If you have a linear ODE, like simply $y'(t)=y(t)$, one method is to assume a solution of the form $e^{ \lambda t}$, find that the characteristic equations is not dependent on $t$ and thus to solve for $ \lambda$.
However, what if you have a system like $$\mathbf{x'} = \begin{bmatrix} x'(t) \\ y'(t) \\ \end{bmatrix} = \begin{bmatrix} y(t) \\ x(t) \\ \end{bmatrix}?$$
Now there is a more general solution as the normalized exponential of the Jordan-form of the matrix, but what if you have a non-trivial system but still want to try a substitution in this higher dimensional case? What would you need to do differently to have the same effect analogous to the one-dimensional case?
In this example, would I say "take $ \textbf{x}(t) = \begin{bmatrix} e^{\lambda_y} \\ e^{\lambda_x} \\ \end{bmatrix}...$"?
If your equation is linear, say, $\mathbf{x}'(t)=\mathbf{A\,x}(t)$ for some constant matrix $\mathbf{A}$, then the general solution takes the form $\mathbf{x}(t)=\exp(\mathbf{A}t)\,\mathbf{x}(0)$, where $$\exp(\mathbf{M}):=\sum_{r=0}^\infty\,\frac{1}{r!}\,\mathbf{M}^r$$ for every square matrix $\mathbf{M}$ (here $\mathbf{M}^0$ is defined to be the identity matrix). In your case, you can write $\mathbf{x}'(t)=\mathbf{A}\,\mathbf{x}(t)$ with $$\mathbf{A}=\begin{bmatrix}0&1\\1&0\end{bmatrix},$$ so that $$\exp(\mathbf{A}t)=\begin{bmatrix}\cosh(t)&\sinh(t)\\\sinh(t)&\cosh(t)\end{bmatrix}.$$ You can actually see that, in this case, $x(t)$ and $y(t)$ are linear combinations of $\exp(+t)$ and $\exp(-t)$. This is because $+1$ and $-1$ are the eigenvalues of the matrix $\mathbf{A}$.
In order to find the eigenvalues of $\mathbf{A}$, you can assume that you have a nonzero solution $\mathbf{x}(t)=\exp(\lambda t)\,\mathbf{x}(0)$ for some $\lambda$. It will turn out that $\lambda$ has to be a root of the characteristic polynomial of $\mathbf{A}$. In your particular example, you can show by plugging in $x(t)=\exp(\lambda t)\,x_0$ and $y(t)=\exp(\lambda t)\,y_0$ that $\lambda^2-1=0$, and $\lambda^2-1$ is precisely the characteristic polynomial of $\mathbf{A}$ in this example.
Even when $\mathbf{A}$ is a matrix function $\mathbf{A}=\mathbf{A}(t)$, you can still write $$\mathbf{x}(t)=\exp\left(\int_0^t\,\mathbf{A}(s)\,\text{d}s\right)\,\mathbf{x}(0).$$ In non-linear cases, unless you are specific about what types of equations you are looking at, I think the question is too broad.