Solve the differential equation y' = sqrt(x) / 2y

I am studying for a quiz and I needed help on this question... can someone answer asap?

Solve the differential equation y' = sqrt(x) / 2y?

y^2=(2x^(3/2))/3 +C

y = (2/3)x^(3/2) + C

rewrite y' so it reads dy/dx
dy/dx = sqrt(x)/2y

Multiply the 2y to the other side and multiply the dx to the other side so it reads

2ydy = sqrt(x)dx

Integrate both sides.

dy/dx = √x/(2y)

2y dy = √x dx

Integrating both sides:

y² = 2/3*x^(3/2) + C

y = +/-√(2/3*x^(3/2) + C)