it seems a pretty easy question,but really confuses me.
anyway,it is a differential equation:
(2xy-y^3)y`+1=0
many thanks
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: $\displaystyle (2xy+ y^3)\frac{dy}{dx}+ 1= y(2x+ y^2)\frac{dy}{dx}+ 1= 0$. Just because it looked like it might simplify things, I let $\displaystyle u= 2x+y^2$. Then $\displaystyle \frac{du}{dx}= 2+ 2y\frac{dy}{dx}$ so that $\displaystyle y\frac{dy}{dx}= \frac{1}{2}\frac{du}{dx}- 1$ and $\displaystyle y(2x+ y^2)\frac{dy}{dx}= uy\frac{dy}{dx}= \frac{1}{2}u\frac{du}{dx}- u$. Putting that into the equation: $\displaystyle (2x+ y^2)\left(y\frac{dy}{dx}\right)+ 1= 0$ becomes $\displaystyle \frac{1}{2}u\frac{du}{dx}- u+ 1= 0$. That's a separable equation and should be easy to solve.
We can make $\displaystyle dx+(2xy-y^3)dy=0$ exact via the integrating factor $\displaystyle e^{y^2}$. Check the book on the $\displaystyle \frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)$ form. This yields:
$\displaystyle e^{y^2}dy+e^{y^2}(2xy-y3)dy=0$
Then the solution is $\displaystyle F(x,y)=xe^{y^2}+T'(y)$
Solving for $\displaystyle T'(y)$ we get $\displaystyle T=-1/2 e^{y^2}(y^2-1)$
The solution is then:
$\displaystyle xe^{y^2}-1/2e^{y^2}(y^2-1)=c$
in which $\displaystyle y(x)$ 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