I don't understand how to solve this. I can solve the left side and get yc, but I have no idea how to solve for the right side.
y''-3y'+2y=(e^2t)/(e^t+1)
3 answers
http://www.wolframalpha.com/input/?i=Solve+y%27%27-3y%27%2B2y%5C:%3D%5C:%5Cfrac%7Be%5E%7B2t%5C:%7D%7D%7B%5Cleft(e%5E%7Bt%5C:%5C:%7D%2B1%5Cright)%7D
ok. I assume you can get the homogeneous solution.
Since x^2-3x+2=0 means x=1 or 2, that would be
y = c1 e^x + c2 e^(2x)
Now for the fun part. The right side indicates we might have success using undetermined coefficients, with something like
u = ln(e^x+1)
since the derivatives will produce powers of (e^x+1) in the denominator. It is kind of like partial fractions. You will wind up with
y = c1 e^x + c2 e^(2x) + xe^(2x) - e^x ln(e^x+1) - e^(2x) ln(e^x+1)
Since x^2-3x+2=0 means x=1 or 2, that would be
y = c1 e^x + c2 e^(2x)
Now for the fun part. The right side indicates we might have success using undetermined coefficients, with something like
u = ln(e^x+1)
since the derivatives will produce powers of (e^x+1) in the denominator. It is kind of like partial fractions. You will wind up with
y = c1 e^x + c2 e^(2x) + xe^(2x) - e^x ln(e^x+1) - e^(2x) ln(e^x+1)
Thanks for the help, Steve. I didn't see that. I'll give it a shot.