Let y(x) be the solution of the differential | vedclass

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(D) The given differential equation is $(x \log x) \frac{dy}{dx} + y = 2x \log x$.Dividing by $(x \log x)$,we get $\frac{dy}{dx} + \frac{1}{x \log x}

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$2$

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(D) The given differential equation is $(x \log x) \frac{dy}{dx} + y = 2x \log x$.Dividing by $(x \log x)$,we get $\frac{dy}{dx} + \frac{1}{x \log x} y = 2$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x \log x}$ and $Q(x) = 2$.The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$.The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.$y \cdot \log x = \int 2 \log x dx = 2(x \log x - x) + C$.Given $y(1) = 0$,substituting $x = 1$ gives $0 \cdot \log(1) = 2(1 \log 1 - 1) + C \Rightarrow 0 = 2(0 - 1) + C \Rightarrow C = 2$.Thus,$y \log x = 2x \log x - 2x + 2$.At $x = e$,$y \log e = 2e \log e - 2e + 2$.Since $\log e = 1$,we have $y(1) = 2e - 2e + 2$,which simplifies to $y = 2$.

Let $y(x)$ be the solution of the differential equation $(x \log x) \frac{dy}{dx} + y = 2x \log x$,$(x \ge 1)$. Then $y(e)$ is equal to: $[y(1) = 0]$

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