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

Answer 1

Answer: (C)

We have
$\frac{dy}{dx}=y\tan x-y^2\sec x$
$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\tan x=-\sec x$
Putting $\frac{1}{y}=V\Rightarrow \frac{-1}{y^2}\frac{dy}{dx}=\frac{dV}{dx}$
we obtain $\frac{dV}{dx}+\tan x\cdot V=\sec x$, which is linear.
I.F. $=e^{\int \tan xdx}=e^{\log \sec x}=\sec x$
Hence, the solution is
$V\sec x-\int {\sec }^{2}xdx+c$ or $\frac{1}{y}\sec x-\tan x+c$
or $\sec x=y(c+\tan x)$