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 $$y = f\left( x \right),\,y = \cos \,x$$ and the $$y$$-axis is :
A.
$$\sqrt 2 - 1$$
B.
$$\sqrt 2 $$
C.
$$1$$
D.
$$\frac{1}{{\sqrt 2 }}$$
Answer :
$$\sqrt 2 - 1$$
Solution :
$$\eqalign{ & {\text{I}}{\text{.F}}{\text{.}} = {e^{ - x}} \cr & \therefore \,y{e^{ - x}} = \int {{e^{ - x}}} \left( {\cos \,x - \sin \,x} \right)dx \cr & {\text{Put }} - x = t \cr} $$
$$\eqalign{ & = - \int {{e^t}\left( {\cos \,t + \sin \,t} \right)dt} \cr & = - {e^t}\sin \,t + c \cr & y{e^{ - x}} = {e^{ - x}}\sin \,x + c \cr} $$
Since, $$y$$ is bounded when $$x \to \infty \Rightarrow c = 0$$
$$\eqalign{ & \therefore \,y = \sin \,x \cr & {\text{Area}} = \int_0^{\frac{\pi }{4}} {\left( {\cos \,x - \sin \,x} \right)dx = \sqrt 2 - 1} \cr} $$
$$\eqalign{ & {\text{I}}{\text{.F}}{\text{.}} = {e^{ - x}} \cr & \therefore \,y{e^{ - x}} = \int {{e^{ - x}}} \left( {\cos \,x - \sin \,x} \right)dx \cr & {\text{Put }} - x = t \cr} $$
$$\eqalign{ & = - \int {{e^t}\left( {\cos \,t + \sin \,t} \right)dt} \cr & = - {e^t}\sin \,t + c \cr & y{e^{ - x}} = {e^{ - x}}\sin \,x + c \cr} $$
Since, $$y$$ is bounded when $$x \to \infty \Rightarrow c = 0$$
$$\eqalign{ & \therefore \,y = \sin \,x \cr & {\text{Area}} = \int_0^{\frac{\pi }{4}} {\left( {\cos \,x - \sin \,x} \right)dx = \sqrt 2 - 1} \cr} $$