**Toolbox:**

- A linear differential equation of the form $\large\frac{dy}{dx}=$$f(x,y)$ is a homogenous differential equation. Such equation can be solved by substituting $y=vx$ and $\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
- $\int \large \frac{dx}{x+a}$$=\log |x+a|+c$
- Slope of the tangent to the curve is $\large\frac{dy}{dx}$

Step 1:

Given : The slope of the tangent to the curve is

$\large\frac{x^2+y^2}{2xy}$ we know slope of the tangent is $\large\frac{dy}{dx}$

Hence $\large\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$

Clearly this is a homologous differential equation

This can be solved by putting $y=vx$ and $\large\frac{dy}{dx} $$=v+x \large\frac{dv}{dx}$

$v+x \large\frac{dv}{dx}$$=\large\frac{x^2+v^2x^2}{2x(vx)}$$=\large\frac{x^2(1+v^2)}{2vx^2}$

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

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

$=\large\frac{1+v^2-2v^2}{2v}$

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

Step 2:

Now seperating the variables we get

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

Let us integrate on both sides,

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

Put $ 1-v^2=t$ on differentiating we get,

$-2vdv=dt=>2vdv=-dt$

Substituting this we get,

$-\int \large\frac{dt}{t}=\int \frac{dx}{x}$

Integrating ob both sides we get,

$-\log |t|=\log x+\log c$

$=>-\log |t|^{-1}=\log cx$

$t^{-1}=cx$

Substituing for t

$(1-v^2)^{-1}=cx$

$=> \large\frac{1}{1-v^2}$$=cx$

Substituting for $v=\large\frac{y}{x}$

$=> \large\frac{1}{1-\Large\frac{y^2}{x^2}}$$=cx$

=>$ \large\frac{x^2}{x^2-y^2}$$=cx$

or $x=c(x^2-y^2)$

Step 3:

It is given that the curve passes through the point (2,1)

Now substitute 2 for x and 1 for y

$2=c(4-1)$

$=>2=3c\qquad or \qquad c=2/3$

Substituting the value of c we get,

$x=\large\frac{2}{3} $$(x^2-y^2)$

or $3x=2(x^2-y^2)$ is the required solution