Question

The general solution of the differential equation $\frac{dy}{dx}=y\tan x-y^2\sec x$ is

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

Answer 1

Answer: (C)

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

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

Put $-\frac{1}{y}=v\Rightarrow \frac{1}{y^2}dy=dv$

$\therefore \frac{dv}{dx}+v\tan x=\sec x$ ...(1)

Here $P=\tan x\Rightarrow \int \tan xdx=\log \sec x$

$\therefore e^{\log \sec x}=\sec x$

Multiplying (1) by $I.F.$ we get

$\sec x\frac{dv}{dx}+v\sec x={\sec }^{2}x$

Integrating both sides w.r.t $x$ we get

$v\sec x=\int {\sec }^{2}xdx\Rightarrow \frac{1}{y}\sec x=\tan x+c$

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