Solve the differential equation
y'= radical(x)/(-3y)
i made y' dy/dx and moved the y's to one side and the x's to the other, and ended up with y^2=(-4/9)x^3/2 + c, but i'm not sure if that's write, and is that the form i'm supposed to leave it in?
y' = √x / -3y
-3y dy = √x dx
-3/2 y^2 = 2/3 x^(3/2) + c
so, yes, you are correct
y^2 = -4/9 x^(3/2) + c
Not much to do with that. I guess you could use a new c. Name the old one k and you have
y^2 = k - 4/9 x^(3/2)
If k = 4/9 c then
y^2 = 4/9 c - 4/9 x^(3/2)
y = 2/3 √(c - x^(3/2))