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.
Method
For $mathbf{x}' = Amathbf{x}$ with $A$ constant, guess $mathbf{x} = mathbf{v}e^{lambda t}$. Substituting gives $lambdamathbf{v}e^{lambda t} = Amathbf{v}e^{lambda t}$, so the guess works exactly when $$Amathbf{v} = lambdamathbf{v}.$$ The differential equation has become an eigenvalue problem.
Real distinct eigenvalues
$$mathbf{x} = c_1mathbf{v}^{(1)}e^{lambda_1t} + c_2mathbf{v}^{(2)}e^{lambda_2t}.$$
- $det(A - lambda I) = lambda^2 - ( ext{tr}A)lambda + det A = 0$ 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 $c_1mathbf{v}^{(1)} + c_2mathbf{v}^{(2)} = mathbf{x}(0)$ — a $2 imes2$ 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 $ ext{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 $mathbf{x}' = egin{pmatrix} 1 & 1 \ 4 & 1end{pmatrix}mathbf{x}$ with $mathbf{x}(0) = egin{pmatrix} 2 \ 3end{pmatrix}$.
- $ ext{tr}A = 2$, $det A = 1 - 4 = -3$, so $lambda^2 - 2lambda - 3 = 0$ and $lambda = -1, 3$.
- For $lambda = -1$: $(A + I)mathbf{v} = egin{pmatrix} 2 & 1 \ 4 & 2end{pmatrix}mathbf{v} = mathbf{0}$, so $2v_1 + v_2 = 0$ and $mathbf{v}^{(1)} = egin{pmatrix} 1 \ -2end{pmatrix}$.
- For $lambda = 3$: $egin{pmatrix} -2 & 1 \ 4 & -2end{pmatrix}mathbf{v} = mathbf{0}$, so $mathbf{v}^{(2)} = egin{pmatrix} 1 \ 2end{pmatrix}$.
- $mathbf{x}(0)$: $c_1 + c_2 = 2$ and $-2c_1 + 2c_2 = 3$, giving $c_1 = frac14$, $c_2 = frac74$.
- $mathbf{x} = frac14egin{pmatrix} 1 \ -2end{pmatrix}e^{-t} + frac74egin{pmatrix} 1 \ 2end{pmatrix}e^{3t}$. Opposite-sign eigenvalues, so the origin is a saddle and the trajectory eventually follows $egin{pmatrix} 1 \ 2end{pmatrix}$.
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: $c_1mathbf{v}^{(1)} + c_2mathbf{v}^{(2)} = mathbf{x}(0)$.
- 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.