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(B) Given the differential equation: $x dy = (\sqrt{x^{2}+y^{2}}+y) dx$.Rearranging the terms,we get: $x dy - y dx = \sqrt{x^{2}+y^{2}} dx$.Dividing b

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$\frac{3}{2}$

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(B) Given the differential equation: $x dy = (\sqrt{x^{2}+y^{2}}+y) dx$.Rearranging the terms,we get: $x dy - y dx = \sqrt{x^{2}+y^{2}} dx$.Dividing both sides by $x^{2}$ (for $x > 0$): $\frac{x dy - y dx}{x^{2}} = \frac{\sqrt{x^{2}+y^{2}}}{x^{2}} dx$.This simplifies to: $d(\frac{y}{x}) = \sqrt{1 + (\frac{y}{x})^{2}} \cdot \frac{dx}{x}$.Integrating both sides: $\int \frac{d(y/x)}{\sqrt{1 + (y/x)^{2}}} = \int \frac{dx}{x}$.Using the standard integral $\int \frac{dt}{\sqrt{1+t^{2}}} = \ln(t + \sqrt{1+t^{2}})$,we get: $\ln(\frac{y}{x} + \sqrt{1 + (\frac{y}{x})^{2}}) = \ln x + C$.This can be written as: $\frac{y + \sqrt{x^{2}+y^{2}}}{x} = kx$,where $k = e^{C}$.So,$y + \sqrt{x^{2}+y^{2}} = kx^{2}$.Given the curve passes through $(1, 0)$,we substitute $x=1, y=0$: $0 + \sqrt{1^{2}+0^{2}} = k(1)^{2} \Rightarrow k = 1$.The equation of the curve is $y + \sqrt{x^{2}+y^{2}} = x^{2}$.For $x = 2, y = \alpha$: $\alpha + \sqrt{4+\alpha^{2}} = 2^{2} = 4$.$\sqrt{4+\alpha^{2}} = 4 - \alpha$.Squaring both sides: $4 + \alpha^{2} = 16 - 8\alpha + \alpha^{2}$.$8\alpha = 12 \Rightarrow \alpha = \frac{12}{8} = \frac{3}{2}$.

Let the solution curve of the differential equation $x dy = (\sqrt{x^{2}+y^{2}}+y) dx$,$x > 0$,intersect the line $x = 1$ at $y = 0$ and the line $x = 2$ at $y = \alpha$. Then the value of $\alpha$ is.

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