tihs is going to sound like a stupid question, but how do i do it when the right side is like this?

x''+4x=1+t+sin(t) with initial conditions x(0)=0 and x'(0)=0

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- Feb 4th 2009, 08:58 PMgambler84solving the differential equation
tihs is going to sound like a stupid question, but how do i do it when the right side is like this?

x''+4x=1+t+sin(t) with initial conditions x(0)=0 and x'(0)=0 - Feb 5th 2009, 12:04 AMGrandadDifferential equation
Hello gambler84Since you talk about the right-hand-side, I assume you know how to solve the Complementary Function to get

$\displaystyle x = A\cos 2t + B \sin 2t$

Next, you need to find the Particular Integral, by saying.

Let $\displaystyle x = P +Qt +R\cos t + S\sin t$

Then $\displaystyle \dot{x} = Q - R\sin t + S\cos t$

$\displaystyle \Rightarrow \ddot{x} = -R\cos t - S\sin t $

$\displaystyle \Rightarrow -R\cos t - S\sin t + 4(P +Qt +R\cos t + S\sin t) \equiv 1 + t + \sin t$

Comparing coefficients gives

$\displaystyle P=Q= \tfrac{1}{4}, R = 0, S = \tfrac{1}{3}$

So the solution to the PI is

$\displaystyle x =\tfrac{1}{4} + \tfrac{1}{4}t +\tfrac{1}{3}\sin t$

The general solution to your equation is the sum of the solutions to the PI and the CF:

$\displaystyle x = A\cos 2t + B \sin 2t + \tfrac{1}{4} + \tfrac{1}{4}t +\tfrac{1}{3}\sin t$

Plug in your initial conditions and you're done.

Grandad