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. $xe^{2\tan^{–1}x}y = e^{\tan^{–1}y} + k$

Answer 1

Answer: (B)

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

Answer 2

$\frac{dx}{dy} + \frac{x}{1 + y^{2}} = \frac{e^{\tan^{–1}y}}{1 + y^{2}}$ Here

P= $\frac{1}{1 + y^{2}},\theta = \frac{e^{\tan^{–1}y}}{1 + y^{2}}$ I.F

= $e^{\int_{}^{}{\frac{1}{1 + y^{2}}dy}} = e^{\tan^{–1}y}$

Solution is given by

$x(e^{\tan^{–1}y}) = \int_{}^{}\frac{(e^{\tan^{–1}y})^{2}}{1 + y^{2}}dy$

Let tan^–1 y = t Ž $\frac{1}{1 + y^{2}}dy = dt$

Ž $x(e^{\tan^{–1}y}) = \frac{e^{2\tan^{–1}y}}{2} + c$