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Integrating factor of the differential equation $\frac{d y}{d x}+y \tan x-\sec x=0$ is
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The correct answer is:
$\frac{1}{\cos x}$
$\frac{d y}{d x}+y \tan x=\sec x$, where $p=\tan x, Q=\sec x$
This is the linear differential equation in $y$, hence
I.F. $=e^{\int \operatorname{ddx}}=e^{\int \tan x d x} ;$ I.F. $=e^{\log \sec x}=\sec x$.
This is the linear differential equation in $y$, hence
I.F. $=e^{\int \operatorname{ddx}}=e^{\int \tan x d x} ;$ I.F. $=e^{\log \sec x}=\sec x$.
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