If frac{dy}{dx} = frac{xy}{x^2 + y^2} and y(1 | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting

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$\sqrt{3} e$

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(D) Given the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + v^2 x^2} = \frac{v}{1 + v^2}$.$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v - v^3}{1 + v^2} = -\frac{v^3}{1 + v^2}$.Separating variables: $\int \frac{1 + v^2}{v^3} dv = -\int \frac{dx}{x}$.$\int (v^{-3} + v^{-1}) dv = -\int \frac{dx}{x}$.Integrating both sides: $-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.Substitute $v = \frac{y}{x}$: $-\frac{x^2}{2y^2} + \ln|\frac{y}{x}| = -\ln|x| + C$.$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C \implies -\frac{x^2}{2y^2} + \ln|y| = C$.Using $y(1) = 1$: $-\frac{1^2}{2(1)^2} + \ln(1) = C \implies C = -\frac{1}{2}$.The equation is $-\frac{x^2}{2y^2} + \ln|y| = -\frac{1}{2}$.For $y = e$: $-\frac{x^2}{2e^2} + \ln(e) = -\frac{1}{2} \implies -\frac{x^2}{2e^2} + 1 = -\frac{1}{2}$.$-\frac{x^2}{2e^2} = -\frac{3}{2} \implies x^2 = 3e^2 \implies x = \sqrt{3}e$.

If $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$ and $y(1) = 1$,then find the value of $x$ that satisfies $y(x) = e$.

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