(dy)/(dt) = - y^(4) + 4 y^(3) + 45 y^(2) What are the constant solutions of this equation? For what values of y is y increasing? (_

Question

`(dy)/(dt) = - y^(4) + 4 y^(3) + 45 y^(2)`

What are the constant solutions of this equation? For what values of y is y increasing? (_

Answer 1

`(\mathrm dy)/(\mathrm dt)=-y^4+4y^3+45y^2=-y^2(y-9)(y+5)`

There are three stationary points at
`y=-5,y=0,y=9`.

When
`y<-5`, you have
`(\mathrm dy)/(\mathrm dt)<0`, so
`y` is decreasing here.

When
`-5<y<0`, you have
`(\mathrm dy)/(\mathrm dt)>0`, so
`y` is increasing here.

When
`0<y<9`, you have
`(\mathrm dy)/(\mathrm dt)>0`, so
`y` is increasing here.

When
`9<y`, you have
`(\mathrm dy)/(\mathrm dt)<0`, so
`y` is decreasing here.

To summarize, when
`(\mathrm dy)/(\mathrm dt)>0`,
`y` will be an increasing function, and this occurs for
`y\in(-5,0)\cup(0,9)`.