The difference between $\frac{\partial^2 y}{\partial x^2}$ and $\frac{\partial y^2}{\partial x^2}$

Question

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?

Answer 1

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

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.

Answer 2

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.

Answer 3

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.