y''-2y'+y=0 ; With Solution y = e^x
So far I got:
let y=f(x)*v = e^xv
y'=e^xv+e^xv'
y''=e^xv+e^xv'+e^xv'+e^xv''
substituting that into y''-2y'+y=0 gets=
e^xv'' and then v''=w' we get
e^xw' not sure where to go from here though : (
Since this is a second order linear constant coefficient ODE, find the characteristic equation
$\displaystyle m^2 - 2m + 1 = 0$
$\displaystyle (m - 1)^2 = 0$
$\displaystyle m = 1$, a repeated root.
So your solution to the DE will have the form
$\displaystyle y = C_1e^{x} + C_2xe^{x}$.
You will need to use some intial conditions to find the constants.