Question

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

`y  dx + x log(y/x)dy - 2x  dy = 0`

Answer 1

`y  dx + x log (y/x)  dy - 2x  dy = 0`

⇒ `y/x + log (y/x) dy/dx - 2 dy/dx = 0`

⇒ `dy/dx = (y/x)/(2 - log (y/x))`           .....(1)

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

Therefore equation (1) is a homogeneous differential equation.

To solve this, put y = vx

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

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

⇒ `x (dv)/dx = v/ (2 - log v) - v`

⇒ `((2 - log v)dv)/(v log v - v) = dx/x`

⇒ `(1 - (log v - 1))/(v (log v - 1)) dv = dx/x`            .....(2)

Integrating (2) both sides, we get

`int (1/ (v (logv - 1)) - 1/v)  dv`

`= int dx/x + C_1`

⇒ `int (1/v)/(log v - 1) dv - int 1/v  dv = int dx /x + C_1`

⇒ `log |log v - 1| - log |v|`

= log |x| + C₁

⇒ `log |(log v - 1)/(v  x)| = C_1`

⇒ `|(log v - 1)/ (v  x)| = e^(C_(1))`

⇒ `(log v - 1)/(v  x) = pm  e^(C_(1)) = C`    (say)

 ⇒ `log (y/x) - 1 = Cy`                   ...`(∵ v = y/x)`

which is the required general solution.