Answer
$y=2e^{2.5x}-8xe^{2.5x}$
Work Step by Step
$4y''-20y'+25y=0$
Use auxiliary equation
$4r^{2}-20r+25=0$
$(2r-5)(2r-5)=0$
$r=\frac{5}{2}$
$y=c_{1}e^{(2.5)x}+c_{2}xe^{(2.5)x}$
$y(0)=2$
$y(0)=2=c_{1}e^{(2.5)x}+c_{2}xe^{(2.5)x}$
$2=c_{1}e^{0}+0c_{2}e^{0}$
$2=c_{1}$
$y'(0)=-3$
$y'(0)=-3=\frac{5}{2}c_{1}e^{0}+\frac{5}{2}c_{2}(0)e^{0}+e^{0}c_{2}$
$-3=\frac{5}{2}c_{1}+c_{2}$
$-3=\frac{5}{2}(2)+c_{2}$
$c_{2}=-8$
Hence, $y=2e^{2.5x}-8xe^{2.5x}$
