Question

A function $y=f(x)$ satisfies the differential equation $\frac{dy}{dx}-y=\cos x-\sin x$, with initial condition that $y$ is bounded when $x\to \infty$. The area enclosed by $y=f(x),y=\cos x$ and the $y$-axis in the $1^{\text{st }}$ quadrant

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

Answer 1

Answer: (A)

$\text{ I.F. }=e^{-x}$
$\therefore y{e}^{-x}=\int e^{-x}(\cos x-\sin x)dx\text{ put }-x=t$

$=-\int e^t(\cos t+\sin t)dt$
$=-e^t\sin t+c$
$y^{-x}=e^{-x}\sin x+c$
$\text{ since }y\text{ is bounded when }x\to \infty \Rightarrow c=0$
$\therefore y=\sin x$
$\text{ Area }=\underset{0}{\overset{\pi /4}{\int }}(\cos x-\sin x)dx=\sqrt{2}-1\Rightarrow \text{ (D) }$