2
$\begingroup$

Question: Solve the system of differential equations

$$\begin{cases}\displaystyle\frac{du}{dt} - 2\Omega v \cos\alpha=0\\ \displaystyle\frac{dv}{dt} + 2\Omega u \cos\alpha = -9.8\sin\alpha\end{cases}$$

with initial conditions $u(0) = 0$, and $v(0) = 0$.

My attempt: I have attempted to solve the system, and I've come up with an answer, but it looks fantastically complicated, and I think I may have done something wrong. I'm out of practice with differential equations, so I'm hoping someone can check it over and tell me if I've made any major mistakes.

Let $f = 2\Omega\cos\alpha$. The general solution to the homogeneous system is $$u_h = c\sin(ft + \phi), \ \ \ v_h = c\cos(ft+\phi),$$ where $c$ and $\phi$ are constants of integration. We can also find a particular solution of the non-homogeneous system, $$u_p = -\frac{9.8}{f}\tan\alpha, \ \ \ v_p = 0,$$ so that the solution to the system is $$u = c\sin(ft + \phi) - \frac{9.8}{f}\tan\alpha, \ \ \ v = c\cos(ft+\phi).$$ Finally, we need to solve for the constants. Notice that at $t = 0$, $u$ and $v$ are both $0$, so $$c = \frac{1}{\cos\phi}, \ \ \ \text{ and } \ \ \ \tan\phi = \frac{9.8}{f}\tan\alpha \implies \phi = \tan^{-1}\left(\frac{9.8}{f}\tan\alpha\right).$$ So, we have $$u = \frac{1}{\cos\left(\tan^{-1}\left(\frac{9.8}{f}\tan\alpha\right)\right)}\sin \left(ft +\tan^{-1}\left(\frac{9.8}{f}\tan\alpha\right)\right) - \frac{9.8}{f}\tan\alpha,$$ and $$v = \frac{1}{\cos\left(\tan^{-1}\left(\frac{9.8}{f}\tan\alpha\right)\right)}\sin \left(ft +\tan^{-1}\left(\frac{9.8}{f}\tan\alpha\right)\right).$$

$\endgroup$
2
$\begingroup$

You can make your life a bit easier by realizing that your equation is equivalent to

$$z'=-ifz-gi$$

where $z=u+iv$, $f$ is as you defined and $g=9.8 \sin \alpha$. You can rewrite it as

$$z'+ifz=-gi$$

then the integrating factor is $e^{ift}$:

$$\frac{d}{dt} \left ( e^{ift} z \right ) = -gi e^{ift}.$$

Then this differential equation is easy to solve.

$\endgroup$
  • $\begingroup$ I might be off here, but I think, after looking at this again, that $z' = -ifz - gi$, since then $z' = -if(u + iv) - gi = -ifu +fv - gi$, which means that $\frac{du}{dt} = fv$, and $\frac{dv}{dt} = -fu-gi$? $\endgroup$ – poppy3345 Feb 17 '16 at 21:21
  • $\begingroup$ @poppy3345 You're right. Let me edit. (The idea still makes the problem easier to solve.) $\endgroup$ – Ian Feb 17 '16 at 21:58
  • $\begingroup$ It definitely works - I've solved the equation using this idea, thank you very much for the idea! Just wanted to be sure all my negatives were correct. $\endgroup$ – poppy3345 Feb 18 '16 at 4:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.