Chapter 7 · Section 7.4

7.4 Homogeneous Constant-Coefficient Systems (2×2)

Section 7.3 showed that the homogeneous system

x^{'} y^{'} = a_{11} x + a_{12} y, = a_{21} x + a_{22} y (7.4.1)

has a general solution built from any two independent solutions. It did not say how to find them. When the $a_{ij}$ are real constants, there is a direct method, and it is the same idea that solved constant-coefficient equations in Section 4.2: look for exponential solutions. The method also explains the "normal modes" of Section 7.2, and it leads to a beautiful geometric picture of all solutions at once, the phase plane.

The operator method of Section 7.1 would also work here. The advantage of the approach below is that it produces both components together and with the correct number of constants, without any matching afterwards.

Trying an exponential

We look for solutions in which both unknowns are multiples of the same exponential:

x = A e^{λ t}, y = B e^{λ t},

with constants $A$ , $B$ , $λ$ to be found and $(A, B) \neq = (0, 0)$ . Then $x^{'} = λ A e^{λ t}$ and $y^{'} = λ B e^{λ t}$ . Substituting into (7.4.1) and dividing by $e^{λ t}$ (which is never zero):

λ A λ B = a_{11} A + a_{12} B, = a_{21} A + a_{22} B, ⟺ (a_{11} - λ) A + a_{12} B a_{21} A + (a_{22} - λ) B = 0, = 0. (7.4.2)

This is a homogeneous linear algebraic system for $A$ and $B$ . It always has the useless solution $A = B = 0$ . It has a solution with $(A, B) \neq = (0, 0)$ exactly when its determinant is zero:

a_{11} - λ a_{21} a_{12} a_{22} - λ = 0.

Expanding the determinant gives a quadratic equation for $λ$ .

Definition 7.4.1 (Characteristic equation of the system)

The characteristic equation of the system (7.4.1) is $λ^{2} - τ λ + δ = 0, where τ = a_{11} + a_{22}, δ = a_{11} a_{22} - a_{12} a_{21} .$ $τ$ is the trace and $δ$ the determinant of the coefficient array. Its roots $λ = \frac{τ \pm τ ^{2} - 4 δ}{2}$ are the characteristic roots of the system.

Two connections are worth noticing. First, the operator determinant of Section 7.1 for this system is $D - a_{11} - a_{21} - a_{12} D - a_{22} = D^{2} - τ D + δ$ , the same polynomial with $D$ in place of $λ$ . Second, in the language of matrices (Sections 7.5–7.6) the numbers $λ$ are the eigenvalues of the coefficient matrix and the pairs $(A, B)$ are its eigenvectors.

Finding $A$ and $B$

When $λ$ is a root, the two equations (7.4.2) are proportional (that is what a zero determinant means), so one equation is enough. If $a_{12} \neq = 0$ , the first equation is solved by

A = a_{12}, B = λ - a_{11},

or by any nonzero multiple of this pair. If $a_{12} = 0$ , use the second equation instead. Always check your pair in the other equation: if it fails, $λ$ was computed wrongly.

Three cases arise, exactly as in Chapter 4: the discriminant $τ^{2} - 4 δ$ is positive, negative, or zero.

Case 1: distinct real roots

Theorem 7.4.1 (Distinct real roots)

If the characteristic roots $λ_{1} \neq = λ_{2}$ are real, and $(A_{1}, B_{1})$ , $(A_{2}, B_{2})$ are nonzero solutions of (7.4.2) for $λ_{1}$ and $λ_{2}$ respectively, then $(A_{1} e^{λ_{1} t}, B_{1} e^{λ_{1} t}), (A_{2} e^{λ_{2} t}, B_{2} e^{λ_{2} t})$ form a fundamental set on $(- \infty, \infty)$ , and the general solution is $x = c_{1} A_{1} e^{λ_{1} t} + c_{2} A_{2} e^{λ_{2} t}, y = c_{1} B_{1} e^{λ_{1} t} + c_{2} B_{2} e^{λ_{2} t} .$

