The solution of the differential equation ( 1 + y 2 ) + ( x − e t a n − 1 y ) d y d x = 0 , is Options: A . &nbsp 2 x e t a n − 1 y , = e 2 t a n − 1 y + k B . &nbsp x e t a n − 1 y , = e t a n − 1 y + k C . &nbsp x e 2 t a n − 1 y , = e − t a n − 1 y + k D . &nbsp ( x − 2 ) k e t a n − 1 y

Answer: Option A : A d x d y + 1 1 + y 2 x = 1 1 + y 2 e t a n − 1 y I . F = e ∫ 1 1 + y 2 d y = e t a n − 1 y ∴ Solution is x . e t a n − 1 y = ∫ e t a n − 1 y . 1 1 + y 2 e t a n − 1 d y = 1 2 e 2 t a n − 1 y + 1 2 k ⇒ 2 x e t a n − 1 y = e 2 t a n − 1 y + k

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

Discussion Forum : Differential Equations

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 .

2 x e^{t a n^{- 1} y}, = e^{2 t a n^{- 1} y} + k

B .

x e^{t a n^{- 1} y}, = e^{t a n^{- 1} y} + k

C .

x e^{2 t a n^{- 1} y}, = e^{- t a n^{- 1} y} + k

D .

(x - 2) k e^{t a n^{- 1} y}

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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