[tex]\dfrac{\mathrm dy}{\mathrm dt}=-y^4+4y^3+45y^2=-y^2(y-9)(y+5)[/tex]
There are three stationary points at [tex]y=-5,y=0,y=9[/tex].
When [tex]y<-5[/tex], you have [tex]\dfrac{\mathrm dy}{\mathrm dt}<0[/tex], so [tex]y[/tex] is decreasing here.
When [tex]-5<y<0[/tex], you have [tex]\dfrac{\mathrm dy}{\mathrm dt}>0[/tex], so [tex]y[/tex] is increasing here.
When [tex]0<y<9[/tex], you have [tex]\dfrac{\mathrm dy}{\mathrm dt}>0[/tex], so [tex]y[/tex] is increasing here.
When [tex]9<y[/tex], you have [tex]\dfrac{\mathrm dy}{\mathrm dt}<0[/tex], so [tex]y[/tex] is decreasing here.
To summarize, when [tex]\dfrac{\mathrm dy}{\mathrm dt}>0[/tex], [tex]y[/tex] will be an increasing function, and this occurs for [tex]y\in(-5,0)\cup(0,9)[/tex].
