# Solve For Exact Differential Equation

1. Jan 30, 2012

### TrueStar

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

Solve the exact equation y^3-(14x+2)dx+3xy^2dy=0

2. Relevant equations

NA

3. The attempt at a solution

I proved these were exact because dM/dy and DN/dx both equal 3y^2

I chose to work with N first and df/dy=3xy^2

Therefore f(x,y)=xy^3+h(x)

I took df/dx of this and got y^3 + h'(x)

I made this equal to the other df/dx so it looks like:
df/dx = y^3+h'(x)=y^3-(14x+2)

h'(x)=-14x+2 and so h(x) is -7x^2+2x (+ constant)

I plugged this into the original problem with h(x) so it now looks like:

f(x,y)=xy^3-7x^2+2x=C

Solving for y, I get y=(-7x^2+2x+C)^(1/3) / x^(1/3)

This is not correct and I don't know what I'm missing here.

2. Jan 31, 2012

### HallsofIvy

Staff Emeritus
You have a basic algebra error! -(14x+ 2)= -14x- 2, not -14x+ 2.

3. Jan 31, 2012

### TrueStar

Ah yes. I did it again from the beginning and was able to solve it correctly. Thanks for pointing out the error. :)