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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 $Q=\cos x$.The integrating fac

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

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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 $Q=\cos x$.The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.Multiplying both sides of the equation by $I.F.$,we get:$e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x} \cos x$$\frac{d}{dx}(y e^{-x}) = e^{-x} \cos x$Integrating both sides with respect to $x$:$y e^{-x} = \int e^{-x} \cos x dx + C$  --- $(1)$Let $I = \int e^{-x} \cos x dx$. Using integration by parts:$I = \cos x (-e^{-x}) - \int (-\sin x)(-e^{-x}) dx$$I = -e^{-x} \cos x - \int e^{-x} \sin x dx$$I = -e^{-x} \cos x - [\sin x (-e^{-x}) - \int \cos x (-e^{-x}) dx]$$I = -e^{-x} \cos x + e^{-x} \sin x - I$$2I = e^{-x} (\sin x - \cos x)$$I = \frac{e^{-x}(\sin x - \cos x)}{2}$Substituting $I$ into equation $(1)$:$y e^{-x} = \frac{e^{-x}(\sin x - \cos x)}{2} + C$Multiplying by $e^{x}$:$y = \frac{\sin x - \cos x}{2} + C e^{x}$.

Find the general solution of the differential equation $\frac{dy}{dx}-y=\cos x$.

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