<s>[INST] How do I handle equations like $f(x)-x = f'(x)$? How do I handle equations like $f(x)-x = f'(x)$?
Integrating both sides gets me nowhere... I have been trying to guess an example, but with no avail. I cannot think of any other approaches ...
Edit: ok i noticed $f(x) =x+1$ ... can someone guide be to how i can solve for the other solutions?
 [/INST] I know that the below solution doesn't work in many situations, but I like to share its idea.
Suppose $f(x) = \sum_{n \geq 0}a_nx^n$ is the power series of $f$. Because $f(x) - x = f'(x)$, term by term differentiating of the power series gives us
\begin{align*}
\sum_{n \geq 0}a_nx^n - x = \sum_{n \geq 1}na_nx^{n-1} \implies a_0 + (a_1 - 1)x + a_2x^2 + \cdots = a_1 + 2a_2x + 3a_3x^2 + \cdots
\end{align*}
Because the above equation is correct for all $x$, so we have:
\begin{align*}
a_1 &= a_0\\
a_1 - 1 = 2a_2 \implies a_2 &= \frac{a_1 - 1}{2} = \frac{a_0 - 1}{2}\\
a_2 = 3a_3 \implies a_3 &= \frac{a_2}{3} = \frac{a_0 - 1}{3!}
\end{align*}
Computing in this way shows that for all $n \geq 2$ we have $a_n = \frac{a_0 - 1}{n!}$. Therefore
\begin{align*}
f(x) = \sum_{n \geq 0}a_nx^n &= a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots\\
&= a_0 + a_0x + (a_0 - 1)\frac{x^2}{2!} + (a_0 - 1)\frac{x^3}{3!} + \cdots\\
&= 1 + x + (a_0 - 1) + (a_0 - 1)x + (a_0 - 1)\frac{x^2}{2!} + (a_0 - 1)\frac{x^3}{3!} + \cdots\\
&= 1 + x + \underbrace{(a_0 - 1)}_Ce^x
\end{align*}
For an arbitrary constant $C \in \mathbb{R}$.
</s>