Question

Apparently, a separable ODE is also exact. However, let’s take a look at this particular example:\[y^{-1}dy+ye^{\cos x}\sin xdx=0.\]We immediately see that\[M=ye^{\cos x}\sin x,M_y=e^{\cos x}\sin x,\]\[N=y^{-1},N_x=0.\]Therefore, it’s not exact.

On the other hand, it’s separable since\[y^{-1}dy+ye^{\cos x}\sin xdx=0,\]\[y^{-2}dy=-e^{\cos x}\sin xdx.\]Therefore, it should also be exact.

I’m totally confused.

Answer 1

are all separable exact?

Answer 2

If an ODE is separable, it is also exact since when you take its partial derivatives, they will turn out to be zero.

Answer 3

Your statement saying that all separable ODE are exact is not true.

Answer 4

I'm curious to know why not?

Answer 5

I don't see how this is exact. Do we not need \[M_y=N_x\]? that does not seem to be the case here.

Answer 6

he said this is not exact

Answer 7

oh, pardon

Answer 8

Oh, I think I can see where your problem is. They are exact after you do the separation.

Answer 9

If that makes sense.

Answer 10

In other words, you have actually multiplied both sides by what is called the integrating factor.

Answer 11

\[y^{-1}dy+ye^{\cos x}\sin xdx=0.\]
\[y^{-2}dy+ye^{\cos x}\sin xdx=0.\]

Answer 12

ignore last one
\[y^{-2}dy+e^{\cos x}\sin xdx=0.\]

Answer 13

This is actually very interesting. :D

Answer 14

You can see the ODE written in last form you gave is exact.

Answer 15

So, as it is presented (above), the ODE is not exact, but it is separable. It being separable does imply it being exact, but we had to multiply the ODE by its integrating factor, as Mr. Math said. I think I see it now. ^^

Answer 16

Thanks, I have never really related separable ODE's to being exact. This is a good question.