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(D) The given differential equation is a linear differential equation of the form $y' + Py = Q$,where $P = 1$ and $Q = 2(\sin x + \cos x)$.The integra

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$y(\pi) = e^{-\pi}$

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(D) The given differential equation is a linear differential equation of the form $y' + Py = Q$,where $P = 1$ and $Q = 2(\sin x + \cos x)$.The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int 1 dx} = e^x$.The general solution is given by $y \cdot IF = \int Q \cdot IF dx + C$.$y \cdot e^x = \int 2(\sin x + \cos x) e^x dx + C$.Using the identity $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$,where $f(x) = \sin x$ and $f'(x) = \cos x$,we get:$y \cdot e^x = 2 e^x \sin x + C$.Dividing by $e^x$,we get $y = 2 \sin x + C e^{-x}$.Given $y(0) = 1$,we substitute $x = 0$ and $y = 1$:$1 = 2 \sin(0) + C e^0 \Rightarrow 1 = 0 + C \Rightarrow C = 1$.Thus,the particular solution is $y = 2 \sin x + e^{-x}$.Now,evaluating at $x = \pi$:$y(\pi) = 2 \sin(\pi) + e^{-\pi} = 2(0) + e^{-\pi} = e^{-\pi}$.

If $y(x)$ satisfies the differential equation $y' + y = 2(\sin x + \cos x)$ and $y(0) = 1$,then

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