and .
Why does it follow that as ?
My attempt:
We can find so that and solve the differential equation to obtain
Clearly, as .
But, how do we know that is a constant? It could well be a variable that is always greater than .
and .
Why does it follow that as ?
My attempt:
We can find so that and solve the differential equation to obtain
Clearly, as .
But, how do we know that is a constant? It could well be a variable that is always greater than .
I don't think your working is correct, because I don't think you can assume the equality holds. Is k positive? If so, why not do this:
and so on. The end result is very like what you got, except that all the steps here are valid (I think), whereas, like I said, I don't think you can assume that the equality holds which gives you your differential equation.
An easier way would be to consider the graphs of dr/dt=k√r, r(0)=R and dr/dt>k√r, r(0)=R. Clearly, the graph of dr/dt>k√r, r(0)=R lies above the graph of dr/dt=k√r, r(0)=R. Since r(t)-->∞ as t-->∞ in the lower graph, r(t)-->∞ as t-->∞ in the upper graph as well.
I'm not seeing where you're going with this. While it's true that the integral on the entire interval doesn't converge for either integrand on either side of the inequality (they are both infinity), that would be true for something as simple as 1 > 1/2, as well. It is true that
for all
So any finite interval integral that exists will preserve inequalities.