Solving separable 2nd order DE

[phantomvommand]

Homework Statement

please see attached pictures.

Relevant Equations

Integration, Substitution

This is a physics problem from Griffith's Electrodynamics. I'm mainly asking about the math here. I found the DE in the box at part (d).
To solve it, I did:
$\sqrt V {d^2 V} = \beta dx^2$
Integrating twice:
$\frac {4} {15} V^{2.5} = \beta x^2/2$

Why is my method wrong?

Thanks for the help.

 

[uart]

Write your second derivative as $\frac{d}{dx} \left( \frac{dV}{dx} \right)$ and it should be clear that when you try to separate it using your method, what you actually get is:

$$\sqrt{V} \, d (\frac{dV}{dx}) = \beta \, dx$$

 

[Delta2]

You just cannot integrate this equation is $$\sqrt{V}d^2V=\beta dx^2$$ because , me at least , doesn't know how an integral of the form $$\int f(V)d^2V$$ is defined or an integral of the form $$\int f(x)dx^2$$ is defined.
That is because you have $d^2V$ there instead of $dV$ and $dx^2$ instead of $dx$.
$$\int f(V)dV$$ and $$\int f(x) dx$$ are well defined, but not the integrals at the starting of my post, those are just not well defined.

What you do is that you assume not only that the integrals are well defined but also that the following equality holds $$\int f(V)d^2V=\ (\int f(V) dV \ )dV$$ (and something similar for the $$\int f(x)dx^2=\ (\int f(x) dx \ )dx$$.

I challenge you to prove the above equalities, there is no way to prove them, simply because the left hand side is not well defined

 

[uart]

BTW, the method the book uses is the standard way to solve DEs of the form $\frac{d^2 y}{dx^2} = f(y)$

Define the first derivative as a new variable, eg $v = \frac{dy}{dx}$.
Now in terms of $v$ the original DE can be written as:
$$\frac{dv}{dx} = f(y)$$
Applying the chain rule you get:
$$\frac{dv}{dy} \, \frac{dy}{dx} = f(y)$$
$$v \frac{dv}{dy} = f(y)$$
Which separates nicely to give:
$$v \, dv = f(y) \, dy$$