Question

The solution of $\frac{dy}{dx}+y\tan x=\sec x$ is:

• A. $y=\sin x+\cos x+c$
• B. $y=\sin x+c\sec x$
• C. $y=\sin x+c\cos x$
• D. $y=\sin x+\tan x+c$

Answer 1

The correct option is C $y=\sin x+c\cos x$

Given, $\frac{dy}{dx}+y\tan x=\sec x$

I.F = $e^{\int p\left(x\right)dx}=e^{\int \tan xdx}=e^{\log \sec x}=\sec x$

$\Rightarrow \sec x\frac{dy}{dx}+y\sec x\tan x={\sec }^{2}x$

$\Rightarrow \frac{d}{dx}\left(y\sec x\right)={\sec }^{2}x$

$\Rightarrow \int d\left(y\sec x\right)=\int {\sec }^{2}xdx+c$

$\Rightarrow y\sec x=\tan x+c$

$\Rightarrow y=\sin x+c\cos x$