# Derivative of square of derivative?

I was trying to solve this differential equation:

$$2yy'' + 3y'^2 = 4y^2$$

And I found this way to solver it: http://eqworld.ipmnet.ru/en/solutions/ode/ode0344.pdf but I don't understand why $w'_y = y''_{xx}$. If $w(y) = (y'_x)^2$, how can I find this:

$$\dfrac{d}{dy}\bigg(\dfrac{dy}{dx}\bigg)^2$$

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Use the Product Rule (or the Chain Rule). –  André Nicolas Oct 10 '12 at 20:33
I have tried it, but when I try to integrate it can't seem to get the original expression. –  hinafu Oct 10 '12 at 20:35
Do you mean $\dfrac{d}{dy}\bigg(\dfrac{dy}{dx}\bigg)^2$ or $\dfrac{d}{dx}\bigg(\dfrac{dy}{dx}\bigg)^2$? –  draks ... Oct 10 '12 at 20:41
The first one: $$\frac{d}{dy}\bigg(\frac{dy}{dx}\bigg)^2$$ –  hinafu Oct 10 '12 at 20:43

By the Chain Rule, $$\frac{d}{dy}\left(\frac{dy}{dx}\right)^2=\frac{dx}{dy}\frac{d}{dx}\left(\frac{dy}{dx}\right)^2.$$

Now use the fact that $\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}.$

Calculate $\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)^2$ using the Product Rule. When we put things together, there is some nice cancellation, which undoubtedly means there is a simple conceptual reason.

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You don't mean $\frac{dz}{dz}$, I guess? Further: are you referring to $\dfrac{d}{dy}\bigg(\dfrac{dy}{dx}\bigg)^2$? –  draks ... Oct 10 '12 at 20:40
This is okay, but I wanted to get is $\frac{d}{dy}z$ not $\frac{d}{dx}z$ –  hinafu Oct 10 '12 at 20:42
@hinafu: I saw what I expected to see, and not what you wrote! Fixed. –  André Nicolas Oct 10 '12 at 20:58
Good job, after only four edits. (I'm only being slightly sarcastic; that's better, I suspect, than my average.) +1. –  Rick Decker Oct 10 '12 at 21:05
@RickDecker: It is only $4$ edits because the post is so short. My average is higher, though usually the edits are obsessive fiddling with wording rather than complete change. –  André Nicolas Oct 10 '12 at 21:14

Here's something that can't possibly be right:

$$\frac{dw}{dy} = \frac{dw}{dx}\frac{dx}{dy} = \frac{2y'y''}{\frac{dy}{dx}} = 2y''$$

The funny thing is this will achieve the result given by the reference if we add $y''+f(y)(y')^2 + g(y) = 0$ to itself and make the substitution.

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Somewhat "surprising," yes, but right. –  André Nicolas Oct 10 '12 at 21:00

Let's set $\ w(y):=(y_x')^2\$ then : $$\frac {dw(y)}{dx}=\frac {dw(y)}{dy}\frac {dy}{dx}=\frac {d\left(\left(\frac {dy}{dx}\right)^2\right)}{dx}=2\frac {dy}{dx}\frac {d^2y}{dx^2}$$ From the second and fourth term we get (if $\frac {dy}{dx}\not = 0$) : $$\frac {dw(y)}{dy}=2\frac {d^2y}{dx^2}$$

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I'll give this a shot. Let $z=\frac{dy}{dx}$

$$\frac{dw}{dx}=\frac{dz^2}{dy}=2z\frac{dz}{dy}=2z\times\frac{\frac{dz}{dx}}{\frac{dy}{dx}}=2z\times\frac{(\frac{d^2y}{dx^2})}z=2\frac{d^2y}{dx^2}$$

This appears to reduce your equation to

$$yw'+3w=4y^2$$

$$w'+\frac{3w}y=4y$$

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Hmm... This is ugly as sin. Let's see if I can clean it up a bit. –  Mike Oct 10 '12 at 21:06
Thanks for the answer! –  hinafu Oct 10 '12 at 21:15
All right, this should be clear and a lot less ugly. I'm starting to hate the chain rule as it applies to second derivatives. :) –  Mike Oct 10 '12 at 21:18