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Question

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

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$y \log x = -\frac{2}{x}(1 + \log x) + C$

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(A) The given differential equation is: $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$. Dividing both sides by $x \log x$,we get: $\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x^2}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x^2}$. The Integrating Factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$. The general solution is $y(I.F.) = \int (Q 	imes I.F.) dx + C$. Substituting the values: $y \log x = \int \left( \frac{2}{x^2} \log x ight) dx + C$. Using integration by parts for $\int \frac{2 \log x}{x^2} dx$: Let $u = \log x$ and $dv = \frac{2}{x^2} dx$. Then $du = \frac{1}{x} dx$ and $v = -\frac{2}{x}$. $\int u dv = uv - \int v du = (\log x)(-\frac{2}{x}) - \int (-\frac{2}{x}) \cdot \frac{1}{x} dx$. $= -\frac{2 \log x}{x} + 2 \int x^{-2} dx = -\frac{2 \log x}{x} + 2(-\frac{1}{x}) = -\frac{2}{x}(\log x + 1)$. Thus,the general solution is $y \log x = -\frac{2}{x}(1 + \log x) + C$.

Find the general solution of the differential equation: $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$.

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