Why independent? At $t = 0$ the Wronskian is $W (0) = A_{1} B_{2} - A_{2} B_{1}$ . If it were zero, the pairs would be proportional, and one nonzero pair $(A, B)$ would satisfy the first equation of (7.4.2) for both roots: $(a_{11} - λ_{1}) A + a_{12} B = 0$ and $(a_{11} - λ_{2}) A + a_{12} B = 0$ . Subtracting gives $(λ_{2} - λ_{1}) A = 0$ , so $A = 0$ . The second equation gives $B = 0$ in the same way. That contradicts $(A, B) \neq = (0, 0)$ . So $W (0) \neq = 0$ .

Example 7.4.1

Solve $x^{'} = - x + 4 y$ , $y^{'} = x - y$ with $x (0) = 0$ , $y (0) = 2$ .

Solution

Characteristic equation. $τ = - 1 + (- 1) = - 2$ and $δ = (- 1) (- 1) - 4 \cdot 1 = - 3$ : $λ^{2} + 2 λ - 3 = (λ + 3) (λ - 1) = 0, λ_{1} = 1, λ_{2} = - 3.$
Pair for $λ_{1} = 1$ . First equation of (7.4.2): $(- 1 - 1) A + 4 B = 0$ , i.e. $A = 2 B$ . Take $(A_{1}, B_{1}) = (2, 1)$ . Check in the second: $1 \cdot 2 + (- 1 - 1) \cdot 1 = 0$ . ✓
Pair for $λ_{2} = - 3$ . $(- 1 + 3) A + 4 B = 0$ , i.e. $A = - 2 B$ . Take $(A_{2}, B_{2}) = (2, - 1)$ . Check: $2 + (- 1 + 3) (- 1) = 0$ . ✓
General solution. $x = 2 c_{1} e^{t} + 2 c_{2} e^{- 3 t}, y = c_{1} e^{t} - c_{2} e^{- 3 t} .$
Initial conditions. $2 c_{1} + 2 c_{2} = 0$ and $c_{1} - c_{2} = 2$ , so $c_{1} = 1$ , $c_{2} = - 1$ : $x = 2 e^{t} - 2 e^{- 3 t}, y = e^{t} + e^{- 3 t} .$
Behaviour. The roots have opposite signs. The $e^{- 3 t}$ part dies out and the $e^{t}$ part grows, so for large $t$ the point $(x, y)$ runs off to infinity along the line $x = 2 y$ . The origin is a saddle (Figure 7.4.1).

Example 7.4.2

Solve $x^{'} = - 5 x + 2 y$ , $y^{'} = x - 4 y$ with $x (0) = 4$ , $y (0) = 1$ .

Solution

Characteristic equation. $τ = - 9$ , $δ = 20 - 2 = 18$ : $λ^{2} + 9 λ + 18 = (λ + 3) (λ + 6) = 0, λ_{1} = - 3, λ_{2} = - 6.$
Pairs. Using $A = a_{12} = 2$ , $B = λ - a_{11} = λ + 5$ : $λ_{1} = - 3$ gives $(2, 2)$ , or more simply $(1, 1)$ ; $λ_{2} = - 6$ gives $(2, - 1)$ . Check in the second equation: $1 + (- 4 + 3) \cdot 1 = 0$ ✓ and $2 + (- 4 + 6) (- 1) = 0$ ✓.
General solution. $x = c_{1} e^{- 3 t} + 2 c_{2} e^{- 6 t}, y = c_{1} e^{- 3 t} - c_{2} e^{- 6 t} .$
Initial conditions. $c_{1} + 2 c_{2} = 4$ and $c_{1} - c_{2} = 1$ . Subtracting, $3 c_{2} = 3$ : $c_{2} = 1$ , $c_{1} = 2$ .
Answer. $x = 2 e^{- 3 t} + 2 e^{- 6 t}$ , $y = 2 e^{- 3 t} - e^{- 6 t}$ . Both roots are negative, so every solution tends to the origin: a stable node (Figure 7.4.2).

