Where does the second term in this harmonic motion come from?

Question

Question: $$ \frac{d^2x}{dt^2} + \omega^2 x = 2\sin \Omega t . $$ Determine the motion of the object, given that $x (0)= -4$ and $x'(0)=0$.

Answer: $$ x (t) = -4\cos\omega t - \frac {\Omega}{\omega}\frac{2}{\omega^2-\Omega^2}\sin \omega t + \frac{2}{\omega^2-\Omega^2}\sin \Omega t . $$

My Attempt:

I was able to find that homogeneous equation is

$h=-4\cos(\omega t)$

and a particular solution is given by

$p=\dfrac{2\sin(\Omega t)}{\omega^2-\Omega^2} $

Working for $p$:

$p$ is in the form:

$p=a\cos(\Omega t)+b\sin(\Omega t)$

$p' = -a\Omega \sin(\Omega t) +b\Omega \cos(\Omega t)$

$p'' = -\Omega^2 p$

Substituting into our original second-order ODE, we get

$p=\dfrac{2\sin(\Omega t)}{\omega^2-\Omega^2} $

My solution

Now,

$x(t) = h+p$

Therefore,

$x(t) =-4\cos(\omega t) +\dfrac{2\sin(\Omega t)}{\omega^2-\Omega^2} $

But as you can see, the Answer at the top of this post is different (there is a middle term). I think I've missed something fundamental. Help appreciated, thanks!

Answer 1

The homogeneous solutions are $$ h (t) = a\cos \omega t + b\sin \omega t , $$ where $a$ and $b$ are reals. The proposed particular solution is correct: $$ p (t) = \frac{2}{\omega^2 - \Omega^2} \sin \Omega t . $$ Now, form $x = h+p$ and apply the initial conditions to determine $a$ and $b$.

Answer 2

The general solution of the homogeneous ODE is : $$x(t)=c_1\sin(\omega t)+c_2\cos(\omega t)$$ Don't take the boundary conditions into account at this stage : this would be a mistake because the conditions are for the non-homogeneous ODE, not for the homogeneous. That is not the same !

The general solution of the non-homogeneous ODE is : $$x(t)=c_1\sin(\omega t)+c_2\cos(\omega t)+\frac{2\sin(\Omega t)}{\omega^2-\Omega^2}$$ Now you can take the conditions into account in order to find $c_1$ and $c_2$. I suppose that you can take it from here.