Question

A function $y=f(x)$ satisfies the differential equation $\frac{d y}{d x}-y=\cos x-\sin x$, with initial condition that $y$ is bounded when $x \rightarrow \infty$. 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 ye ^{- 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 \rightarrow \infty \Rightarrow c=0 $
$\therefore y =\sin x$
$\text { Area } =\int\limits_0^{\pi / 4}(\cos x-\sin x) d x=\sqrt{2}-1 \Rightarrow \text { (D) }$