Answer
$y=4e^{\frac{1}{2}x}-2xe^{\frac{1}{2}x}$
Work Step by Step
$4y''-4y'+y=0$
Use auxiliary equation
$4r^{2}-4r+1=0$
$r^{2}-r+\frac{1}{4}=0$
$(r-\frac{1}{2})^{2}=0$
$r=\frac{1}{2}$
$y=c_{1}e^{r_{1}x}+c_{2}xe^{r_{2}x}$
$y=c_{1}e^{(\frac{1}{2})x}+c_{2}xe^{(\frac{1}{2})x}$
$y(0)=4$
$4=c_{1}e^{(\frac{1}{2})(0)}+c_{2}(0)e^{(\frac{1}{2})(0)}$
$4=c_{1}(1)+0$
$c_{1}=4$
$y(2)=0$
$0=c_{1}e^{(\frac{1}{2})(2)}+c_{2}(2)e^{(\frac{1}{2})(2)}$
$0=c_{1}e+2c_{2}e$
$c_{1}e=-2c_{2}e$
$c_{1}=-2c_{2}$
$4=-2c_{2}$
$c_{2}=-2$
$y=4e^{\frac{1}{2}x}-2xe^{\frac{1}{2}x}$
