Question

The solution of $\frac{dy}{dx}-ytanx=e^x\sec x$ is

• A. ${ye}^x=\cos x+c$
• B. $y\cos x=e^x+c$
• C. $y\sin x=e^x+c$
• D. ${ye}^x=\sin x+c$

Answer 1

Answer: (B)

$y\cos x=e^x+c$

Given, $\frac{dy}{dx}-y\tan x=e^x\sec x$

It is a first-degree linear D.E of the form

$\frac{dy}{dx}+Py=Q$

here $P=-\tan xQ=e^x\sec x$

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

$=e^{\log \cos x}=\cos x$

The solution is given by,

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

$\Rightarrow y\times \cos x=\int e^x\sec x\times \cos xdx+c$

$\Rightarrow y\cos x=\int e^{x}dx+c$

$\Rightarrow y\cos x=e^x+c$