Question

Find the particular solution of the differential equation $\frac{dy}{dx}+2y\tan x=\sin x$, given that $y=0$ when $x=\frac{\pi }{3}$.

Answer 1

The given differential equation $(D.E)$ is $\frac{dy}{dx}+2y\tan x=\sin x$

Comparing it with $\frac{dy}{dx}+Py=Q$, we have

$P=2\tan x\text{and}Q=\sin x$

Now the integral factor $(I.F)$ of this type of $D.E$ is given by
$I.F=e^{\int P.dx}$$=e^{\int 2\tan xdx}=e^{2\log |\sec x|}=e^{\log {\sec }^{2}x}={\sec }^{2}x$

Solution of given differential equation is :

$y(I.F)=\int Q(I.F)dx+C$

$y.{\sec }^{2}x=\int \sin x.{\sec }^{2}xdx+C$

$y.{\sec }^{2}x=\int \tan x.\sec xdx+C$

$y.{\sec }^{2}x=\sec x+C$

When $x=\frac{\pi }{3}$, then $y=0$

$\Rightarrow 0.{\sec }^2\frac{\pi }{3}=\sec \frac{\pi }{3}+C$

$\Rightarrow C=-2$

Hence, the required particular solution of the differential equation is

$y.{\sec }^{2}x=\sec x-2$