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Integrating factor of the differential equation $\large\frac{dy}{dx}$$+y\tan x-\sec x=0$ is \begin{array}{1 1}(A)\;\cos x & (B)\;sec x\\(C)\;e^{\cos x} & (D)\;e^{\sec x} \end{array}

1 Answer

  • A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$ has a general solution $y e^{\large\int pdx}=\int Q.e^{\large \int pdx}.dx+c$. where $e^{\large \int pdx}$ is the integrating factor (I.F)
Given $ \large\frac{dy}{dx}$$+y \tan x-\sec x=0$
This can be written as $ \large\frac{dy}{dx}$$+y \tan x=\sec x$
Clearly this is a linear differential equation which is of the form
$\large\frac{dy}{dx}+Py=Q$ where $P=\tan x$ and $Q=\sec x$
Hence the required solution is
$y e^{\large\int pdx}=\int Q.e^{\large \int pdx}$
Here $ e^{\large\int pdx}$ is the integrating factor
$\int pdx=\int \tan x dx$
$=\log |\sec x|$
Hence $I.F=e^{\large\log |\sec x|}$
But $ e^{\large\log |\sec x|}=\sec x$
Hence $I.F=\sec x$
Hence the option $B$ is correct
answered May 16, 2013 by meena.p

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