Question

The general solution of the differential equation $\frac{dy}{dx}$

= (x³ – 2x tan^–1y) (1 + y²) is-

• A. 2tan^–1x = y² – 1 + 2C $e^{- x^{2}}$
• B. 2tan^–1y = x² – 1 + 2C $e^{- x^{2}}$
• C. 2tan^–1y = y² – 1 + 2C $e^{- x^{2}}$
• D. 2tan^–1x = x² – 1 + 2C $e^{- x^{2}}$

Answer 1

Answer: (B)

2tan^–1y = x² – 1 + 2C $e^{- x^{2}}$

Answer 2

$\frac{1}{1 + y^{2}}.\frac{dy}{dx}$ + 2x(tan^–1y) = x³

Put tan^–1y = z

\ $\frac{1}{1 + y^{2}}.\frac{dy}{dx}$ = $\frac{dz}{dx}$

$\frac{dz}{dx}$ + (2x)z = x³

Ž z. $e^{x^{2}}$ = $\frac{1}{2}\int_{}^{}{2e^{x^{2}}.x^{3}dx + C}$

Ž $2e^{x^{2}}(\tan^{- 1}y)$ = x² $e^{x^{2}}$ – $e^{x^{2}}$ + 2C

Ž 2tan^–1y = x² – 1 + 2C$e^{- x^{2}}$