# Differential Equations: exact equation

1. Sep 12, 2007

### jimmypoopins

i fell asleep when the professor went over how to solve exact equations :-/ i know it's really easy but despite reading the chapter over and over i still can't get it right. please show me where i'm going wrong / what to do next.

1. The problem statement, all variables and given/known data
Determine whether the equation in problem 1 is exact. If it is exact, find the solution.
$$(2x + 3) + (2y - 2)y' = 0$$

2. Relevant equations

3. The attempt at a solution

$$(2x + 3)dx + (2y - 2)dy = 0$$

$$M_{y} = 0 = N_{x} = 0$$ <--- the equation is exact

$$\psi_{x} = 0$$ --> $$\psi = \int^x 0dx = x + h(y)$$

$$\frac{d\psi}{dy} = h'(y) = 2y - 2$$ ---> $$h(y)= y^2 - 2y$$

and then i get stuck. i'm not sure where to go from there. the answer to the problem is $$x^2 + 3x + y^2 - 2y = c$$, which is apparent to me if you turn the original equation into a separable one, but that's not possible with all exact equations.

2. Sep 12, 2007

### Dick

An exact equation is one of the form dF(x,y)=0. So it has solutions F(x,y)=c. The dF equation
becomes F_x*dx+F_y*dy=0, where the underscores are partial derivatives. So F_x is your M and F_y is your N. What you want to do is find F. Since F_x=2x+3, F=x^2+3x+h(y). And sure F_y is then h'(y) which you know is 2y-2. Solve for h(y) (as you did). Now what is F? Hence what are the solutions?

3. Jul 16, 2009

### deezy911

You already solved it. Just plug your c(y) into F. Your F is x^2 + 3x + c(y) after you integrated the M term. Plug c(y) back into F and you get x^2+3x+y^2-2y.

4. Jul 16, 2009

### Dick

It's not really worth while recycling threads from 2007. Now is it?