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 $y{e}^t=e^t(t-1)+c$
$\Rightarrow y=\left(\tan x-1\right)+c{e}^{-\tan x}$
$\Rightarrow y\left(0\right)=0\Rightarrow c=1$
$\Rightarrow y=\tan x-1+e^{-\tan x}$
So $y\left(-\frac{\pi }{4}\right)=e-2$