Case 2: complex roots

If $τ^{2} - 4 δ < 0$ , the roots are $λ = α \pm i β$ with $α = τ /2$ and $β = \frac{1}{2} 4 δ - τ^{2} > 0$ . Everything above still works with complex numbers: for $λ = α + i β$ we find a complex pair $(A, B)$ and a complex solution $(A e^{λ t}, B e^{λ t})$ . We want real solutions, and the following fact delivers them.

Theorem 7.4.2 (Real and imaginary parts)

If $x = u_{1} (t) + i v_{1} (t)$ , $y = u_{2} (t) + i v_{2} (t)$ is a complex solution of the real system (7.4.1), then the real part $(u_{1}, u_{2})$ and the imaginary part $(v_{1}, v_{2})$ are both real solutions.

Indeed, substituting into $x^{'} = a_{11} x + a_{12} y$ gives $u_{1}^{'} + i v_{1}^{'} = (a_{11} u_{1} + a_{12} u_{2}) + i (a_{11} v_{1} + a_{12} v_{2})$ . The coefficients are real, so the real parts and the imaginary parts must agree separately; the same holds for the second equation.

To use it, write $(A, B) = (A_{R} + i A_{I}, B_{R} + i B_{I})$ and expand with Euler's formula $e^{(α + i β) t} = e^{α t} (cos βt + i sin βt)$ :

(A_{R} + i A_{I}) (cos βt + i sin βt) = (A_{R} cos βt - A_{I} sin βt) + i (A_{I} cos βt + A_{R} sin βt),

and similarly for $B$ . Taking real and imaginary parts gives two real solutions:

x_{1} x_{2} = e^{α t} (A_{R} cos βt - A_{I} sin βt), = e^{α t} (A_{I} cos βt + A_{R} sin βt), y_{1} y_{2} = e^{α t} (B_{R} cos βt - B_{I} sin βt), = e^{α t} (B_{I} cos βt + B_{R} sin βt) .

Their Wronskian at $t = 0$ is $A_{R} B_{I} - A_{I} B_{R}$ . It cannot be zero: otherwise $(A, B)$ would be a complex multiple of a real nonzero pair, and that real pair would satisfy (7.4.2) with the non-real number $λ$ ; the imaginary part of $(a_{11} - λ) A + a_{12} B = 0$ would then read $- β A = 0$ , and similarly $- βB = 0$ , which is impossible since $β \neq = 0$ . So the two real solutions form a fundamental set. The conjugate root $α - i β$ gives nothing new: its solutions are the complex conjugates of the ones we already have.

Theorem 7.4.3 (Complex roots)

If the characteristic roots are $α \pm i β$ with $β \neq = 0$ , and $(A, B)$ is a nonzero (complex) solution of (7.4.2) for $λ = α + i β$ , then the real and imaginary parts of $(A e^{λ t}, B e^{λ t})$ , given above, form a fundamental set, and the general solution is $c_{1} (x_{1}, y_{1}) + c_{2} (x_{2}, y_{2})$ .

Example 7.4.3

Solve $x^{'} = x + 5 y$ , $y^{'} = - x - 3 y$ with $x (0) = 5$ , $y (0) = - 1$ .

