# Solution to this Differential Equation $f''(x)=f(x)f'(x)$ needed

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?

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 '18 at 23:02
• 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. – Jameson Sep 25 '18 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 '18 at 23:54
• just evaluate the integral @clathratus – Isham Sep 23 '18 at 0:02
• I added some lines is it more clear now ? @clathratus – Isham Sep 23 '18 at 0:11
• true @WillJagy I solved it for K positive I corrected my answer thanks a lot – Isham Sep 23 '18 at 0:31
• @WillJagy Thanks a lot for your answer I have upvoted it – Isham Sep 23 '18 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 $$-1 < y < 1$$ $$y = - \tanh \frac{x}{2}$$

• @thanks a lot for your answer and the picture...that makes things more clear... – Isham Sep 23 '18 at 1:01