# Thread: solving the differential equation

1. ## solving 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

2. ## Differential equation

Hello gambler84
Originally Posted by gambler84
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
Since you talk about the right-hand-side, I assume you know how to solve the Complementary Function to get

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

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

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

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

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

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

Comparing coefficients gives

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

So the solution to the PI is

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

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

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

Plug in your initial conditions and you're done.