Question

The solution of $\cos x\frac{dy}{dx}+y=\sin x$ is

• A. $y(\sec x+\tan x)=\sec x+\tan x+c$
• B. $y(\sec x+\tan x)=\sec x+\tan x-x+c$
• C. $y(\sec x+\tan x)=\sec x-\tan x-x+c$
• D. $y(\sec x-\tan x)=\sec x+\tan x-x+c$

Answer 1

The correct option is B $y(\sec x+\tan x)=\sec x+\tan x-x+c$

$\cos x\frac{dy}{dx}+y=\sin x$

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

It is a first degree linear differential equation

On comparing it with its standard form,we get

$P=\sec x$ and $Q=\tan x$

$I.F=e^{\int pdx}=e^{\int \sec xdx}=e^{\log \left|\sec x+\tan x\right|}$

$=\sec x+\tan x$

Therefore, solution of given D.E is

$y\times I.F=\int Q\times I.Fdx+c$

$\Rightarrow y\left(\sec x+\tan x\right)=\int \tan x\left(\sec x+\tan x\right)dx+c$

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

$\Rightarrow y\left(\sec x+\tan x\right)={\int }^{}\left(\sec x\tan x+{\sec }^{2}x-1\right)dx+c$

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

$\Rightarrow y\left(\sec x+\tan x\right)=\sec x+\tan x-x+c$