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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!

Differential Equations
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Dang that's a long one...
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.
I think you either wrote it incorrectly or you're not being asked to solve it...?

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@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.
cool. no analytic solutions as far as I can tell, so ...
@Algebraic! Why do you think that?
is this not a quadratic equation in \[y \prime \]? that you should solve the quadratic equation first, then the resulting differential equations?
that y next to the 9 on the first term looks weird
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.
Should I put here an example of solving similar equation?
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.
@Algebraic! take a look.
@mahmit2012!!! That is amazing!!! I'm going to figure it out! Thanx a lot!!!

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