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I have the following differential equation:

$$\ddot y(t) = \frac{10 \cdot a \cdot y(t)}{1 + a \cdot y(t)}$$

With $a \in \Bbb{R^+}$ and $y(t)$ the function I want to solve.

I also know that the function is periodic and symetric ($y(t) = y(-t)$) and that the domain is formed with all the real numbers.

I've tried researching this equation, but I couldn't find anything that was usefull for me (separation of variables, substitution of $y(t) = e^{rt}$, transforming it into a first-order ODE, integrating by parts...).

Help will be appreciated!

Arnau del Río
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3 Answers3

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Hint $$y'' = \frac{10\, a \, y}{1 + a \, y}$$ Switch variables $$-\frac{t''}{[t']^3}= \frac{10\, a \, y}{1 + a \, y}$$ Now, reduction of order $p=t'$ gives a separable equation for $t'(y)$. Integrate again.

Claude Leibovici
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  • Thank you for your answer... I haven't done much of this yet, but I understand your hint and I'm going to try and end it. What I don't get is how $$y\prime\prime = - \frac{t\prime\prime}{[t\prime]^3}$$ Could you explain this equality? And wouldn't this give me $t(y)$ instead of $y(t)$? Thanks in advance, I appreciate your time! – Arnau del Río Oct 31 '22 at 14:41
  • @ArnaudelRío. This is the same as $t'\frac 1 {y'}$. Have a look here https://math.stackexchange.com/questions/566507/prove-that-d2x-dy2-equals-d2y-dx2dy-dx-3 – Claude Leibovici Oct 31 '22 at 14:50
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$$y''(t) = \dfrac{10 a y(t)}{1 + a y(t)}$$ $$ \text { Substitute } p=\dfrac {dy}{dx} \text { and } p'=\dfrac {dp}{dy}$$ $$p\dfrac {dp}{dy} = \dfrac{10 a y(t)}{1 + a y(t)}$$ The DE is a first order separable DE. Maybe the next integration won't be that easy..

user577215664
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  • I used Wolfram|Alpha and I got the following: $\displaystyle \frac{p^2}{2} = \frac{10ay(t) - 10 \cdot \ln{(ay+1)}}{a} + C$. What should I do from here? (By the way, I derivated it back again, and I found that the constant of integration depends on $y$, which is really strange) – Arnau del Río Oct 31 '22 at 16:26
  • Well, I've tried what you've said. I haven't been able to solve it, but I got to this:

    $$t = \int \sqrt{\frac{a}{20ay(t) - 20\ln(ay(t)+1)}} dy$$

    I won't even try to think how to intgrate this, but I can also see that the end result won't be useful to me, as I need to get $y$ in respect with $t$, not the other way around, and I don't think I'll be able to isolate $y(t)$ either. Maybe it's impossible to figure our an exprerssion of $y$ using $t$ and $a$ only.

     – Arnau del Río Oct 31 '22 at 17:21
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For those of you wondering, I ended up solving the differential equation using Euler's method, which states that:

$$f(x)_{n + 1} = f\prime(x)_n \cdot \Delta x + f(x)_n$$

Thank you everyone and hope this helps!

Arnau del Río
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