Question

y" -5y' +4y=0 ;y(0)=1 and y' (0)=1 .

Answer 1

$y'' -5y' +4y=0 ; \space \space\space \space y(0)=1 \space\space \space and \space \space\space y' (0)=1 .$

Setup the auxilary equation

$\lambda^2 - 5\lambda +4 = 0 \to \lambda_1= 1 \space \space\space \space \lambda_2=4$

The general solution is found by

Usually problem like these will have the answer in the form $C_1e^a+C_2e^b...$ where a and b  are the roots of the characteristic equation $e^{rt}$

y = $C_1e^{t} + C_2e^{4t}$ $\to C_1+C_2 = 1$

$y' = C_1e^t+4C_2e^{4t} \to C_1+4C_2 = 1$

$C_1 = 1 \space\space\space\space C_2= 0$

The general solution is: $y = e^t$