Question

Integrating factor problem:
(x^2+y^2) dx - 2xy dy = 0

Answer 1

I got \[e^{-2\ln(x)} = x^{-2}\] for my integrating factor. Plugging this into the equation, it ends up being: \[\frac{ y^2 }{ x^2 } dx - \frac{ 2y }{ x } dy = 0\]
solving this equation gets me the answer \[-xy^2=c \] or \[y=\sqrt{cx} = c \sqrt{x}\]
(confirmed with WolframAlpha)
but this is an incorrect answer...
Accroding to the book the answer is \[\sqrt{x^2+cx}\]

Answer 2

How did you get the integrating factor as such? @yanyaro

Answer 3

-xy^2 = c is what I get as well. I wouldnt owrry about it, you're very likely correct and that the book got a different answer that is equivalent through some other method.

Answer 4

\[R=\frac{ 1 }{ Q }(\frac{ \delta P }{ \delta y }-\frac{ \delta Q }{ \delta x }) = \frac{ (2y+2y) }{ -2xy }=-\frac{ 2y }{ xy }= -\frac{ 2 }{ x }\]
\[e^{\int\limits_{}^{}\frac{ 2 }{ x }}=e^{-2\ln x}\]

Answer 5

You may wish to recheck the step where you introduce your integrating factor into the original DE.
I think you mistakenly put that x^(-2) * x^2 as 0 instead of 1. (Since x^0 = 1)

Answer 6

@AccessDenied I see that now, thanks. Going to retry it.

Answer 7

Yeah, I did the same silly mistake.

Answer 8

@Psymon @yanyaro How did you guys get to the integrating factor? I would like an explanation.

Answer 9

Yeah, sure.

Answer 10

@genius12 I showed that work a few comments above

Answer 11

So, this is an equation that is non-exact, but can be made exact. If the partial derivatives of M with respect to y (M being the part of the DE that has dx tacked on) and N with respect to x (N being the part of the DE that has dy tacked on) do not match, there is a method to make them match by coming up with an integrating factor. Its just this integrating factor creates an exact equation instead of creating a product rule liek it may in other equations. So, the idea is you take one of these two fractions:

\[\frac{ M _{y}-N _{x} }{ N }\]remembering again that M is the function with dx attached and N is the function with dy attached, or you take this fraction:

\[\frac{ N _{x}-M _{y} }{ M } \]The fraction you choose is the one which becomes a function of only one variable. When you simplify that fraction, you come up with your "p(x)", which you integrate and then take the e of like in a normal integrating factor.

So for this problem, the partial of M with respect to y was 2y and the partial of n with respect to x was -2y. From the fractions above, the one which fit the conditions we want was the first one, so this becomes:

\[\frac{ 2y-(-2y) }{ -2xy }= -\frac{ 2 }{ x }\]Then doing:
\[e^{\int\limits_{}^{}\frac{ -2 }{ x }}\], we get the integrating factor we used.

Answer 12

I see. I still need to gain a better understanding however =] But thank you. @Psymon

Answer 13

Yeah, np : )