# Did I solve this DE correctly?

1. Dec 10, 2006

### prace

If someone has a chance out there, could you please check my math here and let me know if I am doing this correctly or not.

Problem:

Solve: (2x-1)dx + (3y+7)dy = 0

I would like to solve this using the "Exact" method for solving DE's, so:

$$\frac{\partial_P}{\partial_y}(2x-1) = 0$$
$$\frac{\partial_Q}{\partial_x}(3y+7) = 0$$

$$\int{2x-1dx} = x^2-x + g(y)$$

$$\frac{\partial_F}{\partial_y}x^2-x+g(y) = g'(y)$$

$$g'(y) = 3y+7 -> g(y) = \frac{3}{2}y^2+7y$$

$$F(x,y) = x^2-x+\frac{3}{2}y^2+7y$$

How does that look?

Last edited: Dec 10, 2006
2. Dec 10, 2006

### prace

Also, one quick question. Does the equation have to be in the form of P(x.y)dx + Q(x,y)dy = 0? Can it be minus instead of plus? The reason I ask is because I vaugly remember hearing something about that at the begining of the school quarter and can't seem to find it in my notes now.

Thanks!

3. Dec 11, 2006

### dextercioby

Incomplete.

$$x^2-x+\frac{3}{2}y^2+7y=C$$

would be better.

Daniel.

4. Dec 11, 2006

### dextercioby

The minus can always be absorbed into Q, so it's always a "+". Kinda simple of you think about, right?

Daniel.

EDIT: Wow, i hit the "back" button and this reply got into another thread...