(x^2 + y^2)dy = xy dx. If y(x_0) = e and y(1) | vedclass

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(A) Given the differential equation $(x^2 + y^2)dy = xy dx$.Rearranging the terms,we get $x^2 dy + y^2 dy = xy dx$.$x^2 dy - xy dx = -y^2 dy$.Dividing

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$\sqrt{3}e$

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(A) Given the differential equation $(x^2 + y^2)dy = xy dx$.Rearranging the terms,we get $x^2 dy + y^2 dy = xy dx$.$x^2 dy - xy dx = -y^2 dy$.Dividing by $xy^2$,we get $\frac{x dy - y dx}{y^2} = -\frac{dy}{x}$.This is equivalent to $d(\frac{x}{y}) = -\frac{dy}{x}$ (Note: The standard form is $x dy - y dx = x^2 d(\frac{y}{x})$ or $y dx - x dy = y^2 d(\frac{x}{y})$).Let us rewrite: $x^2 dy = xy dx - y^2 dy = y(x dx - y dy)$ is not correct. Let's use substitution $y = vx$,$dy = v dx + x dv$.$(x^2 + v^2 x^2)(v dx + x dv) = x(vx) dx$.$x^2(1 + v^2)(v dx + x dv) = vx^2 dx$.$(1 + v^2)(v dx + x dv) = v dx$.$v dx + x dv + v^3 dx + v^2 x dv = v dx$.$x dv(1 + v^2) = -v^3 dx$.$\frac{1 + v^2}{v^3} dv = -\frac{dx}{x}$.Integrating both sides: $\int (v^{-3} + v^{-1}) dv = -\int \frac{dx}{x}$.$-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.$-\frac{x^2}{2y^2} + \ln|\frac{y}{x}| = -\ln|x| + C$.$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C$.$-\frac{x^2}{2y^2} + \ln|y| = C$.Given $y(1) = 1$,so $-\frac{1^2}{2(1)^2} + \ln(1) = C \implies C = -\frac{1}{2}$.So,$-\frac{x^2}{2y^2} + \ln|y| = -\frac{1}{2}$.For $y(x_0) = e$,$-\frac{x_0^2}{2e^2} + \ln(e) = -\frac{1}{2}$.$-\frac{x_0^2}{2e^2} + 1 = -\frac{1}{2}$.$-\frac{x_0^2}{2e^2} = -\frac{3}{2}$.$x_0^2 = 3e^2 \implies x_0 = \sqrt{3}e$.

$(x^2 + y^2)dy = xy dx$. If $y(x_0) = e$ and $y(1) = 1$,then the value of $x_0$ is:

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