#### Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 50

Answer: $\sin x+\log |\sin y|=1$

Hint: Separate the terms of x and y and then integrate them.

Given: $\cos y d y+\cos x \sin y d x=0, y=\frac{\pi}{2} \text { when } x=\frac{\pi}{2}$

Solution:

\begin{aligned} &\cos y d y+\cos x \sin y d x=0 \\\\ &\cos y d y=-\cos x \sin y d y \\\\ &\quad \Rightarrow \frac{-\cos y}{\sin y} d y=\cos x d x \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{-\cos y}{\sin y} d y=\int \cos x d x \\\\ &\Rightarrow-\log |\sin y|=\sin x+c \end{aligned}            ..............(1)

When $x=\frac{\pi}{2}, y=\frac{\pi}{2}$

\begin{aligned} &\Rightarrow-\log \left|\sin \frac{\pi}{2}\right|=\sin \frac{\pi}{2}+c \\\\ &-\log |1|=1+c \Rightarrow 0-1=c \Rightarrow c=-1 \\\\ &{\left[\therefore \log 1=0, \sin \frac{\pi}{2}=1\right]} \end{aligned}

Put in (1) we get

$\begin{gathered} \Rightarrow-\log |\sin y|=\sin x-1 \\\\ \quad \Rightarrow \sin x+\log |\sin y|=1 \end{gathered}$