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 $$v = \frac{1}{\cos\left(\tan^{-1}\left(\frac{9.8}{f}\tan\alpha\right)\right)}\sin \left(ft +\tan^{-1}\left(\frac{9.8}{f}\tan\alpha\right)\right).$$


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


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


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.

  • $\begingroup$ 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

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