Question

A function $y=f\left(x\right)$ 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 $y=f\left(x\right),y=cosx$ and the y-axis in the 1st quadrant is

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

Answer 1

Answer: (A)

$\sqrt{2}-1$

$\frac{dy}{dx}-y=\cos \left(x\right)-\sin \left(x\right)$
$IF=e^{\int -1.dx}$
$=e^{-x}$
Now
$e^{-x}.y=\int e^{-x}\cos \left(x\right)-e^{-x}\sin \left(x\right).dx$
$=e^{-x}\sin \left(x\right)+c$
Now $x\to \infty$ implies $c=0$.
Hence
$y=\sin \left(x\right)$
Therefore
${\int }_0^{\frac{\pi }{4}}\cos \left(x\right)-\sin \left(x\right).dx$
$=[sinx+cosx{]}_0^{\frac{\pi }{4}}$
$=2\left(\frac{1}{\sqrt{2}}\right)-1$
$=\sqrt{2}-1$ sq units.