# Solve differential equation: $f'''(x)=f(x)f'(x)f''(x)$

I came across $$f'''(x)=f(x)f'(x)f''(x)$$ but I don't know how to solve it.

I tried

$$\frac{f'''(x)}{f''(x)}=f(x)f'(x)$$

$$\ln|f''(x)|=\frac{1}{2}f(x)^2+c_{1}$$

But from there I have no idea how to proceed.

After

$$f'' = C_0e^{\frac 12 f^2}$$

we have

$$f'' f' = C_0e^{\frac 12 f^2}f'\Rightarrow \frac 12(f')^2=C_0\sqrt{\frac{\pi}{2}}\phi\left(\frac{f}{\sqrt 2}\right)+C_1$$

with

$$\phi\left(x\right)=\int_0^x e^{\zeta^2}d\zeta$$

and finally we arrive to the solution after integrating

$$\frac{df}{\sqrt{2\left(C_0\sqrt{\frac{\pi}{2}}\phi\left(\frac{f}{\sqrt 2}\right)+C_1\right)}} = dx$$

• Wow. That's pretty elegant. Thank you. – clathratus Sep 29 '18 at 22:21
• Just a clarity question: with respect to what variable does one integrate the last expression? Edit: It's x, right? – clathratus Sep 29 '18 at 22:28
• @clathratus The last expression was obtained due to the variable separation property. So the result should be $F(f) = x + C_2$ which gives an implicit solution for $f(x)$ – Cesareo Sep 29 '18 at 22:35
• Okay thank you. – clathratus Sep 30 '18 at 1:19