Answer
$y=e^{2x}(c_{1}cosx+c_{2}sinx)+\frac{e^{-x}}{10}$
Work Step by Step
$y''-4y'+5y=0$
Use auxiliary equation
$r^{2}-4r+5=0$
$r=\frac{-(-4)±\sqrt ((-4)^{2}-4(1)(5))}{2(1)}$
$r=\frac{4±\sqrt (16-20)}{2}$
$r=\frac{4±\sqrt -4}{2}$
$r=2±i$
$α=2$
$β=1$
$y_{c}=e^{αx}(c_{1}cosβx+c_{2}sinβx)$
$y_{c}=e^{2x}(c_{1}cosx+c_{2}sinx)$
$y_{p}=Ae^{-x}$
$y_{p}'=-Ae^{-x}$
$y_{p}''=Ae^{x}$
Plug back into the original equation
$(Ae^{-x})-4(-Ae^{-x})+5(Ae^{-x})=e^{-x}$
$(A+4A+5A)e^{-x}=e^{-x}$
$(10A)=1$
$A=\frac{1}{10}$
$y_{p}=\frac{e^{-x}}{10}$
$y=y_{c}+y_{p}$
$y=e^{2x}(c_{1}cosx+c_{2}sinx)+\frac{e^{-x}}{10}$
