Systems of Differential Equations

Why straight lines: if $\mathbf u(0)$ lies on an eigenvector's line, then $A\mathbf u$ points along that same line, so the motion never leaves it; it only stretches ($\lambda > 0$) or shrinks ($\lambda < 0$) at the rate $e^{\lambda t}$. Any other starting point is a combination of two eigenvectors, each component evolves on its own, and the trajectory is steered by whichever exponential dominates.

Why rotation: a complex eigenvalue $\alpha + \beta i$ with eigenvector $\mathbf a + i\mathbf b$ gives the complex solution $e^{\alpha t}(\cos\beta t + i\sin\beta t)(\mathbf a + i\mathbf b)$. Its real and imaginary parts are real solutions built from $\cos\beta t$ and $\sin\beta t$, which turn at angular rate $\beta$ while the factor $e^{\alpha t}$ scales them, so the sign of $\alpha$ alone decides between spiraling in, spiraling out, and closed orbits.

Eliminating $t$ from $x = ae^{2t}$, $y = be^{3t}$ gives $y = b\left(\tfrac xa\right)^{3/2}$. Near the origin ($t \to -\infty$) the ratio $\tfrac yx = \tfrac ba e^{t}$ tends to $0$, so every trajectory leaves the origin tangent to the $x$-axis, the eigenvector of the smaller eigenvalue $2$; farther out, the faster growth of $y$ bends it toward the vertical. The axes themselves are the two straight-line solutions.

Because $A$ is real, the equation $\mathbf w' = A\mathbf w$ for a complex solution $\mathbf w$ splits into the same equation for its real part and for its imaginary part, so each part is a real solution. Expanding $e^{\alpha t}(\cos\beta t + i\sin\beta t)(\mathbf a + i\mathbf b)$ and separating real from imaginary gives the two formulas. The conjugate eigenvalue adds nothing new — its solution is the complex conjugate — so one complex eigenpair supplies both real solutions.

Step 3 — the real solutions: $\mathbf u_1 = e^{-t}(\cos 4t, \sin 4t)$ and $\mathbf u_2 = e^{-t}(\sin 4t, -\cos 4t)$.

Step 4 — the data: $c_1(1, 0) + c_2(0, -1) = (1, 0)$, so $c_1 = 1$, $c_2 = 0$, and $\mathbf u = e^{-t}(\cos 4t, \sin 4t)$.

Step 5 — read it: the point turns counterclockwise at angular rate $4$ while its distance $e^{-t}$ from the origin shrinks; each full turn takes time $\tfrac\pi2$ and multiplies the distance by $e^{-\pi/2} \approx 0.208$.

Step 6 — verify: $x' = -e^{-t}\cos 4t - 4e^{-t}\sin 4t = -x - 4y$ ✓ and $y' = -e^{-t}\sin 4t + 4e^{-t}\cos 4t = 4x - y$ ✓.

Step 3 — $\mathbf u_1 = e^{-t}(\cos t,\ 2\cos t + \sin t)$ and $\mathbf u_2 = e^{-t}(\sin t,\ 2\sin t - \cos t)$.

Step 4 — the data: $c_1(1, 2) + c_2(0, -1) = (1, 1)$ gives $c_1 = 1$ and $c_2 = 1$.

Step 5 — so $x = e^{-t}(\cos t + \sin t)$ and $y = e^{-t}(\cos t + 3\sin t)$.

Step 6 — verify: $x' = -2e^{-t}\sin t$ and $x - y = -2e^{-t}\sin t$ ✓; $y' = e^{-t}(2\cos t - 4\sin t)$ and $5x - 3y = e^{-t}(2\cos t - 4\sin t)$ ✓. The loops are tilted and stretched because $\mathbf a$ and $\mathbf b$ are neither perpendicular nor of equal length.

