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$The solution of the differential equation \frac{d y}{d x} = \frac{(1 + y)^{2}}{x (1 + y) - x^{2}} is given as \ln (1 + y) =$

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1+yx+c

1−yx+c

−1−yx+c

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detailed solution

Correct option is A

consider dxdy=x1+y−x2(1+y)2⇒−1x2dydx+1x(1+y)=1(1+y)2 Substituting 1x=t, we get dtdy+t1+y=1(1+y)2 ; Which is linear I.F =e∫11+ydy=(1+y)⇒ solution of differential equation is t1+y=∫11+ydy=c ⇒ c+1+yx=ln1+y

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The solution of the differential equation dydx=(1+y)2x(1+y)−x2 is given as ln⁡(1+y)=

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$The solution of the differential equation \frac{d y}{d x} = \frac{(1 + y)^{2}}{x (1 + y) - x^{2}} is given as \ln (1 + y) =$