Solution

Characteristic equation. $τ = 1 - 3 = - 2$ and $δ = (1) (- 3) - (5) (- 1) = 2$ : $λ^{2} + 2 λ + 2 = 0, λ = \frac{- 2 \pm 4 - 8}{2} = - 1 \pm i .$ So $α = - 1$ , $β = 1$ .
Complex pair for $λ = - 1 + i$ . $A = a_{12} = 5$ , $B = λ - a_{11} = - 2 + i$ , so $(A, B) = (5, - 2 + i)$ . Check in the second equation: $(- 1) \cdot 5 + (- 3 - (- 1 + i)) (- 2 + i) = - 5 + (- 2 - i) (- 2 + i) = - 5 + (4 - 2 i + 2 i - i^{2}) = - 5 + 5 = 0$ . ✓
Real and imaginary parts. $A_{R} = 5$ , $A_{I} = 0$ , $B_{R} = - 2$ , $B_{I} = 1$ . With $α = - 1$ , $β = 1$ the formulas give $x_{1} x_{2} = 5 e^{- t} cos t, = 5 e^{- t} sin t, y_{1} y_{2} = - e^{- t} (2 cos t + sin t), = e^{- t} (cos t - 2 sin t) .$
General solution. $x = 5 e^{- t} (c_{1} cos t + c_{2} sin t), y = e^{- t} [c_{2} (cos t - 2 sin t) - c_{1} (2 cos t + sin t)] .$
Initial conditions. $x (0) = 5 c_{1} = 5$ and $y (0) = - 2 c_{1} + c_{2} = - 1$ , so $c_{1} = 1$ , $c_{2} = 1$ .
Answer. $x = 5 e^{- t} (cos t + sin t), y = - e^{- t} (cos t + 3 sin t) .$ Both components oscillate with angular frequency $β = 1$ inside the decaying envelope $e^{- t}$ : in the phase plane the point spirals into the origin, a stable spiral (Figure 7.4.3).

Case 3: equal roots

If $τ^{2} = 4 δ$ there is one double root $λ = τ /2$ , and the trial solution gives only one family of solutions. Two situations are possible.

(a) The decoupled case. If $a_{12} = a_{21} = 0$ and $a_{11} = a_{22} = λ$ , the system is just $x^{'} = λ x$ , $y^{'} = λ y$ . Every pair $(A, B)$ satisfies (7.4.2), and the general solution is simply

x = c_{1} e^{λ t}, y = c_{2} e^{λ t} .

(b) All other cases. Now the solutions of (7.4.2) are all multiples of one pair $(A_{1}, B_{1})$ , so we have only one solution $(A_{1} e^{λ t}, B_{1} e^{λ t})$ . Chapter 4 suggests multiplying by $t$ . For systems that is not quite enough; we must also allow a second exponential term. We try

x = (A_{1} t + A_{2}) e^{λ t}, y = (B_{1} t + B_{2}) e^{λ t} .

By the product rule $x^{'} = (λ A_{1} t + A_{1} + λ A_{2}) e^{λ t}$ . The first equation of (7.4.1) requires

(λ A_{1} t + A_{1} + λ A_{2}) e^{λ t} = [(a_{11} A_{1} + a_{12} B_{1}) t + (a_{11} A_{2} + a_{12} B_{2})] e^{λ t} .

Compare the coefficients of $t e^{λ t}$ and of $e^{λ t}$ , and do the same for the second equation. We obtain two pairs of algebraic equations:

(t -terms) (constant terms) (a_{11} - λ) A_{1} + a_{12} B_{1} (a_{11} - λ) A_{2} + a_{12} B_{2} = 0, = A_{1}, a_{21} A_{1} + (a_{22} - λ) B_{1} a_{21} A_{2} + (a_{22} - λ) B_{2} = 0, = B_{1} .

The first pair says that $(A_{1}, B_{1})$ is the pair we already found. The second pair has the same coefficients but right-hand side $(A_{1}, B_{1})$ . Its determinant is zero, so it does not have a unique solution, but in case (b) it always has solutions, and any one of them will do. (Different choices of $(A_{2}, B_{2})$ differ by a multiple of $(A_{1}, B_{1})$ , which only changes the constant $c_{1}$ below.)

Theorem 7.4.4 (Equal roots)

Let $λ$ be a double characteristic root and suppose the system is not of the decoupled form (a). Let $(A_{1}, B_{1}) \neq = (0, 0)$ solve (7.4.2) and let $(A_{2}, B_{2})$ be any solution of $(a_{11} - λ) A_{2} + a_{12} B_{2} = A_{1}, a_{21} A_{2} + (a_{22} - λ) B_{2} = B_{1} .$ Then $(A_{1} e^{λ t}, B_{1} e^{λ t})$ and $((A_{1} t + A_{2}) e^{λ t}, (B_{1} t + B_{2}) e^{λ t})$ form a fundamental set, and the general solution is $x = c_{1} A_{1} e^{λ t} + c_{2} (A_{1} t + A_{2}) e^{λ t}, y = c_{1} B_{1} e^{λ t} + c_{2} (B_{1} t + B_{2}) e^{λ t} .$

