You can always turn a linear homogeneous differential equation with constant coefficients (such as the one you are asking about) into a linear system. Here, you have $f''-f=0$, so if we set $v=\begin{pmatrix}f\\ f'\end{pmatrix}$, we have that $v'=\begin{pmatrix}f'\\f''\end{pmatrix}=\begin{pmatrix}f'\\ f\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}v$.
The point of this is that you reduce solving the equation to pure linear algebra. First, call $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, and note that we can diagonalize it as $A=S\Lambda S^{-1}$, where $S=\begin{pmatrix}-1&1\\1&1\end{pmatrix}$ and $\Lambda=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$. (Note that the first column of $A$ is an eigenvector with eigenvalue $-1$, and the second is an eigenvector with eigenvalue $1$.)
The reason why we do this is that $S\Lambda S^{-1}v=v'$ iff $\Lambda S^{-1}v=S^{-1}v'=(S^{-1}v)'$ so that, letting $w=S^{-1}v$, we have that the original equation is equivalent to $\Lambda w=w'$. Now, if $w=\begin{pmatrix}g\\ h\end{pmatrix}$, then this is saying that $g'=-g$, and $h'=h$, so that $g(x)=ae^{-x}$ and $h(x)=be^{x}$ for some constants $a,b$. Since $\begin{pmatrix}f\\f'\end{pmatrix}=v=Sw$, it follows that $f$ is a linear combination of $e^x$ and $e^{-x}$.
The approach is perfectly general, and just as easy as long as the resulting $A$ is diagonalizable. If $A$ is not diagonalizable (which for the matrices we get here, happens precisely when $A$ has repeated eigenvalues), we can still solve the system this way, using now the Jordan form of $A$. Yiorgos's solution can be easily seen to be equivalent to what we are doing here. A nice book explaining all of this is volume II of Apostol's Calculus.
Let me briefly explain the connection. Note first that $A^2-I=0$, and $p(x)=x^2-1$ is the smallest non-zero (monic) polynomial that vanishes when applied to $A$ (it is the minimal polynomial of $A$). This is just the characteristic polynomial of $A$. In fact, given any linear homogeneous differential equation with constant coefficients, rewriting it as a system $Av=v'$ always results in an $A$ whose minimal polynomial is also the characteristic polynomial, and is just the characteristic polynomial of the equation: If we want to solve $f^{(n)}-a_{n-1}f^{(n-1)}-\dots-a_0f=0$, this equation polynomial equation is $x^n-a_{n-1}x^{n-1}-\dots-a_0$. This shows that the eigenvalues of $A$ (the roots of $p$) are closely related to the solutions of the differential equation.
To make the relation more transparent, write $D$ for the derivative operator: $Df=f'$. Then the equation is just $(D^n-a_{n-1}D^{n-1}-\dots-a_0I)f=0$, where $I$ is the identity operator, $If=f$. To solve this, we factor the polynomial in $D$, and solve. For example, starting with $f''-f=0$, we get $(D^2-I)f=0$, or $(D-I)(D+I)f=0$, so if $g=(D+I)f$, then $(D-I)g=0$, or $g'=g$, or $g=ae^x$. Then we need to solve $(D+I)f=g$, or $f'+f=ae^x$, so $f=be^{-x}+ce^x$.
In general, if the resulting polynomial has degree $n$ and $n$ different roots $\lambda_1,\dots,\lambda_n$, the procedure just sketched gives us that the solution has the form $b_1e^{\lambda_1 x}+\dots+b_ne^{\lambda_n x}$ for some constants $\lambda_1,\dots,\lambda_n$. The case where the roots are repeated is slightly more involved, and instead of constants, the $b_i$ are now polynomials in $x$ ($b_i$ of degree $k_i-1$ if $\lambda_i$ appears as a root $k_i$ times). This can either be proved directly, and then used to deduce the general form of the Jordan form of $A$, or can be proved starting with the Jordan form of $A$.
