Question

For the differential equations, find a particular solution satisfying the given condition:\(\frac{dy}{dx}=y tan\,\,x;y=1\) when x=0

Answer 1

\(\frac{dy}{dx}=y tan\,\,x\)
\(⇒\frac{dy}{y}=tanx\, dx\)
Integrating both sides,we get:
\(∫\frac{dy}{y}=-∫tanx\, dx\)
\(⇒logy=log(secx)+logC\)
\(⇒logy=log(C secx)\)
\(⇒y=C secx...(1)\)
Now,y=1 when x=0.
\(⇒1=C\times sec0\)
\(⇒1=C\times1\)
\(⇒C=1\)
Substituting C=1 in equation(1),we get:
\(y=secx.\)