The solution of the differential equ `(x^2 + y^2) dx = 2 xy dy` is-

Explanation

Given, `(x^(2)+y^(2)) dx - 2xy dy = 0 `
or ` (x^(2)+y^(2)) dx = 2xy dy`
or ` (dy)/(dx) = (x^(2)+y^(2))/(2xy) ` ....(i)
Let y = vx
Thus `(dy)/(dx) = v + x (dv)/(dx)`
Thus ` v+ x (dv)/(dx) = (x^(2)+(vx)^(2))/(2x(vx))`
or ` v + x (dv)/(dx) = (1+ v^(2))/(2v)`
or ` x (dv)/(dx) = (1+v^(2))/(2v) - v`
or ` x (dv)/(dx) = (1+v^(2)-2v^(2))/(2v)`
or ` x (dv)/(dx) = (1-v^(2))/(2v)`
or ` (dx)/x = (2v)/(1-v^(2)) dv`
or ` (dx)/x - (2v)/(1-v^(2)) dv = 0` ...(ii)
Integrating both sides,
` log x + log (1-v^(2)) = log C`
or ` log x (1 - v^(2)) = log C`
or ` x (1 - v^(2)) = C`
or ` x (1 - y^(2)/x^(2)) = C`
or ` x ((x^(2)-y^(2))/x^(2)) = C`
or ` x^(2) - y^(2) = Cx`