- #1
DunWorry
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Homework Statement
basically solve [itex]\frac{d^{2}y}{dx^{2}}[/itex] + 4[itex]\frac{dy}{dx}[/itex] + 4y = cos2x
Boundary conditions are y=0, dy/dx =1 at x=0
Homework Equations
The Attempt at a Solution
I am having trouble getting the coefficients to the solution. I got the complementary function as y[itex]_{cf}[/itex] = (A[itex]_{1}[/itex]x+A[itex]_{2}[/itex])e[itex]^{-2x}[/itex]
and the paticular integral is y[itex]_{PI}[/itex] = Bcos2x
So first I find coefficient B I [itex]\frac{dy}{dx}[/itex] = -2Bsin2x
and [itex]\frac{d^{2}y}{dx^{2}}[/itex] = -4Bcos2x
Now I sub back into equation and compare coefficients I get -4Bcos2x -8Bsin2x + 4Bcos2x = cos2x
This simplifies to -8Bsin2x = cos2x. I am not sure how I am supposed to compare coefficients here, in the answers it looks like the particular integral is [itex]\frac{1}{8}[/itex] sin2x. If I rearrange -8Bsin2x = cos2x I get B = - [itex]\frac{1}{8}[/itex]cotx
So not only have I got the trig function wrong but I got the sign wrong too, have I done something wrong elsewhere?
Thanks