Question

The solution of $\frac{dy}{dx} + y \, \tan \, x = \sec \, x, y (0) = 0$ is

• A. y sec x = tan x
• B. y tan x = sec x
• C. tan x = y tan x
• D. x sec x = tan y

Answer 1

Answer: (A)

y sec x = tan x

Answer 2

We have,
$\frac{d y}{d x}+y \tan x=\sec x$
which is a linear differential equation.
$\therefore \text { I. } F=e^{\int \tan \,x \,d x}=e^{\log \sec x}=\sec x$
$\therefore$ The solution is given by
$y \cdot \sec x=\int \sec\, x \cdot \sec\, x \,d x+C$
$y\,\sec x=\tan\, x+C$...(i)
Now, $y=0$, when $x=0$
$\therefore 0=0+c\,\,\,$ [From E(i)]
$\Rightarrow c=0$
Putting $c=0$ in E (i), we get
$y\, \sec\, x=\tan\, x$