Question

The solution to the differential equation $\cos xdy=y(\sin x-y)dx$, $0<x<\frac{\pi }{2}$, is

• A. $\tan x=(\sec x+c)y$
• B. $\sec x=(\tan x+c)y$
• C. $y\sec x=\tan x+c$
• D. $y\tan x=\sec x+c$

Answer 1

Answer: (B)

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

$cosxdy=y(sinx-y)dx$ ___(1)

$\frac{dy}{dx}=\frac{y(sinx-y)}{cosx}=ytanx-y^{2}secx$

$\frac{dy}{dx}-ytanx=-y^{2}secx$ ___(2)

[Divide by $y^2$]

$y^{-2}\frac{dy}{dx}-y^{-1}tanx=-secx$ ___ (3)

Let $+y^{-1}=z\Rightarrow -y^{-2}\frac{dy}{dx}=\frac{dz}{dx}$

$-\frac{dz}{dx}-ztanx=-secx$ ___ (4) $=\frac{dz}{dx}+ztanx=secx$

$I.F=e^{+\int tanx}=e^{+logsecx}=secx$

$z\times I.F=\int secx\ast I.Fdx$

$zsecx=\int se{c}^{2}xdx$

$\frac{1}{y}secx=tanx+c$

$secx=y(tanx+c)$