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Question

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

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$y \log x = (\log x)^2 + c$

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(C) Given differential equation is $x \log x \frac{dy}{dx} + y = 2 \log x$.Dividing both sides by $x \log x$,we get:$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x}$.Integrating factor $(I.F.) = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx}$.Let $\log x = t$,then $\frac{1}{x} dx = dt$.So,$I.F. = e^{\int \frac{1}{t} dt} = e^{\log t} = t = \log x$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.$y \log x = \int \frac{2}{x} \cdot \log x dx$.Let $\log x = u$,then $\frac{1}{x} dx = du$.$y \log x = \int 2u du = u^2 + c$.Substituting $u = \log x$,we get $y \log x = (\log x)^2 + c$.

The solution of the differential equation $x \log x \frac{dy}{dx} + y = 2 \log x$ is

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