For the differential equation (x^2 + y^2) dy | vedclass

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(B) The given differential equation is $(x^2 + y^2) dy = xy dx$, which can be written as $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.This is a homogeneous

Vedclass · Accepted Answer

$\pm \sqrt{3} e$

Vedclass · Answer

(B) The given differential equation is $(x^2 + y^2) dy = xy dx$, which can be written as $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.This is a homogeneous differential equation. Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2(1 + v^2)} = \frac{v}{1 + v^2}$.$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v - v^3}{1 + v^2} = -\frac{v^3}{1 + v^2}$.Separating variables: $\int \frac{1 + v^2}{v^3} dv = -\int \frac{1}{x} dx$.$\int (v^{-3} + v^{-1}) dv = -\ln|x| + C$.$-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C \implies -\frac{x^2}{2y^2} + \ln|y/x| = -\ln|x| + C$.$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C \implies \ln|y| - \frac{x^2}{2y^2} = C$.Given $y(1) = 1$, so $\ln(1) - \frac{1^2}{2(1)^2} = C \implies C = -1/2$.The equation is $\ln|y| - \frac{x^2}{2y^2} = -1/2$.Given $y(x_0) = e$, so $\ln(e) - \frac{x_0^2}{2e^2} = -1/2$.$1 - \frac{x_0^2}{2e^2} = -1/2 \implies \frac{x_0^2}{2e^2} = 3/2 \implies x_0^2 = 3e^2 \implies x_0 = \pm \sqrt{3}e$.

For the differential equation $(x^2 + y^2) dy = xy dx$, it is given that $y(1) = 1$ and $y(x_0) = e$, then the value of $x_0$ is

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