Question

The solution of $\frac { d y } { d x } = \sin ( x + y ) + \cos ( x + y )$is

• A. $\log \left[ 1 + \tan \left( \frac { x + y } { 2 } \right) \right] + c = 0$
• B. $\log \left[ 1 + \tan \left( \frac { x + y } { 2 } \right) \right] = x + c$
• C. $\log \left[ 1 - \tan \left( \frac { x + y } { 2 } \right) \right] = x + c$
• D. None of these

Answer 1

Answer: (B)

$\log \left[ 1 + \tan \left( \frac { x + y } { 2 } \right) \right] = x + c$

Answer 2

Put $x + y = v$ and $1 + \frac { d y } { d x } = \frac { d v } { d x }$

Therefore, the differential equation reduces to

$\frac { d v } { d x } = ( 1 + \cos v ) + \sin v$

⇒ $\int \frac { \sec ^ { 2 } ( v / 2 ) d v } { 2 [ 1 + \tan ( v / 2 ) ] } = \int d x$

⇒ $\log \left[ 1 + \tan \left( \frac { x + y } { 2 } \right) \right] = x + c$.