Question

The solution of the differential equation

$\frac { d y } { d x } = ( 1 + x ) \left( 1 + y ^ { 2 } \right)$ is

• A. $y = \tan \left( x ^ { 2 } + x + c \right)$
• B. $y = \tan \left( 2 x ^ { 2 } + x + c \right)$
• C. $y = \tan \left( x ^ { 2 } - x + c \right)$
• D. $y = \tan \left( \frac { x ^ { 2 } } { 2 } + x + c \right)$

Answer 1

Answer: (D)

$y = \tan \left( \frac { x ^ { 2 } } { 2 } + x + c \right)$

Answer 2

$\frac { d y } { d x } = ( 1 + x ) \left( 1 + y ^ { 2 } \right)$ ⇒ $\frac { d y } { 1 + y ^ { 2 } } = ( 1 + x ) d x$

On integrating both sides, we get

$\tan ^ { - 1 } y = \frac { x ^ { 2 } } { 2 } + x + c$ ⇒ $y = \tan \left( \frac { x ^ { 2 } } { 2 } + x + c \right)$.