(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Solve the ODE with the boundary conditions given

Q''+Q = Sin(2x) where Q(0) = 1 and Q'(0) = 2

So i know i need to solve the general and particular solutions, however, I am a little confused. Any help or advice would be great, Thanks in advance.

2. Relevant equations

Y = Yg +Yp

3. The attempt at a solution

Assume RHS = 0 for general solution

Q'' + Q = 0

Let Q = e^kx

Q' = k e^kx

Q'' = k^2 e^kx

therefore

k^2e^kx + e^kx = 0

e^kx(k^2) = 0

k=0???????

Yg = A e ^kx +Bx e^kx???

Particular Solution

Yp = A Sin 2x + B Cos 2x

Yp' = 2A Cos 2x - 2B Sin 2x

Yp"= -4A Sin 2x - 4B Cos (2x)

-4A Sin 2x - 4B Cos (2x) + A Sin 2x + B Cos 2x = Sin(2x)

-3A Sin 2x = Sin 2x

a = -1/3

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Second Order Differential Equations

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