Question

Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\] 

 

Answer 1

We have,
\[\cos y dy + \cos x \sin y dx = 0\]
\[ \Rightarrow \cos y dy = - \cos x \sin y dx\]
\[ \Rightarrow \cot y dy = - \cos x dx\]
Integrating both sides, we get
\[\int cot y dy = - \int\cos x dx\]
\[ \Rightarrow \log \left| \sin y \right| = - \sin x + C\]
\[ \Rightarrow \log \left| \sin y \right| + \sin x = C . . . . (1)\]
\[\text{ It is given that at }x = \frac{\pi}{2}, y = \frac{\pi}{2} . \]
\[\text{ Substitutuing the values of x and y in }\left( 1 \right),\text{ we get }\]
\[\log \left| \sin\frac{\pi}{2} \right| + \sin\frac{\pi}{2} = C\]
\[ \Rightarrow C = 1\]
Therefore, substituting the value of C in (1), we get 
\[\log \left| \sin y \right| + \sin x = 1\]
\[\text{ Hence, }\log \left| \sin y \right| + \sin x = 1\text{ is the required solution .}\]