Question

The solution of the differential equation $\frac { d y } { d x } = 1 + x + y + x y$ is

• A. $\log ( 1 + y ) = x + \frac { x ^ { 2 } } { 2 } + c$
• B. $( 1 + y ) ^ { 2 } = x + \frac { x ^ { 2 } } { 2 } + c$
• C. $\log ( 1 + y ) = \log ( 1 + x ) + c$
• D. None of these

Answer 1

Answer: (A)

$\log ( 1 + y ) = x + \frac { x ^ { 2 } } { 2 } + c$

Answer 2

$\frac { d y } { d x } = 1 + x + y + x y$

⇒ $\frac { d y } { d x } = ( 1 + x ) ( 1 + y )$ ⇒ $\frac { d y } { 1 + y } = ( 1 + x ) d x$

On integrating, we get $\log ( 1 + y ) = \frac { x ^ { 2 } } { 2 } + x + c$.