The slope of the tangent at $(x, y)$ to the curve passing through $(2,1)$ is $\frac{x^2+y^2}{2 x y}$, then the equation of the curve is

Given the homogeneous differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$

Substitute $y = vx$ so that $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

Substituting into the equation yields:

$v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2vx^2}$

Simplifying the right-hand side gives:

$v + x\frac{dv}{dx} = \frac{1 + v^2}{2v}$

Separating variables:

$x\frac{dv}{dx} = \frac{1 - v^2}{2v}$

$\frac{2v}{1 - v^2} dv = \frac{1}{x} dx$

Integrate both sides:

$\int \frac{2v}{1 - v^2} dv = \int \frac{1}{x} dx$

Let $u = 1 - v^2$, so $du = -2v dv$:

$-\int \frac{1}{u} du = \int \frac{1}{x} dx$
$-\ln|1 - v^2| = \ln|x| + C_1$

Exponentiating both sides and setting $C_1 = \ln|C|$:

$\frac{1}{|1 - v^2|} = C|x|$

Substitute $v = \frac{y}{x}$ and simplify, assuming $x > 0$:

$\frac{x}{x^2 - y^2} = C$
$x^2 - y^2 = \frac{x}{C}$

Let $K = \frac{1}{C}$, so $x^2 - y^2 = Kx$.

Using the point $(2, 1)$:

$2^2 - 1^2 = K \cdot 2$
$3 = 2K$
$K = \frac{3}{2}$

Curve equation: $x^2 - y^2 = \frac{3}{2}x$
Multiply through by 2: $2x^2 - 2y^2 = 3x$

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