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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Integrating factor of the differential equation $\cos x\large\frac{dy}{dx}$$+y\sin x=1$ is:\[(A)\;\cos x\quad(B)\;\tan x\quad(C)\;\sec x\quad(D)\;\sin x\]

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  • 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}$$+Py=Q$
Where $P=\tan x$
The integrating factor is $e^{\large\int pdx}$
$ \int pdx=\int \tan x=\log |\sec |$
Hence $I.F = e^{\large\log |\sec x|}$
But $e^{\large \log x}=x$
Similarly $e^{\large\log |\sec x|}$$=\sec x$
Hence $I.F=\sec x$
The correct option is $C$
answered May 14, 2013 by meena.p

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