Question

The solution of the differential equation

$(1 + y^{2}) + (x - e^{\tan^{- 1}y})\frac{dy}{dx} = 0$is

• A. $(x - 2) = ke^{\tan^{- 1}y}$
• B. $2xe^{\tan^{- 1}y} = e^{2\tan^{- 1}y} + k$
• C. $xe^{\tan^{- 1}y} = \tan^{- 1}y + k$
• D. $xe^{2\tan^{- 1}y} = e^{\tan^{- 1}y} + k$

Answer 1

Answer: (B)

$2xe^{\tan^{- 1}y} = e^{2\tan^{- 1}y} + k$

Answer 2

We have $(x - e^{\tan^{- 1}y})\frac{dy}{dx} = - (1 + y^{2})$

⇒ $\frac{dx}{dy} = - \left( \frac{x - e^{\tan^{- 1}y}}{1 + y^{2}} \right)$

⇒ $\frac{dx}{dy} + \frac{1}{1 + y^{2}}x = \frac{e^{\tan^{- 1}y}}{1 + y^{2}}$ ……..(i)

This is a linear differential equation of the form

$\frac{dx}{dy} + R(y).x = S(y)$

$R = \frac{1}{1 + y^{2}}$, $S = \frac{e^{\tan^{- 1}y}}{1 + y^{2}}$

Integrating factor$= e^{\int_{}^{}{Rdy}} = e^{\int_{}^{}\frac{dy}{1 + y^{2}}} = e^{\tan^{- 1}y}$

Multiplying (i) by I.F. and integrating,$xe^{\tan^{- 1}y} = \int_{}^{}\frac{e^{\tan^{- 1}y}}{1 + y^{2}} \cdot e^{\tan^{- 1}y}dy = \int_{}^{}\frac{(e^{\tan^{- 1}y})^{2}dy}{1 + y^{2}} = \frac{(e^{\tan^{- 1}y})^{2}}{2} + \frac{k}{2}$∴ $2xe^{\tan^{- 1}y} = e^{2\tan^{- 1}y} + k$