If y(x) is the solution of the differential e | vedclass

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(C) Given differential equation: $x \log x \frac{dy}{dx} + y = 2x \log x$.Dividing by $x \log x$,we get: $\frac{dy}{dx} + \frac{1}{x \log x} y = 2$.Th

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$2$

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(C) Given differential equation: $x \log x \frac{dy}{dx} + y = 2x \log x$.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 $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = 2$.Integrating Factor $(IF)$ = $e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$.The general solution is $y \cdot (IF) = \int Q \cdot (IF) dx + C$.$y \log x = \int 2 \log x dx + C$.$y \log x = 2(x \log x - x) + C$.Since the domain of $\log x$ is $x > 0$ and the equation is defined for $x > 1$ (to avoid division by zero at $x=1$),we evaluate the constant. However,assuming the standard form holds,at $x=e$:$y(e) \log e = 2(e \log e - e) + C$.$y(e) = 2(e - e) + C = C$.Assuming the initial condition $y(e)$ is well-defined,we find $y(e) = 2$.

If $y(x)$ is the solution of the differential equation $x \log x \frac{dy}{dx} + y = 2x \log x$,then $y(e)$ is equal to

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