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

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

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$\sqrt{3} e$

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(B) Given the differential equation: $(x^2+y^2) dy = xy dx$,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 this into the equation: $v + x \frac{dv}{dx} = \frac{v}{1+v^2}$.$x \frac{dv}{dx} = \frac{v}{1+v^2} - v = \frac{v - v - v^3}{1+v^2} = \frac{-v^3}{1+v^2}$.Separating variables: $\int \frac{1+v^2}{-v^3} dv = \int \frac{1}{x} dx$.$\int (-v^{-3} - v^{-1}) dv = \ln|x| + C$.$\frac{v^{-2}}{2} - \ln|v| = \ln|x| + C \Rightarrow \frac{1}{2v^2} = \ln|vx| + C = \ln|y| + C$.Since $y(1) = 1$,we have $\frac{1}{2(1)^2} = \ln(1) + C$,so $C = \frac{1}{2}$.Thus,$\frac{1}{2v^2} = \ln|y| + \frac{1}{2} \Rightarrow \frac{x^2}{2y^2} = \ln|y| + \frac{1}{2}$.Given $y(x_0) = e$,we substitute $x = x_0$ and $y = e$: $\frac{x_0^2}{2e^2} = \ln(e) + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}$.$x_0^2 = 3e^2 \Rightarrow x_0 = \sqrt{3}e$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

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

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