Question

(x² – y²)dx + 2xy dy = 0

Answer 1

(x² – y²)dx + 2xy dy = 0

∴ –2xydy = (x² – y²)dx

∴ `(dy/dx) = (x^2 - y^2)/(-2xy)`  ...(1)

As it is on homogeneous equation

Put y = vx

∴ `(dy)/(dx) = v + x(dv)/(dx)`

∴ (1) becomes `v + x(dv)/(dx) = (x^2 - v^2x^2)/(-2x(vx))`

∴ `v + x(dv)/(dx) = (1 - v^2)/(-2v)`

∴ `x(dv)/(dx) = (1 - v^2)/(-2v) - v = (1 - v^2 + 2v^2)/(-2v)`

∴ `x(dv)/(dx) = (1 + v^2)/(-2v)`

∴ `(-2v)/(1 + v^2)dv = 1/xdx`

Integrating both sides, we get

∴ `int (-2v)/(1 + v^2)dv = int1/xdx`

∴ –log|1 + v²| = logx + log c₁  ...`[∵ d/(dx)(1 + v^2) = 2v and int(f^'(x))/(f(x))dx = log|f(x)| + c]`

∴ `log|1/(1 + v^2)|` = log c₁x

∴ `log|1/(1 + (y^2/x^2))|` =  log c₁x

∴ `log|x^2/(x^2 + y^2)|` = c₁x

∴ `x^2/(x^2 + y^2)` = c₁x

∴ x² + y² = `1/c_1x`

∴ x² + y² = cx where c = `1/c_1`

This is the general solution.