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

Vedclass · Accepted Answer

$x(y + \cos x) = \sin x + c$,where $c$ is a constant of integration.

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(B) The given differential equation 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 ($I$.$F$.): $I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$. Substituting the values: $yx = \int (\sin x \cdot x) dx + c$. Using integration by parts $(\int u v dx = u \int v dx - \int (u' \int v dx) dx)$: Let $u = x$ and $v = \sin x$. $yx = x(-\cos x) - \int (1 \cdot -\cos x) dx + c$. $yx = -x \cos x + \int \cos x dx + c$. $yx = -x \cos x + \sin x + c$. Rearranging the terms: $yx + x \cos x = \sin x + c$. $x(y + \cos x) = \sin x + c$.

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

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