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 $\frac{{{x^2} + {y^2}}}{{2xy}}$.

It is given that, the slope of tangent to the curve at point $(x,y)$ is $\frac{{{x^2} + {y^2}}}{{2xy}}$.

$\therefore {\left( {\frac{{dy}}{{dx}}} \right)_{(x,y)}} = \frac{{{x^2} + {y^2}}}{{2xy}}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{x}{y} + \frac{y}{x}} \right)$

……(i)

which is homogeneous differential equation.

Put $\quad y = vx$
$\Rightarrow$ $\quad \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

On substituting these values in Eq. (i),

we get
$v + x\frac{{dv}}{{dx}} = \frac{1}{2}\left( {\frac{1}{v} + v} \right)$

$\Rightarrow$ $v + x\frac{{dv}}{{dx}} = \frac{1}{2}\left( {\frac{{1 + {v^2}}}{v}} \right)$

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

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

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

$\Rightarrow$ $\frac{{2v}}{{1 - {v^2}}}dv = \frac{{dx}}{x}$ On integrating both sides,

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

Put $1 - {v^2} = t$ in LHS,

we get$- 2vdv = dt$
$\Rightarrow$ $\quad - \int {\frac{{dt}}{t}} = \int {\frac{{dx}}{x}}$

$\Rightarrow$ $- \log t = \log x + \log C$

$\Rightarrow$ $- \log \left( {1 - {v^2}} \right) = \log x + \log C$

$\Rightarrow$ $- \log \left( {1 - \frac{{{y^2}}}{{{x^2}}}} \right) = \log x + \log C$

$\Rightarrow$ $- \log \left( {\frac{{{x^2} - {y^2}}}{{{x^2}}}} \right) = \log x + \log C$

$\Rightarrow$ $- \log \left( {\frac{{{x^2} - {y^2}}}{{{x^2}}}} \right) = \log x + \log C$

$\Rightarrow$ $\quad \log \left( {\frac{{{x^2}}}{{{x^2} - {y^2}}}} \right) = \log x + \log C$

$\Rightarrow$ $\frac{{{x^2}}}{{{x^2} - {y^2}}} = Cx$ ……(ii)

Since, the curve passes through the point (2,1)

$\therefore \frac{{{{(2)}^2}}}{{{{(2)}^2} - {{(1)}^2}}} = C(2) \Rightarrow C = \frac{2}{3}$

So, the required solution of the differential equation is $2\left( {{x^2} - {y^2}} \right) = 3x$.