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(A) Given the differential equation: $\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] x \frac{dy}{dx} = x + \left[\frac{x}{\sqrt{x^{2}-y^{2}

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$\frac{1}{2} \exp \left(\frac{\pi}{6}+\sqrt{e}-1\right)$

Vedclass · Answer

(A) Given the differential equation: $\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] x \frac{dy}{dx} = x + \left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] y$.Rearranging the terms,we get: $e^{\frac{y}{x}}(x dy - y dx) + \frac{x}{\sqrt{x^{2}-y^{2}}}(x dy - y dx) = x dx$.Dividing both sides by $x^{2}$,we obtain: $e^{\frac{y}{x}}\left(\frac{x dy - y dx}{x^{2}}\right) + \frac{1}{\sqrt{1-(\frac{y}{x})^{2}}}\left(\frac{x dy - y dx}{x^{2}}\right) = \frac{dx}{x}$.Recognizing the differential form $d(\frac{y}{x}) = \frac{x dy - y dx}{x^{2}}$,the equation becomes: $e^{\frac{y}{x}} d(\frac{y}{x}) + \frac{1}{\sqrt{1-(\frac{y}{x})^{2}}} d(\frac{y}{x}) = \frac{dx}{x}$.Integrating both sides: $\int e^{\frac{y}{x}} d(\frac{y}{x}) + \int \frac{1}{\sqrt{1-(\frac{y}{x})^{2}}} d(\frac{y}{x}) = \int \frac{dx}{x}$.This yields: $e^{\frac{y}{x}} + \sin^{-1}(\frac{y}{x}) = \ln|x| + C$.Since the curve passes through $(1, 0)$,we substitute $x=1, y=0$: $e^{0} + \sin^{-1}(0) = \ln(1) + C \Rightarrow 1 + 0 = 0 + C \Rightarrow C = 1$.The equation is $e^{\frac{y}{x}} + \sin^{-1}(\frac{y}{x}) = \ln x + 1$.Since it passes through $(2\alpha, \alpha)$,we substitute $x=2\alpha, y=\alpha$: $e^{\frac{\alpha}{2\alpha}} + \sin^{-1}(\frac{\alpha}{2\alpha}) = \ln(2\alpha) + 1$.$e^{1/2} + \sin^{-1}(1/2) = \ln(2\alpha) + 1 \Rightarrow \sqrt{e} + \frac{\pi}{6} = \ln(2\alpha) + 1$.$\ln(2\alpha) = \sqrt{e} + \frac{\pi}{6} - 1$.$2\alpha = \exp(\sqrt{e} + \frac{\pi}{6} - 1) \Rightarrow \alpha = \frac{1}{2} \exp(\frac{\pi}{6} + \sqrt{e} - 1)$.

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