Question

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

`x^2 dy/dx = x^2 - 2y^2 + xy`

Answer 1

Given `x^2 dy/dx = x^2 - 2y^2 + xy`

⇒ `dy/dx = (x^2 - 2y^2 + xy)/x^2`

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

∴ put y = vx

⇒ `dy/dx = v.1 + x (dv)/dx`

Substituting these values of y and `dy/dx` in the given equation, we get

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

⇒ `x (dv)/dx = 1 - 2v^2`

⇒ `1/x dx = 1/ (1 - 2v^2)  dv`

On integration, we get

`log |x| = int 1/ (1 - 2v^2) dv + C`

⇒ `log |x| = 1/2 int (dv)/ ((1/ sqrt2)^2 - v^2) + C`

⇒ `log |x| = 1/2 * 1/ (2* (1/sqrt2)) log |((1/sqrt2)+v)/((1/sqrt2) - v)| + C`

⇒ `log |x| = 1/(2sqrt2) log |(1 + sqrt(2v))/(1 - sqrt(2v))| + C`

⇒ `log |x| = 1/ (2 sqrt2) log |(1 + sqrt2* (y/x))/(1 - sqrt2 * (y/x))| + C`

⇒ `log |x| = 1/ (2sqrt2) log | (x + sqrt(2y))/(x - sqrt(2y))| + C`

Where C is an arbitrary constant