## anonymous 3 years ago 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!

1. anonymous

Dang that's a long one...

2. anonymous

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.

3. anonymous

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

4. anonymous

@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.

5. anonymous

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

6. anonymous

@Algebraic! Why do you think that?

7. anonymous

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

8. anonymous

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

9. anonymous

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.

10. anonymous

Should I put here an example of solving similar equation?

11. across

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.

12. anonymous

|dw:1351986536371:dw|

13. anonymous

|dw:1351986790555:dw|

14. anonymous

|dw:1351987002927:dw|

15. anonymous

|dw:1351987163211:dw|

16. anonymous

|dw:1351987461053:dw|

17. anonymous

@Algebraic! take a look.

18. anonymous

@mahmit2012!!! That is amazing!!! I'm going to figure it out! Thanx a lot!!!