This is probably too hard to be the first problem I try for diff eq. I'm trying to learn this stuff. Question, whats the difference between a homogenous and nonhomogenous one? What problem does this pose in solving the problem?(adsbygoogle = window.adsbygoogle || []).push({});

I wanna try this one (jacked from Naeem's post, but I posted separately to avoid hijacking), but I know itll be wrong so here goes:

[tex]y \ dx + (2x - ye^y) \ dy = 0[/tex]

[tex] M(x)dx + N(y)dy = 0 [/tex]

[tex] M(x) = y \mbox{ and } N(y) = 2x - ye^y [/tex]

So then [itex] M_y = 1 \mbox{ and } N_x = 2 [/tex]

We need them to be the same (why?) so we need some factor [itex] \mu[/tex] that will make the partial derivatives [itex] M_y = N_x[/itex].

So I'm going on a limb here:

[tex] M(x)_y\mu = N(y)_x\mu [/tex]

[tex] \int{M(x)_y\mu}{dy} = \int{N(y)_x\mu}{dx} [/tex]

[tex] \int{y\mu}{dy} = \int{(2x-ye^y)\mu}{dx} [/tex]

Im stuck here, but I thought of this, will it work?

[tex]y \ dx + (2x - ye^y) \ dy = 0[/tex]

[tex](2x - ye^y) \ dy = -ydx[/tex]

[tex] \ dy = \frac{-ydx}{(2x - ye^y)}[/tex]

[tex] \int dy = \int \frac{y}{2x-ye^y}dx [/tex]

[tex] y = y\int{\frac{1}{2x-ye^y}dx[/tex]

[tex] u = 2x-ye^y \mbox{ and } du = 2 dx [/tex]

[tex] y = \frac{y}{2}\int{1/u}{du} [/tex]

[tex]y = \frac{ln(2x-ye^y)}{2} [/tex]

I messed something up pretty bad, obviously my second idea didnt work. Can someone look over my first attempt and tell me how to continue? Do I treat mu as a function of x? y? both? cosntant?

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# First attempt at Diff Eq

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