Question

Help Determine if the Equation is Exact and integrate if so.
(2xy+3)dx+(x^2-1)dy=0

Answer 1

How do I know if the equation is exact?

Answer 2

You know the equation is exact if the partial derivative with respect to y of the first parentheses is the same as the partial derivative with respet to x of the second parentheses.

Answer 3

3xy + 3 You already have dx, now do dy

x^2 - 1 You already have dy, now do dx

Answer 4

what do you mean by that?

Answer 5

And the equation is exact because d/dy of (2xy + 3) is 2x and d/dx of (x^2 - 1) is 2x. Since they're both the same, the equation is exact. Do you see what I mean?

Answer 6

that makes sense thank you!

Answer 7

could you please help with integrating as well?

Answer 8

I'm trying to work it out, I haven't done differential equations in a while.

Answer 9

i've never done differential equations, this is my first semester taking it, take your time

Answer 10

The idea of this is that a function w(x,y), where y is also a function of x, has a derivative dw/dx + (dw/dy) * (dy/dx). If you divide your equation up there by dx, you get (2xy + 3) + (x^2-1)dy/dx = 0. If you compare these two, you see that dw/dx = 2xy + 3 and dw/dy = x^2-1, which works if it's an exact equation. So what this is saying is that there's a function of x and y (that we call w) whose derivative is the LHS of the equation you gave.

Answer 11

dw/dx = 2xy + 3
dw = (2xy + 3)dx
w = yx^2 + 3x + f(y)
Where f(y) is kind of like + C, so we have to find f(y) now.

Answer 12

(2xy+3)dx+(x^2-1)dy=0

IF it is exact, we have some wonderful function F(x,y), such that:

\(F_{x} = 2xy + 3\;and\;F_{y} = x^{2}-1\)

We have not yet proven that it is exact, so that is the first thing to do.
Demonstrate that \(F_{xy} = F_{yx}\)

Here we go.

Given \(F_{x} = 2xy + 3\;we\;have\;F_{xy} = 2x\)

Given \(F_{y} = x^{2} - 1\;we\;have\;F_{yx} = 2x\)

Since \(F_{xy} = F_{yx}\), we call it "Exact".

Answer 13

also if determined exact, does that eliminate them from being separable, linear,
or none of these?

Answer 14

Take partial derivative with respect to y.

dw/dy = x^2 + f'(y)

We already know an expression for dw/dy from before so set this equal to that.

x^2 + f'(y) = x^2 - 1
f'(y) = -1
f(y) = -y

So w is HOPEFULLY:

w = yx^2 + 3x - y

Answer 15

Why do we care? It's exact. The solution is relatively simple. It's possible some could be put into other forms, but again, why?

Answer 16

thank you bagajr i also got x^2y+3x-y=C

Answer 17

Just for fun, start with the other one and see if you get the same result!

Answer 18

i did on paper and it is

Answer 19

Super!