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:
$y=secx.$