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

$$f''(x)=f(x)f'(x)$$

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?

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3 Answers 3

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$$y''=y'y \implies y''=\frac 12 (y^2)' $$ 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

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  • $\begingroup$ 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. $\endgroup$
    – clathratus
    Sep 22, 2018 at 23:54
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    $\begingroup$ just evaluate the integral @clathratus $\endgroup$ Sep 23, 2018 at 0:02
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    $\begingroup$ I added some lines is it more clear now ? @clathratus $\endgroup$ Sep 23, 2018 at 0:11
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    $\begingroup$ true @WillJagy I solved it for K positive I corrected my answer thanks a lot $\endgroup$ Sep 23, 2018 at 0:31
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    $\begingroup$ @WillJagy Thanks a lot for your answer I have upvoted it $\endgroup$ Sep 23, 2018 at 0:53
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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

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    $\begingroup$ @thanks a lot for your answer and the picture...that makes things more clear... $\endgroup$ Sep 23, 2018 at 1:01
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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.

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  • $\begingroup$ After I played around with it a little more I reached the same equation. Thanks! $\endgroup$
    – clathratus
    Sep 24, 2018 at 23:02
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    $\begingroup$ 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. $\endgroup$
    – Jameson
    Sep 25, 2018 at 9:50

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