Differential Equations — Class 12 Maths Solution

Given that, $\left( {\frac{{2 + \sin x}}{{1 + y}}} \right)\frac{{dy}}{{dx}} = - \cos x$

$\Rightarrow$ $\frac{{dy}}{{1 + y}} = - \frac{{\cos x}}{{2 + \sin x}}dx$

On integrating both sides,

we get
$\int {\frac{1}{{1 + y}}} dy = - \int {\frac{{\cos x}}{{2 + \sin x}}} dx$

$\Rightarrow$ $\log (1 + y) = - \log (2 + \sin x) + \log C$

$\Rightarrow$ $\log (1 + y) + \log (2 + \sin x) = \log C$

$\Rightarrow$ $\log (1 + y)(2 + \sin x) = \log C$

$\Rightarrow$ $(1 + y)(2 + \sin x) = C$
$\Rightarrow$ $1 + y = \frac{C}{{2 + \sin x}}$

$\Rightarrow$ $y = \frac{C}{{2 + \sin x}} - 1$

……..(i)
When $x = 0$ and $y = 1$,

then
$1 = \frac{C}{2} - 1$

$\Rightarrow$ $C = 4$

On putting $C = 4$ in Eq. (i),

we get
$y = \frac{4}{{2 + \sin x}} - 1$
$\therefore$ $y\left( {\frac{\pi }{2}} \right) = \frac{4}{{2 + \sin \frac{\pi }{2}}} - 1 = \frac{4}{{2 + 1}} - 1$

$= \frac{4}{3} - 1 = \frac{1}{3}$