The solution of the differential equation (1+y2)+(x−etan−1y)dydx=0

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Question

The solution of the differential equation

(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0

, is

Options:

A . 2x etan−1y,=e2tan−1y+k

B . x etan−1y,=etan−1y+k

C . x e2tan−1y,=e−tan−1y+k

D . (x−2)k etan−1y

Answer: Option A
:
A

\frac{d x}{d y} + \frac{1}{1 + y^{2}} x = \frac{1}{1 + y^{2}} e^{t a n^{- 1} y}

I . F = e^{\int \frac{1}{1 + y^{2}} d y} = e^{t a n^{- 1} y}

∴

Solution is

x . e^{t a n^{- 1}} y

= \int e^{t a n^{- 1} y} . \frac{1}{1 + y^{2}} e^{t a n^{- 1}} d y = \frac{1}{2} e^{2 t a n^{- 1} y} + \frac{1}{2} k \Rightarrow 2 x e^{t a n^{- 1} y} = e^{2 t a n^{- 1} y} + k

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