Originally Posted by

**isuckatcalc** I can get this about 3/4ths of the way done, but i'm missing something and i don't know where it comes from.

Solve the initial value problem

y'' +y=cos(x) y(0)=1 y'(0)=-1

My method was to first find the roots of the equation r^2 +1. That gives +-i

Yp= c1cos(x) + c2sin(x)

And the characteristic equation i believe would be (i may be confusing characteristic and particular, but either way, that doesn't impact my problem, i don't believe)

Yc=Acosx + Bsinx

then

Y'c= -Asinx +Bcosx (just so i don't skip a step)

Y''c= -Acosx -Bcosx

If I plug this back into the originial equation of y''+y=cosx, all the terms on the left side cancel each other out, and i get cosx=0. I'm not sure what to do with that. A and B are both 0?

Anyway, I then solve for the initial conditions, and I get c1=1, and c2=-1.

Final solution cosx-sinx. But my answer tells me there is a +1/2xsinx that i am missing. Where does that come from? I tried another problem and had the ssame issue, i got the first two terms right, but then there was a -1/5 sinx. I'm not seeing how these are gotten.