# Second order ODE initial value problem

So the question is y" - y' - 6y = e^-x + 12x, y(0)=1,y'(0)=-2

First I found the general solution which came out to be, Ae^3x + Be^-2x

I then Substituted y=ae^-x + bx + c
y'=-ae^-x + b
y"=ae^-x
Then I just compared the coefficients to get a=-1/4, B=-2 and C=-1/6

So I am getting y = Ae^3x + Be^-2x -(e^-x)/4 - 2x - 1/6
Im not sure if this is right, I have done the rest but I get some funny answers for A and B so I was wondering if someone could verify if this answer is right. Thanks

Your C value isn't correct; try plugging everything back into the equation and see if you get a different C.

Your C value isn't correct; try plugging everything back into the equation and see if you get a different C.
I just tried it again. Got the same answer.

Could you show what you did?

Could you show what you did?
I did ae^-x - (-ae^-x+b) - 6(ae^-x+bx+c) = e^-x + 12x
Then I got e^-x(-4a-1) - b(1+6c) = x(12+6b)

After that I did -4a = 1 so a = -1/4
6c=-1 so c = -1/6
6b=-12 so b = -2

Could you tell me where I am going wrong?

ehild
Homework Helper
I did ae^-x - (-ae^-x+b) - 6(ae^-x+bx+c) = e^-x + 12x
Then I got e^-x(-4a-1) - b(1+6c) = x(12+6b)
the red one. It is -(b+6c)

ehild