it seems a pretty easy question,but really confuses me.
anyway,it is a differential equation:
(2xy-y^3)y+1=0

many thanks

2. ## Integrable Impliciation

$f(x,y)'=f'(x,y)y'$... Abuse of notation perhaps.

$\int (2xy+y^3)y' +1 dy = \int 0 dy$

$xy^2+\frac{y^4}4+y=C$

Now the hard part: isolate y.

3. i'm afraid that you got a wrong answer.. Still,i appreciate your help

4. Originally Posted by Media_Man
$f(x,y)'=f'(x,y)y'$... Abuse of notation perhaps.

$\int (2xy+y^3)y' +1 dy = \int 0 dy$

$xy^2+\frac{y^4}4+y=C$

Now the hard part: isolate y.
No, that's incorrect. x is not a constant and cannot be treated as one in the first integral.

farlist, that is NOT "a pretty easy question" but here is what I came up with: $(2xy+ y^3)\frac{dy}{dx}+ 1= y(2x+ y^2)\frac{dy}{dx}+ 1= 0$. Just because it looked like it might simplify things, I let $u= 2x+y^2$. Then $\frac{du}{dx}= 2+ 2y\frac{dy}{dx}$ so that $y\frac{dy}{dx}= \frac{1}{2}\frac{du}{dx}- 1$ and $y(2x+ y^2)\frac{dy}{dx}= uy\frac{dy}{dx}= \frac{1}{2}u\frac{du}{dx}- u$. Putting that into the equation: $(2x+ y^2)\left(y\frac{dy}{dx}\right)+ 1= 0$ becomes $\frac{1}{2}u\frac{du}{dx}- u+ 1= 0$. That's a separable equation and should be easy to solve.

5. We can make $dx+(2xy-y^3)dy=0$ exact via the integrating factor $e^{y^2}$. Check the book on the $\frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)$ form. This yields:

$e^{y^2}dy+e^{y^2}(2xy-y3)dy=0$

Then the solution is $F(x,y)=xe^{y^2}+T'(y)$

Solving for $T'(y)$ we get $T=-1/2 e^{y^2}(y^2-1)$

The solution is then:

$xe^{y^2}-1/2e^{y^2}(y^2-1)=c$

in which $y(x)$ can be solved for explicitly (how?).

And this is how I checked the solution in Mathematica:

Code:
n[7]:=
ysol = First[Derivative[1][y][x] /.
Solve[D[x*Exp[y[x]^2] -
(1/2)*Exp[y[x]^2]*(y[x]^2 - 1),
x] == 0, Derivative[1][y][x]]]
Simplify[(2*x*y[x] - y[x]^3)*ysol + 1]

Out[7]=
1/(y[x]*(-2*x + y[x]^2))

Out[8]=
0`

6. ok.maybe it is not that EASY.Actually what i figure out is exactly the same as Shawsend's anwser,but i didn't check it in Matlab.