Question

An integrating factor of the differential equation $\frac{dy}{dx}=ytanx-y^{2}secx$ is equal to :

• A. tan x
• B. sec x
• C. cosec x
• D. cot x

Answer 1

Answer: (B)

Consider the differential equation
$\frac{dy}{dx}=ytanx-y^{2}secx$
Divide by $y^2$ on both the sides, we get
$\frac{1}{y^2}\left(\frac{dy}{dx}\right)=\frac{tanx}{y}-secx...\left(1\right)$
Let $\frac{1}{y}=z$
Diff both side, we get
$\frac{-1}{y^2}.\frac{dy}{dx}=\frac{dz}{dx}$
Put value of $\frac{1}{y^2}\frac{dy}{dx}$ in the equation $\left(1\right)$ , we get
$-\left(\frac{dz}{dx}\right)-\left(tanx\right)z=-secx$
$\Rightarrow \left(\frac{dz}{dx}\right)+\left(tanx\right)z=sec$
This is the linear diff equation in 'z' i.e.
This is of the form $\frac{dz}{dx}+P.z=Q$
then integrating factor $=e^{\int Pdx}$
$\therefore$ In the given question
$I.F.=e^{\int tanxdx}=e^{log\left(secx\right)}=secx$