Find this general solution:
y" + 3y' + 2y = 36xe^x
Here's my work:
1. r^2 + 3r + 2 = 0; r=-1, -2
2. yc = c1e^-x + c2^e^-2x
3. yp=Axe^x
y'p=Axe^x + Ae^x
y"p=Axe^x + 2Ae^x
4. (Axe^x + 2Ae^x) + 3(Axe^x + Ae^x) +2(Axe^x) = 36xe^x
5. 6Axe^x + 5Ae^x = 36xe^x
And here's where I get fuzzy. I need to find this value of A and then put in the form y = yc + yp. The answer in the back of the book is:
y= c1e^-x + c2^e^-2x + (6x-5)e^x
Any suggestions? Thanks, JN
Thanks for the input EmptySet. So from this point:
1. yp = (Ax + B)e^x
y'p= Ae^x + (Ax + B )e^x
y"p=2Ae^x + Axe^x + Be^x
2. yp= 2Ae^x + Axe^x + Be^x +3Ae^x + 3xAe^x + 3Be^x + 2Axe^x + 2Be^x 2Be^x
So then 5Ae^x + 6Axe^x + 6Be^x = 36xe^x
how do I make the jump to yp = (6x-5)e^x?