the indicated y1 is a solution of the given differential equation. find a second solution y2(x).

so my problem is: y'' -4y' + 4y = 0 , y1 = e^(2x)

if y=u(x)y1(x) then y= u(x)e^(2x).

after finding y'' and y' i substituted back into the original equation and got,

u''e^(2x) = 0. and this is where i am stuck. I am not sure what to do.

i know that u'' must be 0.

my book goes on to say that, u= c1x + c2? and that c1= 1 and c2=0 and that

y2= xe^(2x)?

a little help please....

thanks in advance.