Answer
$y = (x^2/4 + C_2)^2$
Work Step by Step
$dy/dx = x\sqrt y$
The square root can be written as an exponent (1/2).
$dy/dx = xy^{1/2}$
Now let's separate y on one side and x on the other side:
$ (1/y^{1/2} )dy = x dx $
Now let's integrate both sides:
**Remember that $ 1/y^{1/2}$ can be written as $y^{-1/2}$**
$ \int y^{-1/2} dy = \int x dx $
$ y^{1/2}/(1/2) = x^2/2 + C_1 $
$2y^{1/2} = x^2/2 + C_1$
$y^{1/2} = x^2/4 + C_1/2 $
Let's replace $C_1/2 = C_2$
$y = (x^2/4 + C_2)^2$
