How to Determine if an Analytic Solution Exists for a Differential Equation?

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Homework Statement



I can't figure out how to solve this problem. When i do it straight forward, I get a crazy complex equation. I think I need to play with it a little before I ∫ it, but I am not sure. Everything is real, and it is in my seperable diffy eg section

dy/dx=(4x-4x^3)/(4-y^3)

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The Attempt at a Solution

 
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So why not separate the variable like normal?
(4-y^3)dy = (4x-4x^3)dx
 
I did that and got
4y-(4y^4)/4=2x^2-(x^4)/4+c.
But I think that I need to make it y=F(x). I can't get the y by itself
 
cantdotthis said:
...

I can't get the y by itself
That's pretty common with solutions to differential equations.
 
I'm having a similar problem with simple DE's that are solvable via separation of vars or using an integration factor. Many problems come up with solutions containing y in the form of ye^y. My professor said that it's fine as the solution for her tests and so on, but in a real application, what would you do?
 
In a real application you'd use it as the basis for a numerical solution.
Note: ye^y is not the problem - it is that ye^y=f(x), and f(x) is the problem because it can be anything.

When you can get one as a function of the other you have an analytic solution.
There are a very large number of situations where you don't get one... in fact, for an arbitrary DE it is usually the case.

working out whether an analytic solution exists is a tricky part of number theory.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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