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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. – Pedro Tamaroff Jul 17 '12 at 14:40
up vote 5 down vote accepted


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! $\ddot\smile$ – amWhy Mar 11 '13 at 0:08

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


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


Note that

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




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



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

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

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



We integrate with respect to $x$.


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


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

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Looks good. Might want to remove one line towards the bottom that is the repeat of the one right above it - $y' = \sqrt{y^4+C}$. – Joe Jul 17 '12 at 15:56
@JoeL. Thanks, fixed. – Pedro Tamaroff Jul 17 '12 at 16:00

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