## Calculus: Early Transcendentals 8th Edition

$y = (x^2/4 + C_2)^2$
$dy/dx = x\sqrt y$ 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 $C_1/2 = C_2$ $y = (x^2/4 + C_2)^2$