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General solution of $\frac{dy}{dx} + y\tan x = \sec x$ is

\[(A)\;y\sec x=\tan x+c \quad (B)\;y\tan x=\sec x+c\quad(C)\;\tan x=y\tan x+c \quad(D)\;x\sec x=\tan y+c\]

1 Answer

  • A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$ has general solution as $ye^{\int pdx}=\int Qe^{\int pdx}dx+c$
Given $ \large\frac{dy}{dx}$$+y\tan x=\sec x$
This is a linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$ where $P=\tan x$ and $Q=\sec x$
$\int pdx=\int \tan x$
$=\log |\sec x|$
Hence the integrating factor $ e^{\large\int pdx}=e ^{\large\log |\sec x|}$
$e^{\large\log |\sec x|}=\sec x$
The required solution is
$y(I.F)=\int Q \times I.F dx+c$
$y.\sec x=\int \sec x.\sec x dx+c$
$=\int \sec ^2 x dx+c$
On integrating we get
$y \sec x=\tan x+c$
Hence the correct option is $A$
answered May 21, 2013 by meena.p

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