# Solution to the exact differential equation

1. Sep 22, 2004

### fresh

(x^2)(y^3) + x(1 + y^2)y' = 0

the integrating factor to make the above equation exact is (1)/(xy^3)

i have worked this equation out and have c = .5x^2 as the solution; however, the textbook says the solution is c = x^2 - y^(-2) + 2lnlyl

apparently they got this solution because h'(y) = y^(-3) + y(-1)
i found h'(y) to be equal to zero.

some sort of feedback would be greatly appreciated.

2. Sep 23, 2004

### Tide

I get a positive sign with a factor of 2 for the middle term compared with the textbook solution. Are you sure you typed it correctly?

I just did a direct integration of the ODE since it's separable.

3. Sep 23, 2004

### HallsofIvy

Staff Emeritus
Do you understand that "c= .5x^2" says that x is a constant? What happened to y in your solution?

By the way, this equation is separable. It's easier to do it that way.

Last edited: Sep 23, 2004
4. Sep 23, 2004

### amb123

This is a seperable equation from the get go. I got the book's answer.

It seperates to [(1 + y^2) / y^3 ] dy = -x dx . Once you integrate both sides you get :

ln(y) - 1/2y^2 = -(x^2 / 2) + c. Solve for c gives you :

c = 2ln(y) - y^-2 + x^2 Don't worry about the fact that c would really be 2c, because it is just a constant.

:)
-A