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(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = \cos x - \sin x$.In

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$\sqrt{2} - 1$

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(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = \cos x - \sin x$.Integrating Factor $(I.F.)$ $= e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.The general solution is $y(I.F.) = \int Q(I.F.) dx + C$.$y e^{-x} = \int e^{-x}(\cos x - \sin x) dx + C$.Let $I = \int e^{-x}(\cos x - \sin x) dx$. Using the formula $\int e^{ax}(f(x) + f'(x)) dx = e^{ax}f(x) + C$,where $f(x) = \cos x$ and $f'(x) = -\sin x$,we get:$I = e^{-x} \cos x + C$.Thus,$y e^{-x} = e^{-x} \cos x + C$,which simplifies to $y = \cos x + C e^x$.Since $y$ is bounded as $x ightarrow \infty$,the coefficient of $e^x$ must be zero,so $C = 0$.Therefore,$y = f(x) = \cos x$.Wait,re-evaluating the integral: $\int e^{-x}(\cos x - \sin x) dx$. Let $u = \cos x$,$du = -\sin x dx$. This is not quite right. Let's use integration by parts: $\int e^{-x} \cos x dx - \int e^{-x} \sin x dx$. $int e^{-x} \cos x dx = e^{-x} \sin x - \int -e^{-x} \sin x dx = e^{-x} \sin x + \int e^{-x} \sin x dx$.So,$\int e^{-x}(\cos x - \sin x) dx = e^{-x} \sin x + C$.Then $y e^{-x} = e^{-x} \sin x + C \Rightarrow y = \sin x + C e^x$.For $y$ to be bounded as $x ightarrow \infty$,$C = 0$,so $y = \sin x$.The area enclosed by $y = \sin x$,$y = \cos x$,and the $y$-axis $(x=0)$ is found by finding the intersection point: $\sin x = \cos x \Rightarrow 	an x = 1 \Rightarrow x = \frac{\pi}{4}$.Area $= \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx = [\sin x + \cos x]_{0}^{\frac{\pi}{4}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.

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

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