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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!
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!

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Dang that's a long one...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think you either wrote it incorrectly or you're not being asked to solve it...?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cool. no analytic solutions as far as I can tell, so ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! Why do you think that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is this not a quadratic equation in \[y \prime \]? that you should solve the quadratic equation first, then the resulting differential equations?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that y next to the 9 on the first term looks weird

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Should I put here an example of solving similar equation?

across
 3 years ago
Best ResponseYou've already chosen the best response.0Observe 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351986536371:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351986790555:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351987002927:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351987163211:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351987461053:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! take a look.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012!!! That is amazing!!! I'm going to figure it out! Thanx a lot!!!
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