The Wronskian at $t = 0$ is $A_{1} B_{2} - A_{2} B_{1}$ . If it were zero, $(A_{2}, B_{2})$ would be a multiple $k (A_{1}, B_{1})$ , and the left-hand sides of the equations for $(A_{2}, B_{2})$ would be $k$ times those of (7.4.2), that is zero, while the right-hand side $(A_{1}, B_{1})$ is not zero. So $W (0) \neq = 0$ .

Common mistake

Trying only $x = A t e^{λ t}$ , $y = B t e^{λ t}$ , by analogy with $x e^{r x}$ in Chapter 4, does not work: comparing the terms without $t$ forces $A = B = 0$ . The extra constants $A_{2}$ , $B_{2}$ are essential.

Example 7.4.4

Solve $x^{'} = - 4 x + y$ , $y^{'} = - x - 2 y$ with $x (0) = 2$ , $y (0) = - 1$ .

Solution

Characteristic equation. $τ = - 6$ , $δ = 8 - (1) (- 1) = 9$ : $λ^{2} + 6 λ + 9 = (λ + 3)^{2} = 0, λ = - 3 (double) .$ The system is not decoupled ( $a_{12} = 1 \neq = 0$ ), so we are in case (b).
First pair. $(- 4 + 3) A_{1} + B_{1} = 0$ , i.e. $B_{1} = A_{1}$ . Take $(A_{1}, B_{1}) = (1, 1)$ .
Second pair. Solve $(- 4 + 3) A_{2} + B_{2} = 1$ and $- A_{2} + (- 2 + 3) B_{2} = 1$ . Both say $- A_{2} + B_{2} = 1$ . Choose $A_{2} = 0$ ; then $B_{2} = 1$ .
Fundamental set. $(e^{- 3 t}, e^{- 3 t})$ and $(t e^{- 3 t}, (t + 1) e^{- 3 t})$ .
Quick check of the second: $x = t e^{- 3 t}$ gives $x^{'} = (1 - 3 t) e^{- 3 t}$ , and $- 4 x + y = (- 4 t + t + 1) e^{- 3 t} = (1 - 3 t) e^{- 3 t}$ . ✓
General solution. $x = (c_{1} + c_{2} t) e^{- 3 t}, y = (c_{1} + c_{2} (t + 1)) e^{- 3 t} .$
Initial conditions. $x (0) = c_{1} = 2$ and $y (0) = c_{1} + c_{2} = - 1$ , so $c_{2} = - 3$ .
Answer. $x = (2 - 3 t) e^{- 3 t}$ , $y = (- 1 - 3 t) e^{- 3 t}$ . The solution tends to the origin; for large $t$ it approaches along the line $y = x$ (Figure 7.4.4).

Method in a nutshell

Compute $τ = a_{11} + a_{22}$ and $δ = a_{11} a_{22} - a_{12} a_{21}$ ; solve $λ^{2} - τ λ + δ = 0$ .
For each root find a nonzero pair $(A, B)$ from one equation of (7.4.2), for example $A = a_{12}$ , $B = λ - a_{11}$ ; check it in the other equation.
Distinct real roots: $c_{1} (A_{1}, B_{1}) e^{λ_{1} t} + c_{2} (A_{2}, B_{2}) e^{λ_{2} t}$ .
Complex roots $α \pm i β$ : use $λ = α + i β$ only; take real and imaginary parts of $(A, B) e^{α t} (cos βt + i sin βt)$ .
Double root: if the system is $x^{'} = λ x$ , $y^{'} = λ y$ , the answer is $(c_{1} e^{λ t}, c_{2} e^{λ t})$ . Otherwise find $(A_{1}, B_{1})$ , then any $(A_{2}, B_{2})$ with right-hand side $(A_{1}, B_{1})$ , and use $(A_{1} t + A_{2}, B_{1} t + B_{2}) e^{λ t}$ as the second solution.
For an IVP, set $t = 0$ and solve for $c_{1}, c_{2}$ .

