Every year I'll withdraw $10 from the bank where I've a 2.5% interest rate. I've no money left after 10 years. How much did I have from the beginning? Ok, I will try to explain my problem in plain text:

Every year I'll withdraw $\$10$ from the bank where I have a $2.5\%$ interest rate. I have no money left after $10$ years. How much did I have from the beginning?

I've tried to come up with some kind of differential equation. But I'm probably just making in harder than it is.
 A: If you had $X$ dollar at the beginning you will have
$$(X- c) \cdot p $$ after one year , where p=1 +2.5/100 and c=10.
Iterating this 10 times leaves you with
$$0 = (X-c)\cdot p^{10} - c \cdot \sum_{i=1}^9p^i $$ 
therefore you had
$$X = c \cdot \sum_{i=1}^9 p^{i-10} +c$$
in the beginning.
Edit: This solution works if you take 10$ at the beginning of each year.
A: Note: This calculation assumes interest payment at the end of the year, and one withdrawal at the end of the year.
Calling your unknown initial sum $x_0$, the interest rate $p$, the withdrawal sum $d$, the growth factor $f$ defined as
$$
f = 1 + \frac{p}{100}
$$ 
one gets for the sum after the first year $x_1$:
$$
x_1 = f x_0 - d
$$
and for the sum after the second year $x_2$:
$$
\begin{align}
x_2 &= f x_1 - d \\
    &= f\left(f x_0  - d\right) - d \\
    &= f^2 x_0 - f d - d
\end{align}
$$
and for the sum after the following years:
$$
\begin{align}
x_3 &= f x_2 - d \\
    &= f\left(f^2 x_0 - f d - d \right) - d \\
    &= f^3 x_0 - f^2 d - fd - d \\
\\
    &\vdots \\
x_{10} &= f^{10} x_0 - d \sum_{k=0}^9 f^k \\
    &= f^{10}x_0 - d \frac{f^{10}-1}{f-1} \quad (*)
\end{align}
$$
where the result 
$$
\sum_{k=0}^N q^k = \frac{1 - q^{N+1}}{1-q}
$$
for a geometric sum is used. After 10 years the money is exactly gone: $x_{10} = 0$. We therefore get
$$
x_0 = d \frac{f^{10}-1}{f^{10}(f-1)}
$$
as initial sum. With $p = 2.5$% and $d=10$\$ this results in $x_0 = 87.52$\$.
In regard of Robert's answer:
At what initial sum the sum stays constant? Just checking this with equation $(*)$ gives:
$$
(f^{10} - 1) x_0 = d \frac{f^{10} - 1}{f - 1} \iff
x_0 = \frac{d}{f - 1}
$$
which is independent from the number of iterations and gives $x_0 = 400$\$ too.
A: That $\$10$  would be $2.5\%$ interest on a $\$400$ investment, i.e. if you had $\$400$ in the bank you could go on forever in this manner and your bank account would stay constant.  Since your account is actually depleted, you must have started with less. 
But let's say your friend has an account earning the same interest rate, and together you have $\$400$: you start with $A$ and your friend with $\$400 - A$.  Your friend never withdraws any money.  Since your $\$10$ withdrawal represents the total interest on the two accounts, as time goes on
the sum of your fortune and your friend's stays constant.  After $10$ years your account has $0$ and your friend's has grown to $\$400$.  Now at a $2.5\%$ interest rate, your friend's account was multiplied by $1.025$ every year, 
so your friend started out with $$ \dfrac{400}{1.025^{10}} = \$312.48$$
and so you must have started with
$$ 400 - 312.48 = 87.52$$
This works for any type of compounding as long as the $2.5\%$ is the effective annual rate rather than the nominal rate, and withdrawals happen at the same time as interest is paid.
