Question

A question on exact DE. Suppose I have h'(y) (after solving for Psi_x), but h'(y) depends on x. That cannot happen or should I treat x as a constant as usual? I mean, all examples I saw so far, the x's cancel out, but I am not sure if this should be generalized.

Answer 1

http://www.khanacademy.org/math/differential-equations/v/exact-equations-example-3 for instance, 3:10 of this video. The x cancel out when you take the partial derivative with respect to y and compare with Nx.

Answer 2

Can you please post your problem to see what is going on?

Answer 3

I have this linear equation: \[x^2\frac{dy}{dx} + xy = 1 \]I know this is separable and I have already solved it. But I was trying to practice using an integrating factor to make it exact. I rewrote the equation as: \[x^2dy + (xy - 1)dx = 0 \]I found \(\mu(x) = x^{-1} \) and it seems to work: \[xdy + (y - \frac{1}{x})dx = 0 \]The derivative of the first with respect to x is 1, from the second with respect to y is also 1. That implies that \(\Psi (x,y)_x = x \implies \Psi(x,y) = \frac{1}{2} x^2 + h(y)\). Derivating with respect to y yields: \(\Psi (x,y)_y = h '(y) = y - \frac{1}{x}\). My question is this: can h' depend on x like that or I can't solve the DE like that?

Answer 4

@amistre64 ? :-D

Answer 5

it looks like a couchy

Answer 6

assume x^r is a solution and plug it in

Answer 7

What I did first was I divided by x to get \[x \frac{dy}{dx} + y = \frac{1}{x}\]Then I said that \( \frac{dx}{dx} = 1 \) and rewrote the equation as: \[x \frac{dy}{dx} + y\frac{dx}{dx} = \frac{1}{x} \implies \frac{1}{dx} (xdy + ydx) = \frac{1}{x} \implies \frac{d}{dx}(xy) = \frac{dx}{x}\]And solved it like that :-). But my question is on the exact diff. right below. And thanks for helping me out.

Answer 8

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

Fyx = 2x

Fxy = x

its not exact

Answer 9

I know. The integrating factor is \( \mu (x) = x^{-1} \) :-) At least, the one I found.

Answer 10

im not conversant on all diffyQs so if you found something that works, kudos ;)

Answer 11

But do you remember when you solve an exact diff. equation, you have to derivative the function with respect to y after integrating with respect to x? Supposing the solution is a function of x and y, that is.

Answer 12

i recall exacts yes; and i was reading up the other night on line integrals that looked similar to them. But i still havent wrapped up a good concept for them yet

Answer 13

Hmm. My question is, after you take the derivative with respect to y, do all x's have to disappear from the function? Or can I have something like that above: h'(y) = y - 1/x ?

Answer 14

Mdx + N dy can be seen as a dot product of a partial and another vector

Answer 15

i dont think all xs have to disappear, but that might be my novice views getting in the way

Answer 16

http://archive.org/details/IntroductionToLineIntegrals

and alot of other good stuff there as well

Answer 17

I don't think either, but having the (bad) luck so far of all the examples I did the x's cancelling out after the partial, seeing something like that makes my spider sense starts to tingle

Answer 18

Thanks for the link, mate :-)

Answer 19

\[
\Psi (x,y)_x = y - \frac 1 x \\
\Psi (x,y)= x y -\ln x + h(y)\\
\Psi (x,y)_y= x + h'(y)= x\\
h(y) = k
\\ \text { Hence }
\Psi (x,y)= x y -\ln x + k
\]
\[\Psi (x,y)_x
\]is the coefficient of dx and not of dy

Answer 20

Ah, that makes sense. :-)
In general, h(y) cannot depend on x, right?

Answer 21

yes, if we have exactness.

Answer 22

Thanks for the help :-) This doubt was bothering me for some time.

Answer 23

yw