# First order differential equation problem.

1. Nov 13, 2007

### engineer_dave

1. The problem statement, all variables and given/known data

Find the general solution of 2y(x^3+1)dy + 3x^2(1-y^2)dx = 0

2. Relevant equations

3. The attempt at a solution

So I first grouped the terms with dy or dx

2y/(1-y^2) dy = -3x^2/(x^3 +1) dx

after integrating both sides and solving, I got

ln (1-y^2)= -ln(x^3 +1) + c

and then after simplifying, it becomes 1-y^2= A/(x^3 + 1) and therefore y^2= -A/(x^3+1) + 1.

The answer according to the book was y^2= 1 + A(x^3 +1). How did they get that??

Maybe if i could get rid of the negative sign for ln, it might help....but please if u can help me, it would be appreciated. Thanks.

2. Nov 13, 2007

### Kreizhn

$$\int \frac{2y}{1-y^2} dy$$
$$u = 1-y^2$$ then $$du = -2y$$
$$\int \frac{2y}{1-y^2} dy = -ln(1-y^2)$$