Question

Solve the differential equation: $2\cos x\frac{dy}{dx}+4y\sin x=\sin 2x$

• A. $y{\sec }^{2}x=\int \sec x\tan x+c=\sec x+c.$
• B. $y{\sec }^{2}x=-\int \sec x\tan x+c=-\sec x+c.$
• C. $y{\sec }^{3}x=\int \sec x\tan x+c=\sec x+c.$
• D. None of these

Answer 1

The correct option is A $y{\sec }^{2}x=\int \sec x\tan x+c=\sec x+c.$

Given equation is $2\cos x\frac{dy}{dx}+4y\sin x=\sin 2x$

$\Rightarrow \frac{dy}{dx}+2\tan xy=\sin x$ ...(1)
Here $P=2\tan x\Rightarrow \int Pdx=\int 2\tan xdx=\log \left({\sec }^{2}x\right)$
$\therefore I.F.=e^{\log \left({\sec }^{2}x\right)}={\sec }^{2}x$
Multiplying (1) by $I.F.$, we get
${\sec }^{2}x\frac{dy}{dx}+2\tan x{\sec }^{2}xy=\sin x{\sec }^{2}x$
Integrating both sides, we get
$y{\sec }^{2}x=\int \sec x\tan x+c=\sec x+c$