I came up with this differential equation and I don't know how to solve it.


I attempted to solve it several times, but they were all fruitless. Wolfram Alpha says that the solution is

$$f(x)=\sqrt{2a} \tan\left({\frac{\sqrt{2a}}{2} \cdot (x+b)}\right),$$ where $a$ and $b$ are constants.

How does one get this solution?

up vote 1 down vote accepted

Let us consider your differential equation: $$f''(x)=f(x)\cdot f'(x)$$ Integrate with respect to $x$ on both sides. Recognize that $df'(x)=f''(x)\ dx$ and $df(x)=f'(x)\ dx$: $$\int f''(x)\ dx=\int f(x)\cdot f'(x)\ dx\rightarrow \int df'(x)=\int f(x)\ df(x).$$ It follows that $$f'(x)=\frac{(f(x))^2}{2}+a=\frac{(f(x))^2+2\cdot a}{2}.$$ Divide by $(f(x))^2+2\cdot a$ on both sides: $$\frac{f'(x)}{(f(x))^2+2\cdot a}=\frac{1}{2}.$$ Integrate with respect to $x$ on both sides. Recognize that $df(x)=f'(x)\ dx$: $$\int \frac{f'(x)\ dx}{(f(x))^2+2\cdot a}=\int \frac{dx}{2}\rightarrow \int \frac{df(x)}{(f(x))^2+2\cdot a}=\frac{x}{2}+b=\frac{x+2\cdot b}{2}.$$ Redefine $2\cdot b$ as $b$, since it is a constant: $$\int \frac{df(x)}{(f(x))^2+2\cdot a}=\frac{x+b}{2}.$$ Let $f(x)=\sqrt{2\cdot a}\cdot \tan(s)$ such that $df(x)=\sqrt{2\cdot a}\cdot (\tan^2(s)+1)\ ds$: $$\int \frac{ds}{\sqrt{2\cdot a}}=\frac{x+b}{2}\rightarrow \frac{s}{\sqrt{2\cdot a}}=\frac{x+b}{2}.$$ Isolate $s$ and let $s=\arctan \left (\frac{f(x)}{\sqrt{2\cdot a}}\right)$: $$s=\frac{\sqrt{2\cdot a}}{2}\cdot (x+b)\rightarrow \arctan \left (\frac{f(x)}{\sqrt{2\cdot a}}\right)=\frac{\sqrt{2\cdot a}}{2}\cdot (x+b).$$ Therefore, an expression for your function $f(x)$ can be written as $$f(x)=\sqrt{2\cdot a}\cdot \tan \left (\frac{\sqrt{2\cdot a}}{2}\cdot (x+b)\right).$$ The derived expression is equivalent to what you found with Wolfram Alpha.

  • After I played around with it a little more I reached the same equation. Thanks! – clathratus Sep 24 at 23:02
  • 1
    You're welcome. Whenever you're given a differential equation, try to create a situation in which you are able to obtain expressions on both sides after integrating. With this differential equation that was already the case. – Stijn Dietz Sep 25 at 9:50

$$y''=y'y \implies y''=\frac 12 (y^2)' $$ After integration $$\implies y'=\frac 12 y^2+K$$ $$y'=\frac 12(y^2+2K)$$ For $ K=0 \implies -\frac 1y=\frac x2 +a$ $$\implies y(x)=-\frac 2 {x+c}$$

For K negative $$ \int \frac {dy}{y^2-c^2}=\frac x 2+b$$ $$\ln (\frac {y-c}{y+c})=cx+a$$ $$y(x)=-c \frac {ae^{cx}+1}{ae^{cx}-1}$$

For K positive, substitute $2K=c^2$

$$ \int \frac {dy}{y^2+c^2}=\frac 12\int dx=\frac x 2+b$$ Substitute $z=y/c \implies dz=dy/c$ $$ \int \frac {dz}{z^2+1}=c(\frac x 2+b)$$ $$ \arctan (z)=c(\frac x 2+b)$$ $$ y=c\tan (c(\frac x 2+b))$$ Substitute $c/2=k$ and $bc=a$ $$ y=2k\tan (kx+a)$$

$$y''=y'y$$ Substitute $$y'= p \implies y''=pp'$$ $$p'=y \implies p=\frac 12 y^2 +K$$ $$y'=\frac 12 y^2+K$$ This equation is separable

  • I got to that point by myself, but how do I get to the solution? I do not know the significance of a separate differential equation. – clathratus Sep 22 at 23:54
  • 1
    just evaluate the integral @clathratus – Isham Sep 23 at 0:02
  • 1
    I added some lines is it more clear now ? @clathratus – Isham Sep 23 at 0:11
  • 1
    true @WillJagy I solved it for K positive I corrected my answer thanks a lot – Isham Sep 23 at 0:31
  • 1
    @WillJagy Thanks a lot for your answer I have upvoted it – Isham Sep 23 at 0:53

as an example, in Isham's answer, if we choose $K$ so that $$ y' = \frac{1}{2} (y^2-1), $$ the first thing we notice is two stationary solutions, $y = 1$ and $y = -1$

This is autonomous...

For $y > 1 $ or $y < -1$ we get $$ y = \frac{-1}{\tanh \frac{x}{2}} $$ which blow up in finite time, one in each direction. Also translates.

For $-1 < y < 1$ $$ y = - \tanh \frac{x}{2} $$ enter image description here

  • 1
    @thanks a lot for your answer and the picture...that makes things more clear... – Isham Sep 23 at 1:01

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.