Question

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

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

Answer 1

The correct option is C $e-2$

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