0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Although I edited the $\LaTeX$, I think you still need to clarify your question. $\endgroup$ – picakhu Oct 3 '11 at 17:04
8
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @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
2
$\begingroup$

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:

$dv=\frac{4}{3}y^3dy$

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

$\endgroup$
  • $\begingroup$ +1 But it IS a mess. $\endgroup$ – AD. Oct 3 '11 at 18:50
0
$\begingroup$

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 $F(g)=\frac{1}{3} g^4$...

Cheers,

Paul Safier

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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