1. ## Sample ODE question

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

2. Originally Posted by math2009
How do solve $\displaystyle \displaystyle\frac{dy}{dx}+y^2+2x=0$ ?
Why? Click on this: solve dy&#47;dx &#43; y&#94;2 &#43; 2x &#61; 0 - Wolfram|Alpha

3. Please check my solution, I doubt there are something wrongs in

4. Originally Posted by math2009
Please check my solution, I doubt there are something wrongs in
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 $\displaystyle \frac{dy}{dx} + 4 = 0$ ....

5. I also thought so before.
If a,b are variable but not constants, how about ?

6. Originally Posted by math2009
I also thought so before.
If a,b are variable but not constants, how about ?
That wouldn't seem a very useful 'solution' to me.

7. 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

8. Originally Posted by math2009
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
You are wrong and I have told you why. I am not spending any more time discussing this.

9. mr fantastic , at first , thank you for discussion