Question

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

Answer 1

A\Q $\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$

This is a homogeneous differential equation in $2$ degree.

$y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}v+\frac{xdv}{dx}=\frac{x^2+v^2{x}^2}{2x\times vx}=\frac{x^2(1+v^2)}{2x^{2}v}=\frac{1+v^2}{2v}\frac{xdv}{dx}=\frac{v^2+1}{2v}-1=\frac{v^2+1-2v^2}{2v}=\frac{-v^2+1}{2v}\Rightarrow \int \frac{2v}{-v^2+1}dv=\int \frac{dx}{x}+C$

let ${-v}^2+1=t-2vdv=dt\Rightarrow dv=\frac{-dt}{2v}$

$\Rightarrow -\int \frac{dt}{2v}\times \frac{2v}{({-v}^2+1)}=\int \frac{dx}{x}+C\Rightarrow -\int \frac{dt}{t}=\int \frac{dx}{x}+C\Rightarrow -\ln t=\ln x+C\Rightarrow C=\ln t+\ln x=\ln \left(xt\right)\ln \left(x\right(1-v^2\left)\right)=Cln\left[x\right(1-\frac{y^2}{x^2}\left)\right]=C\Rightarrow \ln \left[\frac{x(x^2-y^2)}{x^2}\right]=C\ln \left[\right(\frac{x^2-y^2}{x}\left)\right]=C$

curve passes through $(2,1)$

$\therefore$ Putting $(2,1)$ in the equation we get

$\ln \left(\frac{x^2-y^2}{x}\right)=C\Rightarrow C=\ln \left[\frac{4-1}{2}\right]=\ln \frac{3}{2}$

$\therefore$ Equation of curve is

$\ln \left(\frac{x^2-y^2}{x}\right)=\ln \frac{3}{2}\Rightarrow \frac{x^2-y^2}{x}=\frac{3}{2}\Rightarrow {2x}^2-{2y}^2-3x=0$