Find the function yyx for x0 which satisfies the separable

Find the function y=y(x) (for x>0 ) which satisfies the separable differential equation
dy/dx=(3+19x)/(xy^2); x>0
with the initial condition y(1)=2.

dy/dx=(3+19x)/(xy^2)

y^2 dy = (3+19x)/x dx

int y^2 dy = int (3/x + 19) dx

1/3 y^3 = 3ln(x) + 19x + C

x = 1 -> 8/3 = 0 + 19 + C -> C = 8/3 - 19 = -49/3

Therefore:

1/3 y^3 = 3ln(x) + 19x - 49/3