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# Integrating factor of the differential equation $\cos x\large\frac{dy}{dx}$$+y\sin x=1 is:$(A)\;\cos x\quad(B)\;\tan x\quad(C)\;\sec x\quad(D)\;\sin x$ Can you answer this question? ## 1 Answer 0 votes Toolbox: • A linear differential equation of the form \large\frac{dy}{dx}$$+Py=Q$ has a general solution $y e^{\large\int pdx}=\int Q.e^{\large \int pdx}.dx+c$. where $e^{\large \int pdx}$ is the integrating factor (I.F)
• $\int \tan xdx=\log |\sec x|+c$
$\cos x. \large\frac{dy}{dx}$$+y \sin x=1 Divide throughout by \cos x \large\frac{dy}{dx}$$+ y \large\frac{\sin x}{\cos x}=\frac{1}{\cos x}$
But $\large\frac{\sin x}{\cos x}$$=\tan x and \large\frac{1}{\cos x}$$=\sec x$
$\large\frac{dy}{dx}$$+y \tan x=\sec x This is a linear differential equation of the form \large\frac{dy}{dx}$$+Py=Q$
Where $P=\tan x$
The integrating factor is $e^{\large\int pdx}$
$\int pdx=\int \tan x=\log |\sec |$
Hence $I.F = e^{\large\log |\sec x|}$
But $e^{\large \log x}=x$