Constant solution and uniqueness of separable differential eq

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mr.tea
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Hi,
I am learning ODE and I have some problems that confuse me.
In the textbook I am reading, it explains that if we have a separable ODE: ##x'=h(t)g(x(t))##
then ##x=k## is the only constant solution iff ##x## is a root of ##g##.
Moreover, it says "all other non-constant solutions are separated by the straight line x=k".

First, why do we do this separation between finding constant and non-constant solutions?

Second, I don't understand the quoted sentence. why is that?

Third, there is an example of finding a solution to the initial value problem ##x'=2tx^3## and ##x(0)=1##. They say that the only constant solution is ##x \equiv 0##, and
"Therefore if ##x(t)## is a solution such that ##x(0)=1##, then, by uniqueness, ##x(t)## cannot assume the value 0 anywhere. Since ##x(0) =1 >0##, we infer that the solution is always positive."
But how can ##x \equiv 0## be a constant solution if the solution should satisfy ##x(0)=1##, and how they got that the solution should be positive?

I am really confused and need some help with this.

Thank you.
 
mr.tea said:
x=k is the only constant solution iff x is a root of g
Are you sure it says only? Clearly any root of g is a constant solution.
mr.tea said:
I don't understand the quoted sentence. why is that
Suppose x > k at some t. hg might be negative here, so as t increases x decreases. But as x approaches k, g(x) approaches zero, so the slope will level off. It can never reach k unless h goes to infinity.
mr.tea said:
But how can x≡0 be a constant solution if the solution should satisfy x(0)=1, and how they got that the solution should be positive
x identically zero is a solution to the differential equation without the initial value. From the foregoing, any other solution cannot cross or meet the line x=0. In particular, a solution such that x(0)=1 cannot anywhere go to x=0.