Thread: Undetermiend Coeffecient Problem

So the problem is
y''+y=cosx y(0)=1 y(0)=-1

The roots are r^2=-1 (+i,-i)

so that would give a homogeneious equation of

y=c1cosx +c2sinx

but how do i figure out the other equation that i need? is it just as simple as y=Acosx?

if i go that route, then
y'=-Asinx
y''=-Acosx.

plugging that into the original equation, i would get -Acosx+Acosx=cosx. cosx=0
that doesn't seem right at all. i'm confused on how to figure out how to pick the second equation.

The forcing function is cos(x), so your particular solution would ordinarily be y = A sin(x) + B cos(x). However, since cos(x) is already a part of the homogeneous solution, you add x's to the particular solution like: y = A x sin(x) + B x cos(x). See if that works.