Question

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

`(1+e^(x/y))dx + e^(x/y) (1 - x/y)dy = 0`

Answer 1

`(1 + e^(x/y))dx + e^(x/y)(1 - x/y) dy = 0`

`=> dx/dy = (e^(x/y)(x/y - 1))/(1 + e^(x/y)) = g(x/y)`     say   ...(i)

∵ The right side of the equation is in the form of `g (x/y)`, so it is a homogeneous differential equation of zero degrees.

∴Putting  x = vy

`dx/dy = v + y (dv)/dy`    ...(from equation (i))

`v + y (dv)/dy = (e^v(v - 1))/(1 + e^v)`

or `y (dv)/dy = (e^v(v - 1))/(1 + e^v) - v`

`=> (ve^v - e^v - v - ve^v)/(1 + e^v)`

`=> ((1 + e^v)/(v + e^v))dv = - 1/y dy`

`= int(1 + e^v)/(v + e^v) dv = - int 1/y dy`

⇒ log |e^v + v| = - log |y| + C₁

⇒ log |(e^v + v)y| = C₁

⇒ |(e^v + v) y| = e^C₁

⇒ (e^v + v)y = ± e^C₁ = C  (say)

⇒ `(e^(x/y) + x/y) y = C`

⇒ `y  e^(x/y) + x = C`

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