Question

If $y = y (x)$ is the solution of the differential equation $\frac{dy}{dx} = ( \tan x -y) \sec^2 x, x \in \left( - \frac{\pi}{2} , \frac{\pi}{2} \right)$, such that $y(0) = 0$, then $y \left( - \frac{\pi}{4} \right)$ is equal to :

• A. $ 2 + \frac{1}{e} $
• B. $ \frac{1}{2} - e $
• C. $e - 2$
• D. $ \frac{1}{2} + e $

Answer 1

Answer: (C)

$\frac{dy}{dx} =\left(\tan x-y\right)\sec^{2}x $
Now, put $ t \Rightarrow \frac{dt}{dx} =\sec^{2}x$
So $ \frac{dy}{dt}+y=t $
On solving, we get $ye^t = e^t (t -1) + c$
$ \Rightarrow y =\left(\tan x-1\right)+ce^{-\tan x} $
$ \Rightarrow y\left(0\right) =0 \Rightarrow c=1 $
$ \Rightarrow y =\tan x - 1+e^{-\tan x} $
So $y\left(- \frac{\pi}{4}\right) =e-2 $