If y+frac{d}{d x}(x y)=x(sin x+log x),then fi | vedclass

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(C) Given equation is $y + \frac{d}{dx}(xy) = x(\sin x + \log x)$.Expanding the derivative: $y + y + x \frac{dy}{dx} = x \sin x + x \log x$.$2y + x \f

Vedclass · Accepted Answer

$y=-\cos x+\frac{2}{x} \sin x+\frac{2}{x^2} \cos x+\frac{x}{3} \log x-\frac{x}{9}+\frac{c}{x^2}$,where $c$ is the constant of integration.

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(C) Given equation is $y + \frac{d}{dx}(xy) = x(\sin x + \log x)$.Expanding the derivative: $y + y + x \frac{dy}{dx} = x \sin x + x \log x$.$2y + x \frac{dy}{dx} = x \sin x + x \log x$.Dividing by $x$: $\frac{dy}{dx} + \frac{2}{x} y = \sin x + \log x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2}{x}$ and $Q(x) = \sin x + \log x$.Integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \log x} = x^2$.The solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$.$y \cdot x^2 = \int x^2(\sin x + \log x) dx + c$.$y x^2 = \int x^2 \sin x dx + \int x^2 \log x dx + c$.Using integration by parts for $\int x^2 \sin x dx$: $-x^2 \cos x + 2x \sin x + 2 \cos x$.Using integration by parts for $\int x^2 \log x dx$: $\frac{x^3}{3} \log x - \frac{x^3}{9}$.So,$y x^2 = -x^2 \cos x + 2x \sin x + 2 \cos x + \frac{x^3}{3} \log x - \frac{x^3}{9} + c$.Dividing by $x^2$: $y = -\cos x + \frac{2 \sin x}{x} + \frac{2 \cos x}{x^2} + \frac{x}{3} \log x - \frac{x}{9} + \frac{c}{x^2}$.This matches option $C$.

If $y+\frac{d}{d x}(x y)=x(\sin x+\log x)$,then find $y$.

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