How Do You Solve This Differential Equation Correctly?

[stosw]

Homework Statement

$$dy/dx$$ = $$y^3/(1-2xy^2)$$

The Attempt at a Solution

I converted it to this form:

(-y^3)dx + (1-2xy^2)dy = 0

M_y = 3y^2
N_x = -2y^2

so

M_y =/= N_x

Finding the integrating factor:

(Nx - My) / M

$$\frac{-2y^2 + 3y^2 }{-y^3}$$ = $$\frac{-1}{y}$$

e^{\int(-1/y)dy\} = e^(-ln(y)) = -y

(After many attempts, I can't seem to get latex to do the integral sign...)

multiplying through by the integrating factor and recalculating the partial dirivatives i still get...

M_y = 4y^3
N_x = 2y^3


Can someone explain where i went wrong?

 

[JThompson]

e^(-ln(y)) = -y

Just an arithmetic error. This should be

$$e^{-ln(y)}=e^{ln(y^{-1})}=y^{-1}$$