Practise with the solver: type the coefficients row by row and an initial point. It shows the characteristic roots, the pairs $(A, B)$ , the general and particular solution, and the phase portrait. The preset is Example 7.4.3; change one coefficient at a time and watch the type of the portrait change.

A first look at the phase plane

A solution $(x (t), y (t))$ can be pictured in two ways: as two graphs against $t$ , or as one curve in the $x y$ -plane traced by the point $(x (t), y (t))$ as $t$ increases. The second picture is called a trajectory (or orbit), the $x y$ -plane is the phase plane, and a drawing of several trajectories is a phase portrait. Time is not visible in the picture, but its direction is, and we mark it with arrows. Because the system (7.4.1) does not contain $t$ explicitly, a trajectory depends only on its starting point, and by uniqueness two trajectories never cross.

The origin is always a solution: $x \equiv 0$ , $y \equiv 0$ . It is an equilibrium point (or critical point). If $δ \neq = 0$ it is the only one, since then $a_{11} x + a_{12} y = 0$ , $a_{21} x + a_{22} y = 0$ has only the zero solution. The type of the characteristic roots determines how the other trajectories behave near it.

Straight-line solutions

For a real root $λ$ with pair $(A, B)$ , the solution $c (A, B) e^{λ t}$ stays on the line through the origin in the direction $(A, B)$ . It moves toward the origin if $λ < 0$ and away from it if $λ > 0$ . These straight lines are the skeleton of the portrait; all other trajectories are combinations and bend between them.

The main types

Saddle (real roots of opposite signs, $δ < 0$ ). Two straight-line solutions come in, two go out; all other trajectories come in along the first line and leave along the second, like hyperbolas. Unstable. Example 7.4.1, Figure 7.4.1.
Node (distinct real roots of the same sign). All trajectories enter the origin ( $λ_{1}, λ_{2} < 0$ , stable) or leave it (both positive, unstable). Near the origin they are tangent to the line of the root closer to zero, because that exponential decays more slowly. Example 7.4.2, Figure 7.4.2.
Spiral (complex roots $α \pm i β$ , $α \neq = 0$ ). Trajectories wind around the origin, inward if $α < 0$ (stable) and outward if $α > 0$ (unstable). Example 7.4.3, Figure 7.4.3.
Center (purely imaginary roots, $α = 0$ ). Trajectories are closed curves (ellipses) around the origin; every solution is periodic with period $2 π / β$ . Stable, but not attracting. Example 7.4.5, Figure 7.4.5.
Degenerate node (double root, case (b)). Only one straight line; trajectories turn around and approach (or leave) the origin tangent to it. Example 7.4.4, Figure 7.4.4. In the decoupled case (a) all trajectories are straight rays, a star.

Stability at a glance

Every solution tends to the origin as $t \to \infty$ exactly when both roots have negative real part. For a quadratic $λ^{2} - τ λ + δ$ this happens if and only if $τ < 0 and δ > 0.$ (The sum of the roots is $τ$ and their product is $δ$ .) For the tank system of Section 7.2, for example, $τ = - \frac{3}{5}$ and $δ = \frac{1}{20}$ , so all the salt eventually leaves, whatever the initial amounts.

For spirals and centers, the direction of rotation is found by checking one point. At $(1, 0)$ the velocity is $(x^{'}, y^{'}) = (a_{11}, a_{21})$ : if $a_{21} > 0$ the trajectory crosses the positive $x$ -axis upward, so the rotation is counterclockwise; if $a_{21} < 0$ it is clockwise.

Figure 7.4.1. Saddle:

x^{'} = - x + 4 y

y^{'} = x - y

(roots

1

and

- 3

). The dashed lines are the straight-line solutions: in along

x = - 2 y

(

λ = - 3

), out along

x = 2 y

(

λ = 1

). Highlighted: the solution of Example 7.4.1 from

(0, 2)

Figure 7.4.2. Stable node:

x^{'} = - 5 x + 2 y

y^{'} = x - 4 y

(roots

- 3

- 6

). The fast part

e^{- 6 t}

along

x = - 2 y

disappears first, so trajectories arrive tangent to the slow line

y = x

. Highlighted: Example 7.4.2.

