Is it 'more rigorous' to perform definite integrations, rather than indefinite integration while solving ODEs?

Question

In the beginning of my ODE course, my professor said something about performing definite integrations being 'more rigorous' than indefinite ones somehow, but also that it really wasn't much important, as both ones worked for the same problems.

I remember he gave an example like the following:

Consider the simple differential equation $$ x'=g(t),\quad x(\tau)=\xi. $$

Then, integrating from $\tau$ to $t$, we have

$$ x(t)-\xi=\int_\tau^tg(s)ds \iff x(t)=\xi+\int_\tau^tg(s)ds $$

And we have our solution.

Note that if we performed an indefinite integration, we wouldn't have gotten a solution, but many 'candidates'.

My questions are:

1. How right is my professor, in terms of the 'rigorousity'?

2. What allows one to change $t\mapsto s$, before integrating?

Answer 1

It is perfectly alright to do an indefinite integral, as long as you don't forget the $+C$ and the initial condition.

For concreteness, take the following equation:

$$x'(t) = 2t, \ \ x(2) = 5$$

Performing an indefinite integral on $x'(t) = 2t$ gives $x(t) = \int 2t\ dt = t^2 + C$, where $C$ is an unkown constant. Now we plug in the initial condition: $x(2) = 2^2 + C = 4 + C = 5$, so $C = 1$. The complete solution is $x(t) = t^2 + 1$, which we could also have gotten via the definite integral route. As far as I can tell, neither method is more rigorous than the other, and neither one is really easier. Arguably the indefinite integral way has the advantage that it gives half a result when you don't have an initial condition, while the definite integral doesn't work.

Answer 2

Your professor's definition of "rigorous" is a personal one.

All his example does is use the explicit condition supplied. It's no more rigorous that integrating the general form with the arbitrary constant and calculating the value of the arbitrary constant from the given condition then.

However I would take the view that such convenient forms of solution do not generally provide themselves with differential equations. In these cases you often have to work out the general form. The arbitrary constant may even give the solution a "symmetry" or "shape" that an attempt to bypass that point will miss, even after you calculate the particular value for the constant in a particular case. Seeing the general shape is often a significant factor in extending and improving physical theories.