Homogeneous Linear Systems
Differential Equations · Section 7.5 · generated practice set
Eigenvalues and eigenvectors to the general solution, initial conditions, and what the phase portrait looks like.
Practice this set → Fresh numbers on every attempt. No account needed.
Try one now
Find the eigenvalues of $$\mathbf{x}' = \begin{pmatrix} -1 & 1 \\ -6 & 4 \end{pmatrix}\mathbf{x}$$ and, for the smaller one, the eigenvector of the form $\begin{pmatrix} 1 \\ k \end{pmatrix}$.
Solution
- $\text{tr}A = -1 + 4 = 3$ and $\det A = (-1)(4) - (1)(-6) = 2$.
- $\lambda^2 - 3\lambda + 2 = 0$ factors as $(\lambda - 1)(\lambda - 2) = 0$.
- For $\lambda = 1$, solve $(A - 1I)\mathbf{v} = \mathbf{0}$. The first row gives $(-1 - 1)v_1 + 1v_2 = 0$, i.e. $-2v_1 + 1v_2 = 0$.
- Taking $v_1 = 1$: $v_2 = 2$, so an eigenvector is $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$.
- The two rows of $A - \lambda I$ are always multiples of one another — if they are not, the eigenvalue is wrong.
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Method
For $\mathbf{x}' = A\mathbf{x}$ with $A$ constant, guess $\mathbf{x} = \mathbf{v}e^{\lambda t}$. Substituting gives $\lambda\mathbf{v}e^{\lambda t} = A\mathbf{v}e^{\lambda t}$, so the guess works exactly when $$A\mathbf{v} = \lambda\mathbf{v}.$$ The differential equation has become an eigenvalue problem.
Real distinct eigenvalues
$$\mathbf{x} = c_1\mathbf{v}^{(1)}e^{\lambda_1t} + c_2\mathbf{v}^{(2)}e^{\lambda_2t}.$$
- $\det(A - \lambda I) = \lambda^2 - (\text{tr}A)\lambda + \det A = 0$ for the eigenvalues.
- For each, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$. The two rows are multiples of one another — use whichever is simpler, set one component to 1.
- Apply initial conditions to $c_1\mathbf{v}^{(1)} + c_2\mathbf{v}^{(2)} = \mathbf{x}(0)$ — a $2\times2$ algebraic system.
The phase portrait
- $\lambda_1, \lambda_2 < 0$ → asymptotically stable node; everything decays to the origin.
- $\lambda_1, \lambda_2 > 0$ → unstable node; everything runs away.
- opposite signs → saddle point, always unstable; trajectories come in along one eigendirection and leave along the other.
Quick test: $\det A < 0$ means a saddle, full stop. If $\det A > 0$, the sign of $\text{tr}A$ decides stability.
For large $t$ the term with the largest eigenvalue dominates, so trajectories become parallel to its eigenvector — that is the direction a phase portrait's arrows line up with far from the origin.
Definitions and theorems
Worked example
Solve $\mathbf{x}' = \begin{pmatrix} 1 & 1 \\ 4 & 1\end{pmatrix}\mathbf{x}$ with $\mathbf{x}(0) = \begin{pmatrix} 2 \\ 3\end{pmatrix}$.
- $\text{tr}A = 2$, $\det A = 1 - 4 = -3$, so $\lambda^2 - 2\lambda - 3 = 0$ and $\lambda = -1, 3$.
- For $\lambda = -1$: $(A + I)\mathbf{v} = \begin{pmatrix} 2 & 1 \\ 4 & 2\end{pmatrix}\mathbf{v} = \mathbf{0}$, so $2v_1 + v_2 = 0$ and $\mathbf{v}^{(1)} = \begin{pmatrix} 1 \\ -2\end{pmatrix}$.
- For $\lambda = 3$: $\begin{pmatrix} -2 & 1 \\ 4 & -2\end{pmatrix}\mathbf{v} = \mathbf{0}$, so $\mathbf{v}^{(2)} = \begin{pmatrix} 1 \\ 2\end{pmatrix}$.
- $\mathbf{x}(0)$: $c_1 + c_2 = 2$ and $-2c_1 + 2c_2 = 3$, giving $c_1 = \tfrac14$, $c_2 = \tfrac74$.
- $\mathbf{x} = \tfrac14\begin{pmatrix} 1 \\ -2\end{pmatrix}e^{-t} + \tfrac74\begin{pmatrix} 1 \\ 2\end{pmatrix}e^{3t}$. Opposite-sign eigenvalues, so the origin is a saddle and the trajectory eventually follows $\begin{pmatrix} 1 \\ 2\end{pmatrix}$.
Common mistakes
- Pairing an eigenvector with the wrong eigenvalue. Label them as you go; swapping them produces something that is not a solution at all.
- Row-reducing $(A - \lambda I)$ too far. It is singular by construction, so you should reach a row of zeros. If you do not, recheck $\lambda$.
- Applying initial conditions to each term separately. The condition applies to the sum: $c_1\mathbf{v}^{(1)} + c_2\mathbf{v}^{(2)} = \mathbf{x}(0)$.
- Reading stability off the eigenvectors. Eigenvectors give directions; only the eigenvalues give stability.
- Calling a saddle "stable along one direction". A saddle is unstable, full stop — almost every initial condition leads away.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.