Properties of the solution of the Ordinary Differential Equation $y' = y(y-1)(y-2)$ as per the Initial conditions? Consider the Ordinary Differential equation $y' = y(y-1)(y-2)$.
Then from the different initial conditions, can we derive properties of the function $y$ ?

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*I thought of finding the solution to this differential equation, which I tried to use Partial fractions!, (any other easier or efficient method to solve this ODE?). After doing Partial fractions  I got

$0.5 \ln|y| - \ln|y-1| +0.5 \ln|y-2| = x + c$.
Now if $y(0) = 0.5$, is the function $y$ decreasing?, well I thought of substituting value of $y(0)$ into the ode to get $y' = 0.5(0.5-1)(0.5-2)>0$, implying $y$ is decreasing (is this the correct way?).
Also if $y(0) =1.2$ then using the above criteria I think $y$ is increasing?
Also if $y(0) = 2.5$ then can we say anything about the boundedness of $y$ ?
If $y(0)<0$, can we say $y$ is bounded below?.
 A: This is just a different way to state the other answer.
We can use a direction field (also called slope or vector field) plot, see Drawing vector field plots has never been so easy, for a qualitative analysis of your DEQ.
For $y' = y(y-1)(y-2)$, we immediately notice three critical points (where $y'=0$) at $y = 0, 1, 2$.
Using those three critical points, we can choose points within the ranges $y \in (-\infty, 0), (0, 1), (1, 2), (2, \infty)$ and determine the slope values. 
Notice that if you choose values $y \gt 2$, the slope is always positive and the solution becomes unbounded to positive infinity. If you choose $1 \lt y \lt 2$, the slope is always negative and approaches the next critical point. If you choose $0 \lt y \lt 1$, the slope is always positive and approaches the next critical point. If you choose values $y < 0$, the slope is always negative and the solution approaches negative infinity.
From all these points, we can draw the direction field plot (the two colored plots are actual initial conditions based solutions superimposed on the direction field plot) and see all these qualitative behaviors.

You can see other examples at Paul's Online Notes and Direction Field Examples.
A: Some hints:
Draw an $(x,y)$-plane. On the $y$-axis mark the points $y_k$ where the right hand side of the given ODE is zero. This gives you a certain number of special solutions for free. Draw them. No other solution can intersect the curves you have drawn. Between two such curves the slope prescribed by the ODE has a constant sign, and this slope is almost zero near the special solutions. 
And the most important thing about ODEs of this kind: If you have a solution curve all horizontal translates of this curve are solutions as well.
The described principles should allow you to sketch the complete solution portrait without any calculations.
