Originally Posted by
zxcv Hi,
I would like to know how to solve the following:
Solve the differential equation using the method of varaition of parameters.
y" - 3y' + 2y = 1/(1+e^-x)
What I have done till now (not really able to continue):
- solution to y" - 3y' + 2y = 0 is c1e^x+c2e^2x
-y1 = e^x & y2 = e^2x
-yp = u1e^x + u2e^2x
However, I am not sure if I have u2 correctly:
u1 = ln (1+e^-x)
u2 = -(1+e^-x) + ln (1+e^-x) where I have integral of (e^-2x)/(1+e^-x)
can someone please check with me and help me solve the following please.
Thanks in advance!