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Question

The solution of the differential equation cosx dy=y(sin(x)y)dx, 0<x<π2 is
(Here, C is a constant of integration)

A
ysecx=tanx+C
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B
ytanx=secx+C
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C
tanx=(secx+C)y
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D
secx=(tanx+C)y
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Solution

The correct option is D secx=(tanx+C)y
dydx=y(sin(x)y)cosxdydx(tanx)y=(secx)y2y2dydx+(tanx)y=secx (1)
which is Bernoulli's differential equation.
Put 1y=t
1y2dydx=dtdx

From equation (1),
dtdx+(tanx)t=secx
I.F. =etanx dx=elnsecx=secx
So, the general solution is
tsecx=sec2x dxtsecx=tanx+C
Hence, 1ysecx=tanx+C
secx=(tanx+C)y

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