Let y=y(x) be the solution curve of the diffe | vedclass

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(A) The given differential equation is $\frac{dy}{dx} = \frac{y}{x} + y^3(1 + \log_e x)$.Rearranging,we get $\frac{dy}{dx} - \frac{1}{x}y = (1 + \log_

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$\frac{x^2}{5 - 2x^3(2 + \log_e x^3)}$

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(A) The given differential equation is $\frac{dy}{dx} = \frac{y}{x} + y^3(1 + \log_e x)$.Rearranging,we get $\frac{dy}{dx} - \frac{1}{x}y = (1 + \log_e x)y^3$.Divide by $y^3$: $y^{-3}\frac{dy}{dx} - \frac{1}{x}y^{-2} = 1 + \log_e x$.Let $t = y^{-2} = \frac{1}{y^2}$. Then $\frac{dt}{dx} = -2y^{-3}\frac{dy}{dx}$,so $y^{-3}\frac{dy}{dx} = -\frac{1}{2}\frac{dt}{dx}$.Substituting into the equation: $-\frac{1}{2}\frac{dt}{dx} - \frac{1}{x}t = 1 + \log_e x$,which simplifies to $\frac{dt}{dx} + \frac{2}{x}t = -2(1 + \log_e x)$.The integrating factor is $I.F. = e^{\int \frac{2}{x} dx} = e^{2\log_e x} = x^2$.The solution is $t \cdot x^2 = \int -2(1 + \log_e x)x^2 dx$.Using integration by parts,$\int x^2(1 + \log_e x) dx = \frac{x^3}{3}(1 + \log_e x) - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3}(1 + \log_e x) - \frac{x^3}{9}$.So,$\frac{x^2}{y^2} = -2[\frac{x^3}{3} + \frac{x^3}{3}\log_e x - \frac{x^3}{9}] + C = -2[\frac{2x^3}{9} + \frac{x^3}{3}\log_e x] + C = -\frac{4x^3}{9} - \frac{2x^3}{3}\log_e x + C$.Given $y(1) = 3$,$\frac{1}{9} = -\frac{4}{9} - 0 + C \Rightarrow C = \frac{5}{9}$.Thus,$\frac{x^2}{y^2} = \frac{5 - 4x^3 - 6x^3\log_e x}{9} = \frac{5 - 2x^3(2 + 3\log_e x)}{9} = \frac{5 - 2x^3(2 + \log_e x^3)}{9}$.Therefore,$\frac{y^2}{9} = \frac{x^2}{5 - 2x^3(2 + \log_e x^3)}$.

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \log_e x))$ for $x > 0$ and $y(1) = 3$. Then $\frac{y^2(x)}{9}$ is equal to:

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