# Did I solve this DE correctly?

• prace

#### 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?

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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 beginning of the school quarter and can't seem to find it in my notes now.

Thanks!

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?

Incomplete.

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

would be better.

Daniel.

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 beginning of the school quarter and can't seem to find it in my notes now.

Thanks!

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...