Question

The solution of the differential equation

$( 1 + \cos x ) d y = ( 1 - \cos x ) d x$is

• A. $y = 2 \tan \frac { x } { 2 } - x + c$
• B. $y = 2 \tan x + x + c$
• C. $y = 2 \tan \frac { x } { 2 } + x + c$
• D. $y = x - 2 \tan \frac { x } { 2 } + c$

Answer 1

Answer: (A)

$y = 2 \tan \frac { x } { 2 } - x + c$

Answer 2

Here $\frac { d y } { d x } = \frac { 1 - \cos x } { 1 + \cos x } = \tan ^ { 2 } \frac { x } { 2 }$

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

Now on integrating both the sides, we get $y = 2 \tan \frac { x } { 2 } - x + c$.