Question

Please, solve this differential equation
9*y*(y')^2 + 4*(x^3)*y' - 4*(x^2)*y = 0

PS I'm a foreigner, so I might you ask more questions during the problrm solving!

Thanx a lot!

Answer 1

Dang that's a long one...

Answer 2

I know. And I could not solve it. I suppose, that there must be a parameter. But which one?
Should I put y' = p or y' = px ???
I've already done both but I hadn'd any result.

Answer 3

I think you either wrote it incorrectly or you're not being asked to solve it...?

Answer 4

@Algebraic! I DID wrote this equation correctly (I checked it 3 times).
And I didn't understand, do you have doubts about my homework? You may trust me, I really have to solve this for my Math classes.

Answer 5

cool. no analytic solutions as far as I can tell, so ...

Answer 6

@Algebraic! Why do you think that?

Answer 7

is this not a quadratic equation in \[y \prime \]?
that you should solve the quadratic equation first, then the resulting differential equations?

Answer 8

that y next to the 9 on the first term looks weird

Answer 9

This is special differential Equation. What I know is that (y') should replace with p(x) or p(x)*x and then dy = pdx or dy=pdx+xdp 1) accordingly. And I solved this equation with both replacements and as a result a had really huge equations. I thought, I had done something wrong that is why I asked you for help.
But I have just understood that you had not even seen such equations.

Answer 10

Should I put here an example of solving similar equation?

Answer 11

Observe how\[y(x)=\frac13ix^2\]is a solution of this ODE:\[\begin{align}
9y(y')^2&=-\frac43ix^4,\tag{1}\\
4x^3y'&=\frac83ix^4,\tag{2}\\
4x^2y&=\frac43ix^4,\tag{3}
\end{align}\]\((1)+(2)-(3)=0\).

This is also true for its conjugate.