Question

The solution of the differential equation $x\cos x\frac{dy}{dx}+(x\sin x+\cos x)y=1$ is

• A. $x\sec x-y\tan x=C$
• B. $x^{2}y\cos x-\tan x=C$
• C. $xy\sec x+y\tan x=C$
• D. $xy\sec x-\tan x=C$

Answer 1

Answer: (D)

We have
$x\cos x\frac{dy}{dx}+(x\sin x+\cos x)y=1$
$\frac{dy}{dx}+\left(\tan x+\frac{1}{x}\right)y=\frac{\sec x}{x}$
IF $=e^{\int \left(\tan x+\frac{1}{x}\right)dx}=e^{\log \sec x+\log x}$
$=e^{\log x\sec x}=x\sec x$
Solution of differential equation is
$xy\sec x=\int \frac{\sec x}{x}\cdot x\sec xdx+C$
$\Rightarrow xy\sec x=\int {\sec }^{2}xdx+C$
$\Rightarrow xy\sec x=\tan x+C$
$\Rightarrow xy\sec x-\tan x=C$