Question

y'=ty(3-y)

Describe how solutions appear to behave as t increases and how their behavior depends on the initial value y0 ("0" is a subscript) when t=0.

How do I get the solutions in order to plot them?

Answer 1

y'=dy/dt
so you can simplify above equation as
\[\frac{ dy }{ y(3-y)}=tdt\]
now integrate this by using partial fraction
\[\frac{ 1 }{ 3} [\frac{ 1 }{ y }+ \frac{ 1 }{ 3-y }]dy= tdt\]
i think you can integrate it now

Answer 2

I got that far. After that point this is what I have:
Integrating gets me to:
\[\frac{ 1 }{ 3 }\left( \ln \frac{ y }{3-y} \right)=1/2t ^{2}+C\]
then simplifying and trying to solve for C, I get stuck at:
\[\frac{ y }{ 3-y }=Ce ^{(3/2)t ^{2}}\]
Not sure how to proceed from here to get C and then the solutions based on what the question is asking for.