Figure 7.4.3. Stable spiral:

x^{'} = x + 5 y

y^{'} = - x - 3 y

(roots

- 1 \pm i

). Left: the components of Example 7.4.3 against time. Right: the same solution as a trajectory; since

a_{21} = - 1 < 0

it turns clockwise.

Figure 7.4.4. Degenerate node:

x^{'} = - 4 x + y

y^{'} = - x - 2 y

(double root

- 3

). The only straight-line solutions lie on

y = x

(dashed); every trajectory enters the origin tangent to it. Highlighted: Example 7.4.4.

Example 7.4.5 (A center)

Find the general solution of $x^{'} = 2 x - 5 y$ , $y^{'} = x - 2 y$ and describe its trajectories.

Solution

Characteristic equation. $τ = 0$ , $δ = - 4 + 5 = 1$ : $λ^{2} + 1 = 0$ , so $λ = \pm i$ ( $α = 0$ , $β = 1$ ).
Pair for $λ = i$ . $A = a_{12} = - 5$ , $B = i - 2$ ; multiplying by $- 1$ : $(A, B) = (5, 2 - i)$ . Check: $5 + (- 2 - i) (2 - i) = 5 + (- 4 + i^{2}) = 0$ . ✓
Real solutions. With $A_{R} = 5$ , $A_{I} = 0$ , $B_{R} = 2$ , $B_{I} = - 1$ : $x = 5 (c_{1} cos t + c_{2} sin t), y = c_{1} (2 cos t + sin t) + c_{2} (2 sin t - cos t) .$
Trajectories. Every solution is periodic with period $2 π$ , so each trajectory is a closed curve. In fact the combination $x^{2} - 4 x y + 5 y^{2}$ is constant along solutions: its derivative is $2 x x^{'} - 4 x^{'} y - 4 x y^{'} + 10 y y^{'}$ , which, after substituting the system, equals $0$ . The curves $x^{2} - 4 x y + 5 y^{2} = C$ are ellipses (the discriminant $16 - 20$ is negative), centred at the origin. Since $a_{21} = 1 > 0$ , they are traversed counterclockwise (Figure 7.4.5).

Figure 7.4.5. Center:

x^{'} = 2 x - 5 y

y^{'} = x - 2 y

(roots

\pm i

). The trajectories are the ellipses

x^{2} - 4 x y + 5 y^{2} = C

, all traversed counterclockwise in time

2 π

. There is no damping, so nothing spirals in or out.

The phase plane will come back in Section 7.6, where the same ideas are written with matrices and eigenvectors, and in numerical work (Section 8.8), where systems are rarely solvable by formulas but their trajectories can always be computed and drawn.

Exercises

In Exercises 1–6 find the general solution.

