How Do You Solve an Exact Differential Equation?

[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?

 

[mathmike]

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

after that you would intergate and combine terms

can you do this

 

[Pengwuino]

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

Which part do i integrate?

 

[mathmike]

you would integrate the original functions M and N

M = -2xy^2 + 2y dx

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

 

[Pengwuino]

Ok they both integrate to be:

-x^2y^2+2xy

now what?

 

[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, I am just playing along here. Don't take what I post seriously. Do you have an idea of what the solution should look like? I don't see how you can get a specific value of c without any boundary conditions.

 

[HallsofIvy]

Your original equation was
$$( - 2xy^2 + 2y)dx + ( - 2x^2 y + 2x)dy = 0$$
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.

 

[Pengwuino]

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