Question

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

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

Answer 1

The correct option is B $e-2$

$\frac{dy}{dx}+y\left({\sec }^{2}x\right)=\tan x{\sec }^{2}x$
$\Rightarrow I.F.=e^{\int {\sec }^{2}x}=e^{\tan x}$

$\Rightarrow y{e}^{\tan x}=\int e^{\tan x}\tan x{\sec }^{2}xdx$
$I=\int e^{\tan x}\tan x{\sec }^{2}xdx$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\Rightarrow I=\int t{e}^{t}dt$
$=t{e}^t-e^t+c=e^t(t-1)+c=e^{\tan x}(\tan x-1)+c$

$\Rightarrow y{e}^{\tan x}=e^{\tan x}(\tan x-1)+c\because y\left(0\right)=0\Rightarrow c=1$

$\Rightarrow y{e}^{\tan x}=e^{\tan x}(\tan x-1)+1$
$\therefore y(-\frac{\pi }{4})=e-2$