$x^{'} = x + 2 y$ , $y^{'} = 4 x - y$ .
Answer
$λ = 3, - 3$ . $x = c_{1} e^{3 t} + c_{2} e^{- 3 t}$ , $y = c_{1} e^{3 t} - 2 c_{2} e^{- 3 t}$ .
$x^{'} = - 2 x + 2 y$ , $y^{'} = x - 3 y$ .
Answer
$λ = - 1, - 4$ . $x = 2 c_{1} e^{- t} + c_{2} e^{- 4 t}$ , $y = c_{1} e^{- t} - c_{2} e^{- 4 t}$ .
$x^{'} = 3 x - 2 y$ , $y^{'} = 4 x - y$ .
Answer
$λ = 1 \pm 2 i$ . $x = e^{t} (c_{1} cos 2 t + c_{2} sin 2 t)$ , $y = e^{t} [c_{1} (cos 2 t + sin 2 t) + c_{2} (sin 2 t - cos 2 t)]$ .
$x^{'} = x - 2 y$ , $y^{'} = 5 x - y$ .
Answer
$λ = \pm 3 i$ . $x = 2 c_{1} cos 3 t + 2 c_{2} sin 3 t$ , $y = c_{1} (cos 3 t + 3 sin 3 t) + c_{2} (sin 3 t - 3 cos 3 t)$ .
$x^{'} = 6 x - y$ , $y^{'} = x + 4 y$ .
Answer
$λ = 5$ (double). $x = (c_{1} + c_{2} t) e^{5 t}$ , $y = (c_{1} + c_{2} (t - 1)) e^{5 t}$ .
$x^{'} = - x - y$ , $y^{'} = 4 x - 5 y$ .
Answer
$λ = - 3$ (double). $x = (c_{1} + c_{2} t) e^{- 3 t}$ , $y = (2 c_{1} + c_{2} (2 t - 1)) e^{- 3 t}$ .
Solve the system of Exercise 1 with $x (0) = 3$ , $y (0) = 0$ .
Answer
$x = 2 e^{3 t} + e^{- 3 t}$ , $y = 2 e^{3 t} - 2 e^{- 3 t}$ .
Solve $x^{'} = - x - 4 y$ , $y^{'} = 2 x - 5 y$ with $x (0) = 2$ , $y (0) = 1$ .
Answer
$λ = - 3 \pm 2 i$ . $x = 2 e^{- 3 t} cos 2 t$ , $y = e^{- 3 t} (cos 2 t + sin 2 t)$ .
Solve the system of Exercise 6 with $x (0) = 1$ , $y (0) = 0$ .
Answer
$x = (1 + 2 t) e^{- 3 t}$ , $y = 4 t e^{- 3 t}$ .
Classify the equilibrium at the origin (type and stability): (a) $x^{'} = 2 x + y$ , $y^{'} = x + 2 y$ ; (b) $x^{'} = x + 4 y$ , $y^{'} = x + y$ ; (c) $x^{'} = - x - 2 y$ , $y^{'} = 4 x - y$ ; (d) $x^{'} = 3 x - 5 y$ , $y^{'} = 2 x - 3 y$ .
Answer
(a) $λ = 1, 3$ : unstable node. (b) $λ = 3, - 1$ : saddle. (c) $λ = - 1 \pm 22 i$ : stable spiral, counterclockwise. (d) $λ = \pm i$ : center.
Solve $x^{'} = - x$ , $y^{'} = 3 x + 2 y$ (a) one equation at a time, (b) by the method of this section, and compare.
Answer
$x = c_{1} e^{- t}$ , $y = - c_{1} e^{- t} + c_{2} e^{2 t}$ . Pairs: $(1, - 1)$ for $λ = - 1$ and $(0, 1)$ for $λ = 2$ .
Solve the tank system of Example 7.2.3, $x^{'} = - \frac{1}{5} x + \frac{3}{20} y$ , $y^{'} = \frac{1}{5} x - \frac{2}{5} y$ , $x (0) = 8$ , $y (0) = 0$ , by the method of this section.
Answer
$λ = - \frac{1}{10}$ with $(3, 2)$ , $λ = - \frac{1}{2}$ with $(1, - 2)$ ; $x = 6 e^{- t /10} + 2 e^{- t /2}$ , $y = 4 e^{- t /10} - 4 e^{- t /2}$ .
For the system $x^{'} = - 2 x + k y$ , $y^{'} = x - 2 y$ , find the values of the constant $k$ for which the origin is a spiral, a node, or a saddle.
Answer
$τ = - 4$ , $δ = 4 - k$ , discriminant $4 k$ . $k < 0$ : stable spiral; $k = 0$ : double root $- 2$ and, since $a_{21} = 1 \neq = 0$ , a degenerate stable node; $0 < k < 4$ : stable node; $k > 4$ : saddle. ( $k = 4$ gives the root $0$ and a whole line of equilibria.)

Trying an exponential

Finding A and B

Case 1: distinct real roots

Case 2: complex roots

Case 3: equal roots

A first look at the phase plane

Straight-line solutions

The main types

Exercises

Finding $A$ and $B$