Calculus 10th Edition

$4y^2= x^3$ passes through the point (4,4)
Use the general solution of the differential equation and plug in (4,4). $y^2= Cx^3$ $16= 64C$ $C=\frac{1}{4}$ Substitute C back into the general solution $y^2= \frac{1}{4}x^3$ $4y^2= x^3$