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Question

(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = \sin x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py

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$x(y + \cos x) = \sin x + c$

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(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = \sin x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = \sin x$. First,we find the integrating factor $(I.F.)$: $I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$. The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$. Substituting the values,we get $y \cdot x = \int x \sin x dx + c$. Using integration by parts for $\int x \sin x dx$: Let $u = x$ and $dv = \sin x dx$. Then $du = dx$ and $v = -\cos x$. $\int x \sin x dx = -x \cos x - \int (-\cos x) dx = -x \cos x + \sin x$. Thus,$xy = -x \cos x + \sin x + c$. Rearranging the terms,we get $xy + x \cos x = \sin x + c$,which simplifies to $x(y + \cos x) = \sin x + c$.

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

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