I am trying to solve a system of differential equations (for the full system see below) and I am stuck the following 2nd order ODE (with $a$ and $b$ being constants and $\dot x = \frac{dx}{dt}$):

$$\ddot x - \frac34 a\dot x^2 -b \dot x + 2 a b = 0$$

I tried to substitute $v := \dot x$, which leads me to a different ODE which I was not able to solve. And now I have no idea what else I could try here and would love to get some pointers.

And here the original system / IVP I want to solve: $$ \dot x = 2 a + 2 e^{ax}y \\ \dot y = \frac32 e^{-ax}a^3+(a^2+b)y-\frac12 a e^{ax}y^2 \\ x(0)=0 \\ y(T)=0$$

In which I solved the first equation for $y$ and substituted $y$ and $\dot y$ in the 2nd which gave me the equation above.

  • $\begingroup$ What if you define $z=x'$ first ? $\endgroup$ – Claude Leibovici Sep 27 '14 at 11:22
  • $\begingroup$ @ClaudeLeibovici Do you mean first as in before reducing the system to the 2nd order ODE or before solving the 2nd order ODE? If you are talking about the second: I have done that, but I still was unable to solve the resulting 1st order ODE. $\endgroup$ – user155 Sep 30 '14 at 9:31
  • $\begingroup$ Have a look to my answer. I hope I did not make any mistake; in any manner, this is the idea as Semsem suggested from beginning. Let me know. $\endgroup$ – Claude Leibovici Sep 30 '14 at 9:57

As I said in a comment, start defining $z=x'$; so the differential equation becomes $$\frac{dz}{dt} - \frac34 a z^2 -b z + 2 a b = 0$$ that is to say $$\frac{dz}{dt} = \frac34 a z^2 +b z - 2 a b $$ then, as Semsem suggested, it is separable; so $$\frac{dt}{dz} = \frac{1}{\frac34 a z^2 +b z - 2 a b}$$ so $$t+C=-\frac{2 \tanh ^{-1}\left(\frac{3 a z+2 b}{2 \sqrt{b} \sqrt{6 a^2+b}}\right)}{\sqrt{b} \sqrt{6 a^2+b}}$$ from which $$z=-\frac{2 \left(\sqrt{b} \sqrt{6 a^2+b} \tanh \left(\frac{1}{2} \sqrt{b} \sqrt{6 a^2+b} (C+t)\right)+b\right)}{3 a}$$ Now, one more unpleasant integration leads to $$x=-\frac{2 \left(2 \log \left(\cosh \left(\frac{1}{2} \sqrt{b} \sqrt{6 a^2+b} (C+t)\right)\right)+b t\right)}{3 a}+D$$

  • $\begingroup$ Thanks a lot! Too bad that I can not accept 2 answers. $\endgroup$ – user155 Sep 30 '14 at 10:04


you can use $\ddot x=\frac{\dot x d\dot x}{dx}$ and then use separation of variables

  • $\begingroup$ Thanks for the answer! However, I am not able to get your hint. If I use $\ddot x$ as you suggested, I get an equation with $\frac{d \dot x}{d x}$ and $\frac{d x}{d t}$ how can I use separation here? $\endgroup$ – user155 Sep 30 '14 at 9:29
  • $\begingroup$ Thanks to @Claude, I finally managed to understand your hint. Too bad that I can not accept both of your answers. So, thanks again! $\endgroup$ – user155 Sep 30 '14 at 10:05
  • $\begingroup$ You are very welcome. I hope and wish you understood the mechanism. You will need it a lot of times. Cheers :) $\endgroup$ – Claude Leibovici Sep 30 '14 at 10:11

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