## Calculus: Early Transcendentals 8th Edition

$y = -1/(x^3 + C)$
$dy/dx = 3x^2y^2$ Let's separate the y on one side and the x on the other side: $(1/y^2 )dy = 3x^2dx$ Now let's integrate both sides: $\int (1/y^2 )dy = \int 3x^2dx$ $-1/y = x^3 + C$ Isolate y: $y = -1/(x^3 + C)$