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

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$y = \frac{1}{16} x^2 (4 \log x - 1) + Cx^{-2}$

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(A) The given differential equation is: $x \frac{dy}{dx} + 2y = x^2 \log x$. Dividing by $x$,we get: $\frac{dy}{dx} + \frac{2}{x} y = x \log x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x \log x$. The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \log x} = e^{\log x^2} = x^2$. The general solution is $y(IF) = \int (Q 	imes IF) dx + C$. Substituting the values: $y \cdot x^2 = \int (x \log x \cdot x^2) dx + C = \int x^3 \log x dx + C$. Using integration by parts: $\int x^3 \log x dx = \log x \cdot \frac{x^4}{4} - \int (\frac{1}{x} \cdot \frac{x^4}{4}) dx = \frac{x^4 \log x}{4} - \frac{1}{4} \int x^3 dx = \frac{x^4 \log x}{4} - \frac{x^4}{16} + C$. Thus,$x^2 y = \frac{x^4}{16} (4 \log x - 1) + C$. Dividing by $x^2$,we get the general solution: $y = \frac{1}{16} x^2 (4 \log x - 1) + Cx^{-2}$.

List $I$ (Differential Equation)	List $II$ (Integrating Factor)
$(P)$ $(x^3+1)\frac{dy}{dx}+x^2y=3x^2$	$(1)$ $x^3$
$(Q)$ $x^2\frac{dy}{dx}+3xy=x^6$	$(2)$ $(x^3+1)^2$
$(R)$ $(x^3+1)^2\frac{dy}{dx}+6x^2(x^3+1)y=x^2$	$(3)$ $(x^2+1)^2$
$(S)$ $(x^2+1)\frac{dy}{dx}+4xy=\ln x$	$(4)$ $x^2+1$
	$(5)$ $(x^3+1)^{1/3}$
	$(6)$ $(x^3+1)^{1/2}$

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

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