# Homework Help: Exact DE's

1. Feb 5, 2006

### Pengwuino

How do you determine a solution to a DE if you know the DE is exact? My professor (or possibly my notes) did a horrible job of explaining how to do it and the book does an even worse job explaining how to do it. For example, how would i go about solving this equation? I need to find c, the constant.

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

I know that I differentiate each part by dx/dy and dy/dx to attain:

$$- 2xy + 2\frac{{dx}}{{dy}} = - 2xy + 2\frac{{dy}}{{dx}}$$

But where do I go from that point?

2. Feb 5, 2006

### mathmike

first you took the derivative wrong it should be -4xy+2

after that you would intergate and combine terms

can you do this

3. Feb 5, 2006

### Pengwuino

haha oops oh yah... missed that twice it seems.

Which part do i integrate?

4. Feb 6, 2006

### mathmike

you would integrate the original functions M and N

M = -2xy^2 + 2y dx

N = -2x^2y +2x dy

5. Feb 6, 2006

### Pengwuino

Ok they both integrate to be:

-x^2y^2+2xy

now what?

6. Feb 6, 2006

### Cyrus

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

$$(-x^2y^2 + 2yx + c_1) + ( - x^2y^2 + 2xy + c_2 ) = 0$$

$$-2x^2y^2 + 4yx + k = 0$$

Edit: OOPS should be this:

$$k = 2x^2y^2 -4xy$$

$$k= 2xy(xy - 2)$$

I have no clue, Im just playing along here. Don't take what I post seriously. Do you have an idea of what the solution should look like? I dont see how you can get a specific value of c without any boundary conditions.

Last edited: Feb 6, 2006
7. Feb 6, 2006

### HallsofIvy

$$( - 2xy^2 + 2y)dx + ( - 2x^2 y + 2x)dy = 0$$
That is an exact equation because (-2xy2+ 2y)y= -4xy+ 2 and (-2x2y+ 2x)x= -2xy+ 2 also.

What that means is that there exist a function F(x,y) such that
$$dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= (-2xy^2+ 2y)dx+ (-2x^2y+ 2x)dx$$
(Above, we were checking the equality of the mixed second derivatives,
$$\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial^2 F}{\partial y\partial x}$$)

That is, we must have
$$\frac{\partial F}{\partial x}= -2xy^2+ 2x$$
Integrate that, with respect to x since in a partial derivative you are treating y as a constant. Of course, the "constant of integration" might, in fact, be a function of y. Once you get that expression for F, differentiate with respect to y and set it equal to
$$\frac{\partial F}{\partial y}= -2x^2y+ 2y$$ to determine what that "constant" must be.
In terms of that F, the equation just says "dF= 0" so F= constant is the general solution.

8. Feb 7, 2006

### Pengwuino

So I set $$- x^2 y^2 + x^2 + c = - 2x^2 y + 2y$$ ??