An initial value problem for an ordinary differential equation

[AlfredFlaßhaar]

I recently learned about the following problem, whose solution function u(t) requires determining:
u´(t) = -u(t)/t + 1/(t^3+t), u(0) = 0
I intentionally called this problem a riddle 😉 .
Please answer in MC14-readable format.

[AlanStevens]

There is a singularity at t=0, so:

(Done in M15, but I think M14 can read it).

(Actually, the limit, as t tends to 0, of atan(t)/t is 1, so more sensible to set u(0) = 1.)

[AlfredFlaßhaar]

The limit eps-->0, like t-->0, yields the ordinate u(0) = 1. Therefore, the differential equation with the required initial

value u(0) = 0 has no solution?
The general solution (without an initial value) is:
u(t) = C*1/t + 1/t*arctan(t).

[Werner_E]

From a math point of view you can't demand u(0)=... because of the singularity mentioned by Alan.

I am not sure if you could demand

and how we would handle this constraint.

Prime is able to solve some simple ODEs and so it can come up with the correct general solution (valid of course only for t <>0)

When asked to solve the ODE with the (forbidden) IC u(0)=0 it happily comes up with a wrong solution 🙂

but

And here a last 'interesting' result given by Prime

Hmmm ....

[AlfredFlaßhaar]

This exercise is from a book that, according to the title, is an "introduction" to the world of ordinary differential equations. I don't want to mention the title or author. It is written in a conversational tone and describes a great deal of interesting information (including from physics and the history of mathematics). It seems to be intended for interested beginners without any proof of cited theorems. Therefore, it is reasonable to conclude that the initial value u(0) = 0 is a misprint. After all, what beginner (who doesn't know the theorems of Peano, Picard/Lindelöf, etc.) would think that the initial value lies not in the continuity region of the "right-hand side" of u'(t) but on the boundary of the open interval? Therefore, the attempt eps-->0 also fails. Thus, there is no real solution for this initial value.