Differentiate the proposed solution: $\mathbf u_2' = \lambda e^{\lambda t}(t\mathbf v + \mathbf w) + e^{\lambda t}\mathbf v$, while $A\mathbf u_2 = e^{\lambda t}(t\lambda\mathbf v + A\mathbf w)$. They agree exactly when $A\mathbf w = \lambda\mathbf w + \mathbf v$, that is, $(A - \lambda I)\mathbf w = \mathbf v$. This is the system version of the $xe^{rx}$ rule for a double root; the extra vector $\mathbf w$ is needed because $te^{\lambda t}\mathbf v$ alone falls short by exactly $e^{\lambda t}\mathbf v$, as the note below shows.

Step 3 — the ratio $\tfrac yx = -\tfrac13$ never changes, so the trajectory is the straight ray from $(3, -1)$ into the origin. Every trajectory is such a ray: a star node, stable because $\lambda < 0$. Here two independent eigenvectors exist, so no $\mathbf w$ is needed.

Step 3 — solve $(A + 2I)\mathbf w = \mathbf v$: both rows say $-w_1 + w_2 = 1$, so $\mathbf w = (0, 1)$ works.

Step 4 — $\mathbf u = c_1e^{-2t}(1, 1) + c_2e^{-2t}\left(t(1, 1) + (0, 1)\right)$, and at $t = 0$, $c_1(1, 1) + c_2(0, 1) = (1, 0)$ gives $c_1 = 1$, $c_2 = -1$.

Step 5 — $\mathbf u = e^{-2t}(1 - t,\ -t)$.

Step 6 — verify: $x' = (-3 + 2t)e^{-2t}$ and $-3x + y = (-3 + 3t - t)e^{-2t}$ ✓; $y' = (-1 + 2t)e^{-2t}$ and $-x - y = (-1 + t + t)e^{-2t}$ ✓.

Step 7 — read it: for large $t$ the term $-te^{-2t}(1, 1)$ dominates, so the trajectory crosses the $y$-axis at $t = 1$ and enters the origin along the line $y = x$ from the third quadrant.

The two agree, so every eigenpair yields a straight-line solution; independence of eigenvectors makes their combinations the general solution. $\blacksquare$

The picture follows from the signs alone: both $\lambda < 0$ — every trajectory slides to the origin (stable node); both $> 0$ — everything escapes (source); opposite signs — a saddle: trajectories sweep in along the negative eigenline and out along the positive one, and $\det A = \lambda_1\lambda_2 < 0$ detects this without computing anything else. Complex $\alpha \pm \beta i$: spirals, inward if $\alpha < 0$, outward if $\alpha > 0$, closed ellipses if $\alpha = 0$. Trace and determinant of a $2\times2$ thus classify the whole portrait before any solving happens.

Step 1 — eigenpairs (computed in the eigenvalue topic): $\lambda = 3$ with $(1,1)$; $\lambda = 1$ with $(1,-1)$.

Step 2 — general solution: $\mathbf u = c_1e^{3t}(1,1) + c_2e^{t}(1,-1)$.

Step 3 — initial data: $(3, 1) = c_1(1,1) + c_2(1,-1)$ means $c_1 + c_2 = 3$, $c_1 - c_2 = 1$: $c_1 = 2$, $c_2 = 1$.

Step 4 — the answer: $\mathbf u(t) = 2e^{3t}(1,1) + e^{t}(1,-1)$, i.e. $x = 2e^{3t} + e^t$, $y = 2e^{3t} - e^t$.

Step 5 — behavior: both eigenvalues are positive, so the origin is a source; as $t$ grows the $e^{3t}$ term dominates and the trajectory hugs the line $y = x$ — the fast eigendirection — while for $t \to -\infty$ it aligned with $(1,-1)$. Eigenvectors are the skeleton every trajectory drapes itself on.

For $A$: $\det = 1 - 4 = -3 < 0$ — eigenvalues have opposite signs (their product is negative): a saddle, unstable, no further computation required; indeed $\lambda = 3, -1$ with eigenlines $y = x$ (outflow) and $y = -x$ (inflow). For $B$: $\mathrm{tr} = 0$, $\det = 1$: eigenvalues satisfy $\lambda^2 + 1 = 0$, so $\lambda = \pm i$ — pure imaginary: closed orbits, a center; and indeed $x' = y, y' = -x$ conserves $x^2 + y^2$, so trajectories are literal circles. Two matrices, four numbers total examined, both portraits fully determined.