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Note. I know that this non-linear differential equation is generalized in other posts. I wanted to check my proof, different from others.

$f’(x)=c^2-\{f(x)\}^2, \ c>0, \ f: \Bbb{R}\to\Bbb{R}, \ f(x) \in [-c, c]$. Solve $f(x)$.

The normal generalization in domain of reals goes following:

\begin{align} & \dfrac{f’(x)}{c^2-\{f(x)\}^2}=1=\dfrac{1}{2c}\left(\dfrac{f’(x)}{f(x)+c}-\dfrac{f’(x)}{f(x)-c}\right) \\ \ \\ \rightarrow \; & \ln\left| \dfrac{f(x)+c}{f(x)-c} \right| = 2cx+d, \dfrac{c+f(x)}{c-f(x)}=e^{2cx+d}=ke^{2cx}. \\ \ \\ \therefore \; & f(x)=-\dfrac{2c}{ke^{2cx}+1}+c = c\left(\dfrac{ke^{2cx}-1}{ke^{2cx}+1}\right). \end{align}

And my idea was, expanding the domain and range to complex numbers.

We already have the generalization of $k’(x)=c^2+\{k(x)\}^2$ as $k(x)=c\tan(cx+\alpha)$.

Setting $f(x)=-ig(ix)$, we have $g’(ix)=c^2+\{g(ix)\}^2$, so we have $g(ix)=c\tan(cix+\alpha)$.

Therefore, we can write $f(x)$ as $-ci\tan(cix+\alpha)$, then apply $\tan(ix)=i\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$:

$f(x)=-ci \cdot i\dfrac{e^{cx+\beta}-e^{-cx-\beta}}{e^{cx+\beta}+e^{-cx-\beta}} = c\dfrac{ke^{2cx}-1}{ke^{2cx}+1}.$ ($\alpha=i\beta, e^{2\beta}=k.$)

And now I’m claiming that, the function we were finding is just a part of the function we solved. Is my claim valid?

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  • $\begingroup$ Is it not evident from expansion of the direct solution $ y = c~ \tanh ( c x) + c_{1} ?$ $\endgroup$ Commented Aug 14 at 13:11
  • $\begingroup$ @Narasimham Yes, it is, but I realized that later. (I’m quite newbie for ODE…) Before knowing the solution, I solved as above, and I’m checking if my solution is still valid. $\endgroup$ Commented Aug 14 at 13:21
  • $\begingroup$ What do the braces $\{\cdot\}$ mean? $\endgroup$ Commented Aug 14 at 13:22
  • $\begingroup$ @ThomasAndrews Just parentheses. $\endgroup$ Commented Aug 14 at 13:49
  • $\begingroup$ Be careful with that - mathematicians rarely use braces as parens, and most commonly when dealing with large complex formulas. Here, you could actually just write $f^2(x).$ @rdk $\endgroup$ Commented Aug 14 at 14:09

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This is a Riccati DE. Moreover, if $F$ is an anti-derivative of $f$, then $$ (e^F)''=e^F(f'+f^2)=c^2e^F, $$ so that $e^F=Ae^{cx}+Be^{-cx}$ and $$ f=e^{-F}(e^F)'=c\frac{Ae^{cx}-Be^{-cx}}{Ae^{cx}+Be^{-cx}}. $$ This trick should help illuminate the solution structure of this DE and closely related ones.

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  • $\begingroup$ Thanks for simpler proof(+1), but I’m wondering if my solution is also valid. :) $\endgroup$ Commented Aug 14 at 12:25
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    $\begingroup$ Yes, everything you wrote is correct. $\endgroup$ Commented Aug 14 at 17:03
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If you start switching variables, you have $$\frac 1 {x'(f)}=c^2-f^2 \quad \implies \quad x(f)=\frac{\tanh ^{-1}\left(\frac{f}{c}\right)}{c}+C_1$$ and inversing $$f(x)=c \tanh (c x+C_1)$$

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  • $\begingroup$ Sorry but I’m not that familiar with the switching variable notation… Does $x(f)$ mean the inverse function of $f$, and $x’(f)$ is the derivative of inverse $f$? $\endgroup$ Commented Aug 15 at 4:26
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    $\begingroup$ @RDK. Yes; this is correct. It is quite often a good trick $\endgroup$ Commented Aug 15 at 5:37

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