1. y(dy/dx) = (y^3 - 1) (x/y) e^(ax^2) ; y(0) = -1

a > 0 is fixed constant

I solved to split up the x and y components and got:

(y^2)/(y^3 -1) dy = (xe^ax^2) dx

n(y) = m(x) (using equation above i made this one)

so then i made it

m(x) - n(x) = 0

then dm/dx - dn/dy = C where C=constant

soo...

3Ln(y^3 -1) - (e^(ax^2))(1- 1/4a^2) = C

.... solve for y....

y = cube rt (e^(1/3)((e^(ax^2))(x-1/4a^2)+C)

sorry for all the "( )" i know it makes it hard to read.

then when i plugged in y(0) = -1 i couldnt get it to work because i ended with...

-1 = e^-(1/12a^2) + C

and the e function cant be negative right? help!