Question

Find the general solution of $dy=y\sec xdx$.

• A. $y=C(\sec x-2\tan x)$
• B. $y=C(\sec x+\tan x)$
• C. $y=C(2\sec x+\tan x)$
• D. None of these

Answer 1

Answer: (B)

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

Given Differential Equation is $dy=y\sec xdx$

Integrating with respect to $x$ gives,

$\int \frac{1}{y}dy=\int \sec xdx$

$\ln y=\ln (\sec x+\tan x)+\ln c$

Where $c$ is the integration constant,

$\therefore y=c(\sec x+\tan x)$.