# Find the equation of a curve passing through (2,1) if the slope of the tangent to the curve at any point (x,y) is $\Large \frac{x^2+y^2}{2xy}.$

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