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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 $Q = \cos x$.First,we fin

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$y = \frac{1}{2}(\cos x + \sin x) + ce^{-x}$

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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 $Q = \cos x$.First,we find the integrating factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x$.The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.Substituting the values,we get $y e^x = \int \cos x \cdot e^x dx + c$.To solve $\int e^x \cos x dx$,we use the integration by parts formula $\int u v dx = u \int v dx - \int (u' \int v dx) dx$:Let $I = \int e^x \cos x dx$.$I = e^x \cos x - \int e^x (-\sin x) dx = e^x \cos x + \int e^x \sin x dx$.Using integration by parts again for $\int e^x \sin x dx$:$I = e^x \cos x + (e^x \sin x - \int e^x \cos x dx) = e^x \cos x + e^x \sin x - I$.$2I = e^x(\cos x + \sin x) \implies I = \frac{1}{2} e^x(\cos x + \sin x)$.Thus,$y e^x = \frac{1}{2} e^x(\cos x + \sin x) + c$.Dividing by $e^x$,we get $y = \frac{1}{2}(\cos x + \sin x) + ce^{-x}$.

The solution of the differential equation $\frac{dy}{dx} + y = \cos x$ is

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