I am trying to solve the following:

$$y''(x)=\frac{4}{3} y(x)^3 y'(x)$$

given that $y(0)=1$ and $y'(0)=1/3$. This is a link to Wolfram Alpha.

My idea was that because when $y=1$, $y^3 = 1$ it can be solved for the private case only by putting $1$ instead of $y^3$

the question is to find y I need the way as I have no clue what to do.

  • $\begingroup$ Although I edited the $\LaTeX$, I think you still need to clarify your question. $\endgroup$ – picakhu Oct 3 '11 at 17:04

Hint: By the chain rule, the right-hand side equals $\frac{d}{dx}\frac{y^4}{3}$. Now integrate both sides...

  • $\begingroup$ @makholm , you are the bomb! $\endgroup$ – Nahum Oct 4 '11 at 15:51
  • 4
    $\begingroup$ Gee, thanks, @Nahum. Now I'll never fly again. $\endgroup$ – Henning Makholm Oct 4 '11 at 18:39

let's make substitution $v=y'$

$y''=\frac{dv}{dx}=\frac{dv}{dy}\frac{dy}{dx}=v\frac{dv}{dy}$ , so we may write:

$v\frac{dv}{dy}=f(y,v)$ , which is in your specific case equal to:

$v\frac{dv}{dy}=\frac{4}{3}y^3v$ , which is separable first order differential equation:


After you find $v$ you have to solve $y'=\frac{dy}{dx}=v$ ,which is also separable first order equation.

  • $\begingroup$ +1 But it IS a mess. $\endgroup$ – AD. Oct 3 '11 at 18:50

For this exact case, the route suggested by makholm is easiest. I just want to mention the general solution for a more general equation, $ g''=f(g)g'$, where $f$ is some function of $g(x)$, is: $$ \int \frac{dg}{F(g)+C_1}=C_2+x$$ where $F(g)=\int f(g)dg$ and $C_1$ and $C_2$ are integration constants. In this case $f(g)=\frac{4}{3} g^3$. Thus $F(g)=\frac{1}{3} g^4$...


Paul Safier


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