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Q.
The curve that satisfies the differential equation $y' = \frac{x^{2}+y^{2}}{2xy}$ and passes through (2, 1) is a hyperbola with eccentricity :
Differential Equations
Solution:
Putting $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$, the given equation reduces to the form
$v + x \frac{dv}{dx} = \frac{x^{2}+v^{2}x^{2}}{2x\left(vx\right)} = \frac{1+v^{2}}{2v} \Rightarrow x \frac{dv}{dx} = \frac{1+v^{2}}{2v}- v = \frac{1-v^{2}}{2v}$
$\Rightarrow \frac{2v dv}{1-v^{2}} = \frac{dx}{x} \Rightarrow -log \left(1-v^{2}\right) = log \,x - a$
$\Rightarrow log\left(1-v^{2}\right) x = a \Rightarrow x\left(1-v^{2}\right) = c$, where $c = e^{a}$
When $x = 2,\, v = \frac{1}{2}$, [when $x = 2\, y =1,\, v = \frac{1}{2}$]
so that $c = \frac{3}{2}$
So the equation of the curve is $x^{2} - y^{2} = \frac{3x}{2}$
$\Rightarrow \left(x-\frac{3}{4}\right)^{2}-y^{2} = \frac{9}{16}$
which is a rectangular hyperbola with eccentricity $\sqrt{2}$.
[Eccentricity of a rectangular hyperbola $= \sqrt{2}$ ]