(a) What can you say about a solution of the equation y' = −(1/6)y2 just by looking at the differential equation? The function y must be increasing (or equal to 0) on any interval on which it is defined. The function y must be strictly increasing on any interval on which it is defined. The function y must be decreasing (or equal to 0) on any interval on which it is defined.

Respuesta :

[tex]y^2\ge0[/tex] for any [tex]y[/tex], and so [tex]y'<0[/tex] for all [tex]y[/tex], meaning [tex]y[/tex] is decreasing.