Question

Solution of $\frac{dy}{dx}+y\,\sec\,x=\tan\,x$ is

• A. $y = \sec x + \tan x - x + C$
• B. $y\, (\sec x + \tan x) = \sec x + \tan x - x + C$
• C. $y\, (\sec x + \tan x) = \sec x + \tan x + x + C$
• D. none of these

Answer 1

Answer: (B)

$y\, (\sec x + \tan x) = \sec x + \tan x - x + C$

Answer 2

$\frac{dy}{dx}+y\,sec\,x = tan\,x$. $P= sec \,x, Q = tan \,x$ $e^{\int Pdx} = e^{log\left(sec\,x+tan\,x\right)}= sec\,x+tan\,x$ $\therefore$ Sol. is $y \left(sec\,x+tan\,x\right) = \int \left(sec\,x+tan\,x+tan^{2}\,x\right)dx$ $= \int sec\,x\,tan\,xdx+\int sec^{2}\,xdx-\int dx$ $= sec\,x+tan\,x-x+C$