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March 30, 2015

Posted by **anonymous** on Tuesday, November 1, 2011 at 9:50am.

- university maths -
**drwls**, Tuesday, November 1, 2011 at 11:06amSee if this works:

http://www.math.colostate.edu/~gerhard/M345/CHP/ch2_6.pdf

It does not look like it meets the "exact differential" requirements

- university maths -
**Steve**, Tuesday, November 1, 2011 at 12:23pmGood reference. Try looking at the last example:

Substitute y = xv ⇒ dy = v dx+x dv

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

(x^2v - x^2v^2)dx - (x^2 + x^2v)(vdx + xdv) = 0

2/x dx + (1+v)/v^2 dv = 0

2lnx + (lnv - 1/v) = C

ln(x^2v) - 1.v = C

vx^2 = Cexp(1/v)

y = vx

xy = Ce^{x/y}

Check my math, as always.

- university math -
**drwls**, Tuesday, November 1, 2011 at 10:04pmI want to congratulate and thank Steve for that very impressive solution.

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