Why solving a differentiated integral equation might eventually lead to erroneous solutions of the original problem?

Why solving a differentiated integral equation might eventually lead to erroneous solutions of the original problem?

The reason is that taking a derivative is not an invertible operation. So the new equation you obtain is true, but not equivalent to the original one -- the set of solutions has increased.

The simplest example is trying to solve an ordinary equation, say, $$ x=1 $$ The obvious solution is $x=1$. But if you square both sides, you obtain $x^2=1$, which now has two solutions, $x=\pm 1$. The new "wrong" solution appeared because taking a square is not invertible (the kernel is the negatives).

Similarly, taking a derivative is not invertible. Consider the equation $$ f(t)=t $$ The obvious solution is $f(t)=t$. But if you take a derivative, you get $f'(t)=1$, whose general solution is $f(t)=t+c$, for an arbitrary $c$. The "wrong" solutions, those with $c\neq0$, appeared because taking a derivative is not invertible (the kernel is the constants).

The problem is that $f(t) =-1$ is not the unique solution of the integral equation below

$$ -\frac{1}{r} \int_0^r \frac{f(t)t \, \mathrm{d}t}{\sqrt{r^2-t^2}} = 1 \, , $$

For example,

$$ f(t) =- \frac 2r \sqrt{r^2-t^2}$$ would be another valid solution. In fact, there are numerous functions $f(t)$ that satisfy the integral equation above.

Edit: an alternative approach

Similarly, the original integral equation also admits multiple solutions. One particular solution can be derived by assuming that $f(t)=a$ is a simple flat function, where $a$ is a constant. Then, we have

$$a\int_0^r \arcsin\left( \frac tr \right)dt +\frac{\pi}{2}a(R-r)=r$$

or,

$$a\left[ \left( \frac {\pi}{2} -1 \right)r+ \frac{\pi}{2}(R-r)\right]=r$$

and the flat function solution is

$$f(t)=a= \frac{1}{ \frac{\pi}{2}\frac Rr -1}$$