Answer
$y=\frac{1}{3}(e^{4x}-e^{x})$
Work Step by Step
$y''-5y'+4y=0$ which has the characteristic equation $t^2-5t+4=0$ with solutions $t=1$ and $t=4$
General solution: $c_1e^{4x}+c_2e^{x}$
Plug in the values: $y(0)=0$ and $y'(0)=1$ giving $c_1=1/3$, $c_2=\frac{-1}{3}$
Hence, $y=\frac{1}{3}(e^{4x}-e^{x})$
