I have been trying for days to figure this one out by myself, but I am stuck. Here is the equation:

$$\dfrac{a x^2}{ b x^2 + c} = \dfrac{dx}{dt}$$

I can get a solution, BUT, I cannot recover the following condition for $t = 0$

$$x(0) = x_0$$

Any help is greatly appreciated. Thank you.

  • $\begingroup$ Did you mean $x(0)=x_0$? $\endgroup$ – John Doe Oct 29 '17 at 23:20
  • $\begingroup$ I separate it and end up getting a quadratic. This gives me an equation for x(t). But at t = 0 I do not recover x(0) = x(0) $\endgroup$ – bsmith144 Oct 29 '17 at 23:35
  • $\begingroup$ yes John Dee, and thank you for the edit, Moo $\endgroup$ – bsmith144 Oct 29 '17 at 23:35
  • $\begingroup$ what are you calling K ? $\endgroup$ – bsmith144 Oct 29 '17 at 23:53
  • $\begingroup$ as you have it K = 0 $\endgroup$ – bsmith144 Oct 30 '17 at 0:04

The general solution of this separable equation is

$$ - \frac{c}{ax} + \frac{b x}{a} = t + k $$

where $k$ is an arbitrary constant. Plugging in the initial condition,

$$ - \frac{c}{a x_0} + \frac{b x_0}{a} = k $$

Thus $$ - \frac{c}{ax} + \frac{b x}{a} = t - \frac{c}{a x_0} + \frac{b x_0}{a} $$

Now multiply both sides by $x$, and solve the resulting quadratic, if you want an explicit expression for $x$. Caution: the quadratic will have two roots, and which one is valid will depend on the sign of $b x_0^2 + c$.

EDIT: Namely, the two roots are

$$ x = \frac{a t}{2b} + \frac{x_0}{2} - \frac{c}{2 b x_0} \pm \frac{\sqrt{a^2 t^2 x_0^2 + (2 a b x_0^3 - 2 a c x_0) t + b^2 x_0^4 + 2 b c x_0^2 + c^2}}{2 b x_0} $$

For $t = 0$, $x =x_0$, this is

$$ \frac{x_0}{2} - \frac{c}{2 b x_0} \pm \frac{\sqrt{ b^2 x_0^4 + 2 b c x_0^2 + c^2}}{2 b x_0} = \frac{x_0}{2} - \frac{c}{2 b x_0} \pm \frac{\sqrt{(b x_0^2 + c)^2}}{2 b x_0}$$

You want to make $\pm \sqrt{(b x_0^2 + c)^2} = b x_0^2 + c$. Thus if $b x_0^2 + c \ge 0$ you can use $+$, but if $b x_0^2 + c < 0$ you must use $-$.

  • $\begingroup$ I actually did this but then when you get the resulting quadratic and set t = 0 i do not recover x(0) = x_initial $\endgroup$ – bsmith144 Oct 30 '17 at 1:23
  • $\begingroup$ Very clear, thank you very much. I will examine this closely, and respond with my results / potential further questions. Thank you again, I really needed your help. $\endgroup$ – bsmith144 Oct 30 '17 at 3:00
  • $\begingroup$ Updating you that I successfully understand this now. Thank you again. A side question I have is the following: Could you break up the initial equation up into two sections? For example. One case dealing with simply the dx, and the other case dealing with the (1/x^2) dx ? $\endgroup$ – bsmith144 Oct 30 '17 at 21:18

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