The solution of the differential equation cosxdy=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 Dsecx=(tanx+C)y dydx=y(sin(x)−y)cosx⇒dydx−(tanx)y=−(secx)y2⇒y−2dydx+(−tanx)y=−secx…(1)
which is Bernoulli's differential equation.
Put 1y=t ⇒−1y2dydx=dtdx
From equation (1), dtdx+(tanx)t=secx
I.F. =e∫tanxdx=elnsecx=secx
So, the general solution is tsecx=∫sec2xdx⇒tsecx=tanx+C
Hence, 1ysecx=tanx+C ⇒secx=(tanx+C)y