Question

Show that the given differential equation is homogeneous and solve them.

(x² – y²) dx + 2xy dy = 0

Answer 1

(x² - y²) dx + 2xy dy = 0

Which can be written as

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

`= ((y/x)^2 - 1)/(2 (y/x))`                 ....(1)

Since R.H.S is of the form `g(y/x)`, and so it is a homogeneous function of degree zero

Therefore equation (1) is a homogeneous differential equation.

⇒ `dy/dx = v + x (dv)/dx`, then (1) become

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

⇒ `x (dv)/dx = (v^2 - 1)/(2v) - v`

⇒ `(2vdv)/(v^2 + 1) = -dx/x`                 ....(2)

Integrating (2) both sides, we get

log |v² + 1| = - log |x| + C

⇒ log |(v² + 1) x | = C

⇒ `log |(y^2 + x^2)/x| = C_1`                 ...`(∵ v = y/x)`

⇒ `|(y^2 + x^2)/x| = e^(C_(1))`

⇒ `(x^2 + y^2)/x =pm  e^(C_(1)) = C`    (say)

⇒ `x^2 + y^2 = Cx`

which is the required general solution of the given differential equation.