Question

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

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

Answer 1

Answer: (D)

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