Let $\omega_0 \ne \omega$, solve $y'' + \omega_0^2 y = \sin(\omega t)$ by Laplace transform using the initial conditions $y(0)=0$, and $y'(0)=0$.

I have gotten down as far as the partial fraction decomposition being this, $\frac{\omega}{(s^2 + \omega_0^2)(s^2+\omega^2)}$ but am confused as how to proceed from there.

  • $\begingroup$ Posting this to see if there's a better or easier solution for this particular partial fraction decomposition. My method works, but it seems very complicated. $\endgroup$ – OmnipotentEntity Aug 15 '18 at 4:50

From $$A(s^2 + \omega^2) + B(s^2 + \omega_0^2) =(A+B)s^2 + (A\omega^2 + B\omega_0^2) = 1,$$ you get $$A+B=0 ,\qquad A \omega^2 + B \omega_0^2=1 , $$ where you can solve for $A$ and $B$. Only $s$ is considered as a variable here; $\omega$ and $\omega_0$ are constants.


We have $\frac{\omega}{(s^2 + \omega_0^2)(s^2 + \omega^2)}$ and we want to find $\frac{A}{(s^2 + \omega_0^2)} + \frac{B}{(s^2 + \omega^2)}$, ignoring the factor of $\omega$ that we can multiply through later, we have that we want to find $A$ and $B$ such that $A(s^2 + \omega^2) + B(s^2 + \omega_0^2) = 1$. Therefore, $As^2 + Bs^2 = 0s^2$ and $A\omega^2 = 0$ and $B\omega_0^2 = 0$ and we have a mysterious 1 left over. That means the expressions $A$ and $B$ must be more complex than simple natural numbers or expressions in $\omega$, $s$ and $\omega_0$. (In fact, Mathematica flatly refuses to use partial fractions at all on this particular expression.)

Instead you need to be a bit tricky to get the partial fraction for this expression. Consider what would happen if you multiplied through by $\frac{\omega^2 - \omega_0^2}{\omega^2 - \omega_0^2}$ and held the denominator away as a factor for later.*

Now we have $A(s^2 + \omega^2) + B(s^2 + \omega_0^2) = (\omega^2 - \omega_0^2)$, which gives:

$$\begin{split} As^2 + Bs^2 &= 0s^2 \\ A\omega^2 &= \omega^2 \\ B\omega_0^2 &= -\omega_0^2 \\ \end{split}$$

so $A = 1$ and $B = -1$. So our partial fractions are:

$$\frac{\frac1{\omega^2 - \omega_0^2}}{s^2 + \omega_0^2} - \frac{\frac1{\omega^2 - \omega_0^2}}{s^2 + \omega^2} = \frac1{(s^2 + \omega_0^2)(s^2 + \omega^2)}$$

And let's not forget our factor of $\omega$ that we left behind initially:

$$\frac{\omega}{(\omega^2 - \omega_0^2)(s^2 + \omega_0^2)} - \frac{\omega}{(\omega^2 - \omega_0^2)(s^2 + \omega^2)} = \frac{\omega}{(s^2 + \omega_0^2)(s^2 + \omega^2)}$$

The inverse Laplace transform of the second term is easy, you get simply $\frac{\sin(\omega t)}{\omega^2 - \omega_0^2}$.

But the first term is a bit trickier, where can I get a $\omega_0$? Well, let's just make one by multiplying by $\frac{\omega_0}{\omega_0}$!

$$\frac{\omega}{(\omega^2 - \omega_0^2)(s^2 + \omega_0^2)} = \frac{\omega \omega_0}{\omega_0(\omega^2 - \omega_0^2)(s^2 + \omega_0^2)}$$

And now we can take the inverse transform:

$$\frac{\omega \sin(\omega_0 t)}{\omega_0(\omega^2 - \omega_0^2)}$$

And finally we can combine everything by multiplying the first term through by $\frac{\omega_0}{\omega_0}$ as well:

$$\frac{\omega \sin(\omega_0 t) - \omega_0 \sin(\omega t)}{\omega_0 (\omega^2 - \omega_0^2)}$$

And check to ensure that our result satisfies our original differential equation:

$$ \begin{split} y &= \frac{\omega \sin(\omega_0 t) - \omega_0 \sin(\omega t)}{\omega_0 (\omega^2 - \omega_0^2)} \\ y' &= \frac{\omega (\cos(\omega_0 t) - \cos(\omega t))}{\omega^2 - \omega_0^2} \\ y'' &= \frac{\omega (\omega \sin(\omega t) - \omega_0 \sin(\omega_0 t))}{\omega^2 - \omega_0^2} \end{split}$$


$$\begin{split} y'' + \omega_0^2 y &= \frac{\omega (\omega \sin(\omega t) - \omega_0 \sin(\omega_0 t))}{\omega^2 - \omega_0^2} + \omega_0^2 \frac{\omega \sin(\omega_0 t) - \omega_0 \sin(\omega t)}{\omega_0 (\omega^2 - \omega_0^2)} \\ &= \frac{\omega (\omega \sin(\omega t) - \omega_0 \sin(\omega_0 t)) + \omega_0 (\omega \sin(\omega_0 t) - \omega_0 \sin(\omega t))}{\omega^2 - \omega_0^2} \\ &= \frac{\omega^2 \sin(\omega t) - \omega \omega_0 \sin(\omega_0 t)) + \omega \omega_0 \sin(\omega_0 t) - \omega_0^2 \sin(\omega t)}{\omega^2 - \omega_0^2} \\ &= \frac{(\omega^2 - \omega_0^2) \sin(\omega t)}{\omega^2 - \omega_0^2} \\ y'' + \omega_0^2 y &= \sin(\omega t) \\ \end{split}$$

Which was our original equation.

* A point on how I arrived at this seemingly arbitrary factor. I noted before that I had a mystery 1, and how everything was coming up zeros. So I needed to "create" something solvable. Which meant I needed to come up with some way having some of the variables in my expression. This was an easy way of producing a $\omega^2$ and a $\omega_0^2$ while still having $A+B = 0$ as required by the $s^2$ equation. The fact that the problem also specifies $\omega - \omega_0 \ne 0$ can clue you in to the fact that the solution almost certainly involves dividing by $\omega - \omega_0$ at some point.

  • 2
    $\begingroup$ The conclusion “$A \omega^2=B \omega_0^2=0$ and a mysterious $1$ is left over” is wrong, since $\omega$ and $\omega_0$ are just constants, not independent variables. The correct conclusion is $A \omega^2+B \omega_0^2=1$, and this together with $A+B=0$ immediately gives you $A$ and $B$. $\endgroup$ – Hans Lundmark Aug 15 '18 at 8:28
  • $\begingroup$ If you enter this as an answer I will mark it as accepted. This was the simple step I was missing. $\endgroup$ – OmnipotentEntity Aug 15 '18 at 13:05
  • $\begingroup$ OK, as you wish. :-) $\endgroup$ – Hans Lundmark Aug 15 '18 at 15:01

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