The general solution of the differential equation (1+y^2) dx=( tan^-1 ⁡y-x) dy is

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Accepted Answer

Correct answer: D. (d) Given that, (1+y^2) dx=( tan^-1 ⁡y-x) dy ⇒ dx dy= tan^-1 ⁡y-x1+y^2 ⇒ dy dx+ x1+y^2= tan^-1 ⁡y1+y^2 This is in the form linear differential equation of ∴ P= 11+y^2 and Q= tan^-1 ⁡y1+y^2 IF=e^ Pdy=e^ 11+y^2 dy+ Px=Q IF=e^ tan^-1 ⁡x Now, solution of linear differentiate equation is x IF= Q IF+Cx e^ tan^-1 ⁡y= tan^-1 ⁡y1+y^2 e^ tan^-1 ⁡y dy Let tan^-1 ⁡y=t 11+y^2 dy= dt= t e^t dt=t e^t- (e^t-1) dt = te^t-e^t+c=e^t (t-1 )+c Putting t= tan^-1 ⁡yx e^ tan^-1 ⁡y=e^ tan^-1 ⁡y( tan^-1 ⁡y-1)+c

$The general solution of the differential equation (1 + y^{2}) dx = (tan^{- 1} y - x) dy is$

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