Question

Solution of $\frac{dy}{dx} - y\tan x = - y^{2}\sec x$ is

• A. ysec x = tan x + c
• B. $\frac{\sec x}{y} = \tan x + c$
• C. ycosx = tan x + c
• D. None of these

Answer 1

Answer: (B)

$\frac{\sec x}{y} = \tan x + c$

Answer 2

Re-writing the given equation,

$y^{- 2}\frac{dy}{dx} - y^{- 1}\tan x = - \sec x$

Let $y^{- 1} = v$ ⇒ $- y^{- 2}\frac{dy}{dx} = \frac{dv}{dx}$

∴ $\frac{dv}{dx} + \tan x.v = \sec x$ ……..(i)

I.F.$= e^{\int_{}^{}{\tan x}} = e^{{lnsec}x} = \sec x$

Multiplying (i) by sec x and integrating,

$v\sec x = \int_{}^{}{\sec^{2}xdx} = \tan x + c$

∴ $\frac{\sec x}{y} = \tan x + c$