if xy + yx = ab find dy/dx

xy+ yx= ab, here ab is const .

let xy= u ; take log on both sides == y log x = log u == y*1/x + logx * dy/dx (by products rule) = 1/u du / dx

==du/dx = xy(y/x + logx dy/dx)

let yx= v ; take log on both sides == x log y = logv == x*1/ydy/dx + logy*1 (by products rule) = 1/v dv/dx

== dv/dx = yx(x/y dy/dx + logy)

== u + v = ab

differentiating wrt x we get , du/dx + dv/dx = 0 == xy(y/x + logx dy/dx) + yx(x/y dy/dx + logy) = 0

== xylogx dy/dx + yxx/y dy/dx = - xy*y/x - yxlogy

=== dy/dx = - (xy* y/x + yxlogy )/ xylogx + yxx/y

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general solution of the differential equations dy÷dx-y=sinx
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Find dy/dx

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Hope it helps.................
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