4B=1$
Multiply
$-2B-4A+4A-8B=2$
$-10B=2$
$B=-\frac{1}{5}$
Substitute for B
$-2(-\frac{1}{5})-4A=0$
$A=\frac{1}{10}$
Therefore,
$y_{p}=\frac{1}{10}e^{x}sinx-\frac{1}{5}e^{x}cosx$
$y=y_{c}+y_{p}$
$y=c_{1}e^{-2x}+c_{2}e^{2x}+\frac{1}{10}e^{x}sinx-\frac{1}{5}e^{x}cosx$
First conditon is $y(0)=1$
$1=c_{1}e^{-2(0)}+c_{2}e^{2(0)}+\frac{1}{10}e^{(0)}sin(0)-\frac{1}{5}e^{(0)}cos(0)$
$1=c_{1}+c_{2}-\frac{1}{5}$
$c_{1}+c_{2}=\frac{6}{5}$
Differentiate the general solution
$y'=-2c_{1}e^{-2x}+2c_{2}e^{2x}+\frac{1}{10}e^{x}sinx+\frac{1}{10}e^{x}cosx-\frac{1}{5}e^{x}cosx+\frac{1}{5}e^{x}sinx$
Second condition $y'(0)=2$
$2=-2c_{1}e^{0}+2c_{2}e^{0}+\frac{1}{10}e^{0}sin(0)+\frac{1}{10}e^{0}cos(0)-\frac{1}{5}e^{0}cos(0)+\frac{1}{5}e^{0}sin(0)$
$2=-2c_{1}+2c_{2}+\frac{1}{10}-\frac{1}{5}$
$c_{1}-c_{2}=-\frac{21}{20}$
Add $c_{1}+c_{2}$ and $c_{1}-c_{2}$
$2c_{1}=\frac{3}{20}$
$c_{1}=\frac{3}{40}$
Substitute $c_{1}=\frac{3}{40}$
$c_{2}=\frac{6}{5}-\frac{3}{40}$
$c_{2}=\frac{9}{8}$
Solution to initial value problem is
$y=\frac{3}{40}e^{-2x}+\frac{9}{8}e^{2x}+\frac{1}{10}e^{x}sinx-\frac{1}{5}e^{x}cosx$
