Solving Systems with Real Eigenvalues

For one equation $y' = ay$ the solution is $ce^{at}$. For a system $\mathbf{x}' = A\mathbf{x}$ we look for solutions of the same shape, $\mathbf{x} = e^{\lambda t}\mathbf{v}$, a fixed vector whose length grows or shrinks exponentially. This is where eigenvalues and diagonalization meet differential equations.

Idea clave

$\mathbf{x} = e^{\lambda t}\mathbf{v}$ solves $\mathbf{x}' = A\mathbf{x}$ exactly when $A\mathbf{v} = \lambda\mathbf{v}$. If a $2\times 2$ matrix $A$ has real eigenvalues $\lambda_1 \ne \lambda_2$ with eigenvectors $\mathbf{v}_1, \mathbf{v}_2$, then $$\mathbf{x}(t) = c_1e^{\lambda_1t}\mathbf{v}_1 + c_2e^{\lambda_2t}\mathbf{v}_2.$$

Por qué funciona

Substituting $\mathbf{x} = e^{\lambda t}\mathbf{v}$: the left side is $\lambda e^{\lambda t}\mathbf{v}$ and the right side is $e^{\lambda t}A\mathbf{v}$. Cancelling the nonzero scalar $e^{\lambda t}$ leaves $A\mathbf{v} = \lambda\mathbf{v}$. The eigenvalues solve $\det(A - \lambda I)$ $= \lambda^{2} - (\operatorname{tr}A)\lambda + \det A = 0$. Superposition combines the two straight-line solutions, and since $\mathbf{v}_1, \mathbf{v}_2$ are independent, the constants can match any $\mathbf{x}(0)$. In the eigenvector coordinates the system splits into two separate equations, $u_1' = \lambda_1u_1$ and $u_2' = \lambda_2u_2$.

Ejemplo resuelto: Eigenvalues and eigenvectors

Find the general solution of $\mathbf{x}' = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\mathbf{x}$.

$\lambda^{2} - 2\lambda - 3 = (\lambda - 3)(\lambda + 1)$ $= 0$: $\lambda = 3, -1$.
$\lambda = 3$: $A - 3I = \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}$, so $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
$\lambda = -1$: $A + I = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$, so $\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.

Respuesta: $\mathbf{x} = c_1e^{3t}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2e^{-t}\begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Ejemplo resuelto: Fit the initial condition

Solve the same system with $\mathbf{x}(0) = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$.

$c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$: $c_1 + c_2 = 3$, $c_1 - c_2 = -1$.
$c_1 = 1$, $c_2 = 2$.

Respuesta: $x_1 = e^{3t} + 2e^{-t}$, $x_2 = e^{3t} - 2e^{-t}$

Ejemplo resuelto: A zero eigenvalue

Solve $\mathbf{x}' = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}\mathbf{x}$, $\mathbf{x}(0) = \begin{bmatrix} 4 \\ 0 \end{bmatrix}$ (two connected tanks).

$\lambda^{2} + 2\lambda = 0$: $\lambda = 0, -2$, with $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
$c_1 + c_2 = 4$, $c_1 - c_2 = 0$: $c_1 = c_2 = 2$.

Respuesta: $x_1 = 2 + 2e^{-2t}$, $x_2 = 2 - 2e^{-2t}$: both settle at 2, sharing the total equally

Error común

Scaling an eigenvector is fine (any nonzero multiple works), but mixing up which eigenvector goes with which eigenvalue is not: each $e^{\lambda_it}$ must multiply its own $\mathbf{v}_i$. Check one column: $A\mathbf{v}_1$ should equal $\lambda_1\mathbf{v}_1$.

Trajectories of $\mathbf{x}' = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\mathbf{x}$. Along the purple eigen-directions solutions move in straight lines: outward along $(1, 1)$ (growth $e^{3t}$) and inward along $(1, -1)$ (decay $e^{-t}$). Every other path bends from one direction toward the other.

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