Question

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

(x² + xy) dy = (x² + y²) dx

Answer 1

Let `dy/dx = (x^2 + y^2)/(x^2 + xy) = f (xy)`       ..... (i)

Now, f `(lamda x, lambda y) = (lambda^2 (x^2 + y^2))/(lambda^2 (x^2 + xy)) = lambda^0` f (x, y)

`therefore` F(x,y) is an exponential function of degree zero.

Hence the given differential equation is a homogeneous differential equation.

Now y = vx

`dy/dx  = v + x  (dv)/dx`

Then, from equation (i)

V + x `(dv)/dx = (x^2 + v^2 x^2)/(x^2 + vx^2)`

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

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

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

`=> (1 + v)/(1 - v)  dv = dx/x`

On integrating,

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

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

⇒ - v - 2 log (1 - v) = log x + log C

⇒ - v = log Cx + 2 log (1 - v)

⇒ - v = log Cx + log (1 - v)²

Cx . (1 - v)² = e^-v

On substituting `y/x` in place of v,

`C. x ((x - y)^2)/x^2 = e^(-y/x)`

`=> (x - y)^2 = Cxe^(-y/x)`