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The question is $y''=2y^3$. I know I can substitute $y'=p$. My question is if I can seperate x and y and integrate both sides twice?

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    Not really. What you should better do is multiply the equation by$y'$, integrate by the chain rule and integrate again.2012-07-17

3 Answers 3

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$\dfrac{d^2y}{dx^2}=\dfrac{d(\frac{dy}{dx})}{dx}$

So you have

$\frac{d(\frac{dy}{dx})}{dx}=2y^3\implies d(\frac{dy}{dx})=2y^3 dx$

Separation of variables won't work directly here. However if you multiply each side by $\frac{dy}{dx}$:

$\frac{dy}{dx}d(\frac{dy}{dx})=2y^3 dx\cdot(\frac{dy}{dx})=2y^3dy$

and now you can integrate all you like. Keep in mind that looking at separation of variables as multiplying and/or cancelling is a bit 'handwavy', though it does work.

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Hint: As Peter noted that you can use the chain rule here. Put $y'=p$. Then you have $y''=\frac{d\big(\frac{dy}{dx}\big)}{dx}=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p\frac{dp}{dy}$ Now, you equation becomes $p\frac{dp}{dy}=2y^3$ which is separable equation.

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    well done, Babak! $\ddot\smile$2013-03-11
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You seem to be wondering about some notation, namely $\frac{d^2y}{dx^2}$ versus $\frac{dy^2}{dx^2}$

The last expression doesn't make much sense, notationally, and I would expect it to be

$\left(\frac{dy}{dx}\right)^2$

The general notation for the $n$-th derivative $\frac{d^n y}{dx^n }$ is a suggestive notation because we can think about it as applying $\dfrac{d}{dx}$ $n$ times, so we might say

$y^{(n)}=\left(\frac{d}{dx}\right)^ny=\frac{d^n }{dx^n }y=\frac{d^n y}{dx^n }$

I'm just abusing the notation trying to make you understand why we put the $^n$ before the $y$ and not after the $y$. So your equation can be written as

$\frac{d^2y}{dx^2}=2y^3$

Note that

$\frac{dy^2}{dx^2}$ is rather ambiguous, and might mean

$\frac{d}{dx}(y^2)=2yy'$

or

$\left(\frac{dy}{dx}\right)^2=(y')^2$

In any case you can solve you ODE by multypling by $y'$ to get

$y''y'=2y^3y'$

$y'y''=2y^3y'$

$\frac 1 2[ (y')^2]'=\frac 1 2[ y^4]'$

$[ (y')^2]'=[ y^4]'$

$ (y')^2=y^4+C$

$y'=\sqrt{y^4+C}$

$\frac{y'}{\sqrt{y^4+C}}=1$

We integrate with respect to $x$.

$\int\frac{y'(x)}{\sqrt{y(x)^4+C}}dx=x+K$

Let $y(x)=u$ and $y'(x)dx=du$, so you get

$\int\frac{du}{\sqrt{u^4+C}}=x+K$

This last integral has no nice closed form, so you might want to aim for an implicit solution.

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    @JoeL. Thanks, fixed.2012-07-17