Question

The solution of the differential equation

(1 + y²) + (x –$e^{\tan^{- 1}y}$) $\frac{dy}{dx}$= 0 is –

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

Answer 1

Answer: (B)

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

Answer 2

Given (1 + y²) + (x –$e^{\tan^{- 1}y}$) $\left( \frac{dy}{dx} \right)$ = 0

This can be rewritten as

(1 + y²) $\frac{dx}{dy}$ + x = $e^{\tan^{- 1}y}$

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

It is a linear equation, comparing with the standard equation $\frac{dx}{dy}$ + Px = Q Ž P = $\frac{1}{1 + y^{2}}$, Q = $\frac{e^{\tan^{- 1}y}}{1 + y^{2}}$ = $e^{\int_{}^{}{pdy}}$

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

Thus solution is

x($e^{\tan^{- 1}y}$)= $e^{\tan^{- 1}y}$dy

Put $e^{\tan^{- 1}y}$ = t Ž $\frac{e^{\tan^{- 1}y}}{1 + y^{2}}$dy = dt

\ x($e^{\tan^{- 1}y}$) = $\int_{}^{}t$dt = $\int_{}^{}\frac{t^{2}}{2}$ + c = $\frac{e^{2\tan^{- 1}y}}{2}$ + c

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

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