Answer 1

(x² − y² ) dx + 2xy dy = 0

∴ 2xy dy = (y² - x²) dx

∴ `dy/dx = (y^2 - x^2)/(2xy) ......(i)`

Put y = tx  ...(ii)

Differentiating w.r.t. x, we get

`dy/dx = t +x dt/dx  ...(iii)`

Substituting (ii) and (iii) in (i), we get

`t + x dt/dx = (t^2 x^2-x^2)/(2tx^2)`

∴ `x dt/dx = (t^2 - 1)/(2t )- t = (-(1+t^2))/(2t)`

∴ `2t/(1+t^2) dt = - dx/x`

Integrating on both sides, we get

`int 2t/(1+t^2) dt = - int dx/x`

∴ log |1 + t²| = -log |x| + log |c|

∴`log | 1+y^2/x^2| = log |c/x|`

∴ `(x^2 + y^2)/x^2 = c/x`

∴  x² + y² = cx