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differential equation of cheese

Name: Anonymous 2007-09-23 17:57 ID:XM4X3M1H

(y+x) dy/dx = y       use y=xu to solve

save me from evil problem lords!

Name: Anonymous 2007-09-23 18:05 ID:n5cvFPom

um

dy/dx X = 1 so

2?

Name: Anonymous 2007-09-23 18:12 ID:nO83moQo

(ux + x)(u + x du/dx) = xu
x(u+1)(u + x du/dx) = xu
(u+1)(u+x du/dx) = u
(u + x du/dx) = u/(u+1)
x du/dx = u/(u+1) - u
x du/dx = (u - u^2 - u)/(u+1)
1/x dx = (u+1)/(-u^2) du

Name: Anonymous 2007-09-23 18:15 ID:koOqfoww

I was gonna solve that but the way you wrote it out is confusing and ambiguous as fuck

Name: Anonymous 2007-09-23 18:19 ID:XM4X3M1H

>>3
i got to there but then what

Name: Anonymous 2007-09-23 18:20 ID:nO83moQo

>>4
(y+x) => (xu + x)
dy/dx => (u + xdu/dx)
y => xu

Plug it all in

(xu + x)(u + xdu/dx) = xu
Reduce it all to
(1/x)dx = (u+1)/(-u^2)  du

wth is ambiguous about that?

Name: Anonymous 2007-09-23 18:22 ID:nO83moQo

>>5

Integrate both sides and plug in y/x for u.  Then solve for y

Name: 4tran 2007-09-24 0:40 ID:Heaven

Right side is probably easier to integrate if reexpressed as
-[(1/u) + (1/u2)] du
The result of that is probably a large mess...

Name: Anonymous 2007-09-25 1:37 ID:vqhvgHJp

That's what you get when you make cheese with homogenized milk!

Name: McPeePants 2007-09-26 0:06 ID:Ll7lFx+r

42

Name: Anonymous 2007-09-26 1:29 ID:gXp6f4GK

9000

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