y' = dy/dx =[ 1 - (y/x)？ ]/(y/x)

Let y/x = u, then y = x * u, y' = u+x * u'. Substituting the above formula, we get:

y' = u + x * u' = ( 1-u？ )/u

Simplify:

x * u' = ( 1-u？ )/u - u = ( 1-2u？ )/u

x * du/dx = ( 1-2u？ )/u

u * du/( 1-2u？ )= dx/x

Integrating both sides of the equation at the same time, we get:

1/2 * ∫(2u * du)/( 1-2u？ )= ∫dx/x

1/2 * ∫d(u？ )/( 1-2u？ )= ∫dx/x

- 1/2 * 1/2 *∫d( 1-2u？ )/( 1-2u？ )= ∫dx/x

- 1/4 * ln( 1-2u？ )= lnx + c

ln( 1-2u？ )= -4lnx - 4c

1-2u？ = x^(-4) * e^(-4c) = K * x^(-4)

1 - 2y？ /x？ = K * x^(-4)

x？ - 2y？ = K * x^(-2) = K/x？

Because y(2) = 5, substitute the above formula to get:

2? - 2 * 5? = K/2？ = -46

Therefore, K =-164.

So the solution is: x? - 2y？ = - 164/x？