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(A) Given the differential equation: $(1+x^2) dy = y(x-y) dx$.Dividing by $(1+x^2) dx$,we get: $\frac{dy}{dx} = \frac{xy - y^2}{1+x^2} = \frac{x}{1+x^

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$e^{3\beta^{-1}} = e(3+2\sqrt{2})$

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(A) Given the differential equation: $(1+x^2) dy = y(x-y) dx$.Dividing by $(1+x^2) dx$,we get: $\frac{dy}{dx} = \frac{xy - y^2}{1+x^2} = \frac{x}{1+x^2}y - \frac{1}{1+x^2}y^2$.This is a Bernoulli differential equation. Rearranging gives: $\frac{dy}{dx} - \frac{x}{1+x^2}y = -\frac{1}{1+x^2}y^2$.Divide by $y^2$: $y^{-2} \frac{dy}{dx} - \frac{x}{1+x^2}y^{-1} = -\frac{1}{1+x^2}$.Let $t = y^{-1}$,then $\frac{dt}{dx} = -y^{-2} \frac{dy}{dx}$.Substituting this into the equation: $-\frac{dt}{dx} - \frac{x}{1+x^2}t = -\frac{1}{1+x^2} \implies \frac{dt}{dx} + \frac{x}{1+x^2}t = \frac{1}{1+x^2}$.This is a linear differential equation. The integrating factor $I.F. = e^{\int \frac{x}{1+x^2} dx} = e^{\frac{1}{2} \ln(1+x^2)} = \sqrt{1+x^2}$.The solution is $t \cdot \sqrt{1+x^2} = \int \frac{1}{1+x^2} \cdot \sqrt{1+x^2} dx = \int \frac{1}{\sqrt{1+x^2}} dx = \ln(x + \sqrt{1+x^2}) + C$.Since $t = \frac{1}{y}$,we have $\frac{\sqrt{1+x^2}}{y} = \ln(x + \sqrt{1+x^2}) + C$.Given $y(0)=1$,we have $\frac{\sqrt{1}}{1} = \ln(0+1) + C \implies 1 = 0 + C \implies C=1$.So,$\frac{\sqrt{1+x^2}}{y} = \ln(x + \sqrt{1+x^2}) + 1 = \ln(x + \sqrt{1+x^2}) + \ln e = \ln(e(x + \sqrt{1+x^2}))$.For $x = 2\sqrt{2}$,$y = \beta$: $\frac{\sqrt{1+(2\sqrt{2})^2}}{\beta} = \ln(e(2\sqrt{2} + \sqrt{1+8})) = \ln(e(2\sqrt{2} + 3))$.$\frac{3}{\beta} = \ln(e(3+2\sqrt{2})) \implies e^{3\beta^{-1}} = e(3+2\sqrt{2})$.

Let $y=y(x), y>0$,be a solution curve of the differential equation $(1+x^2) dy = y(x-y) dx$. If $y(0)=1$ and $y(2\sqrt{2})=\beta$,then

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