How do solve $\displaystyle \displaystyle\frac{dy}{dx}+y^2+2x=0$ ?

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- Jul 6th 2010, 05:26 PMmath2009Sample ODE question
How do solve $\displaystyle \displaystyle\frac{dy}{dx}+y^2+2x=0$ ?

- Jul 6th 2010, 06:52 PMmr fantastic
Why? Click on this: solve dy/dx + y^2 + 2x = 0 - Wolfram|Alpha

- Jul 6th 2010, 07:50 PMmath2009
Please check my solution, I doubt there are something wrongs in Attachment 18118

- Jul 6th 2010, 10:06 PMmr fantastic
Your method is wrong and your solution is wrong (did you even bother to check the solution by substituting it into the original DE?)

Although $\displaystyle y = \frac{1}{2x + a}$ satisfies $\displaystyle \frac{dy}{dx} + 2y^2 = 0$, it is also required to satisfy $\displaystyle \frac{dy}{dx} + 4 = 0$ and it doesn't. Similarly, the other part of your 'solution' has to satisfy both $\displaystyle \frac{dy}{dx} + 2y^2 = 0$ and $\displaystyle \frac{dy}{dx} + 4 = 0$ .... - Jul 6th 2010, 10:48 PMmath2009
I also thought so before.

If a,b are variable but not constants, how about ? - Jul 6th 2010, 11:11 PMmr fantastic
- Jul 6th 2010, 11:33 PMmath2009
My opinion is : $\displaystyle \begin{bmatrix}x \\y \end{bmatrix} \rightarrow \begin{bmatrix}x(a,b) \\ y(a,b) \end{bmatrix} $

If equation#1 's solution is $\displaystyle E_1 $ and equation#2 's is $\displaystyle E_2 $ , then $\displaystyle E_1\cap E_2 $ must fit both equations, is it right ?

And F(x, a, b) = 0 , it may simply solution - Jul 7th 2010, 04:56 AMmr fantastic
- Jul 8th 2010, 09:48 PMmath2009
mr fantastic , at first , thank you for discussion

please ignore this post.

There are two cases in this issue.

Case #1(constant) , as above mention, that's wrong.

Case #2(variables) , is there any proof ?