For solving systems of autonomous differential equations we use Jordan normal forms as typically they give us an easy way to calculate the solution to our system.
If we have the IVP:
$$\left\{\begin{matrix}
\frac{dx}{dt} = Ax\\
x(0) = x_0
\end{matrix}\right.$$ where $A$ is a constant $n \times n$ matrix and $x_0 \in \mathbb{R}^n$, then the solution is $x(t) = \text{exp}(tA)x_0.$
For a general matrix $A$, this may be a pain to calculate, but using Jordan normal forms we can simplify this process. A property of the matrix exponential is that:
If $A$, $B$ and $T$ are $n \times n$ matrices and $T$ is invertible, then if $B = T^{-1}AT$, we have $\text{exp}(B) = T^{-1}\text{exp}(A)T$.
This is where Jordan normal forms become useful, because there is a specific "formula" for taking the exponential matrix of Jordan normal forms.
(See here for a good explanation of the formula and here for an example of someone calculating the exponential.)
For $\dot{x} = Ax$, we can make a linear change of coordinates, $x = Ty$ where $T$ is invertible which transforms our differential equation into $$\dot{y} = T^{-1}ATy.$$ From linear algebra, we can choose $T$ such that $T^{-1}AT$ is in Jordan normal form $B$. Then using the "formula" for the matrix exponential for Jordan normal forms we can easily calculate $\text{exp}(tB).$ Finally, using the property above, the solution to our IVP will be $$x(t) = \text{exp}(tA)x_0 = T\text{exp}(tB)T^{-1}x_0.$$
So in short, Jordan normal forms are used for calculating the solutions of systems of differential equations because they make the calculations easier.
Alternative methods would be calculating the exponential term-by-term, not a good idea unless you have a very simple matrix. Or it's possible to get the solution just by looking at the eigenvalues and eigenvectors, but as far as I'm aware that is just an equivalent version of the Jordan normal form method.
