Chapter 7 · Section 7.3

7.3 Basic Theory of Linear Systems in Normal Form (2×2)

The operator method of Section 7.1 produces answers, but it says little about why the answers have the form they do, or when a solution exists at all. In Chapter 4 we answered such questions for a single linear equation: existence and uniqueness, superposition, the Wronskian, fundamental sets. This section does the same for systems. We work with two unknowns, where everything can be written out in full, but almost every statement carries over word for word to $n$ unknowns (Sections 7.7 and 7.8).

The theory is cleanest for systems in which each equation is solved for one derivative. That is the normal form, and it is the form used throughout the rest of the chapter, in numerical methods (Section 8.8) and in engineering software.

Systems in normal form

Definition 7.3.1 (Linear system in normal form)

A linear system of two first-order equations in normal form is

x^{'} y^{'} = a_{11} (t) x + a_{12} (t) y + F_{1} (t), = a_{21} (t) x + a_{22} (t) y + F_{2} (t), (7.3.1)

where the coefficients $a_{ij}$ and the forcing terms $F_{1}, F_{2}$ are given functions on an interval $I$ . If $F_{1} (t) \equiv 0$ and $F_{2} (t) \equiv 0$ on $I$ the system is homogeneous; otherwise it is nonhomogeneous. A solution on $I$ is a pair of differentiable functions $x = f (t)$ , $y = g (t)$ satisfying both equations at every $t \in I$ .

We will often write a solution as a column, $(f (t) g (t))$ , or as an ordered pair $(f (t), g (t))$ . Think of it as a point that moves in the $x y$ -plane as time goes on. Section 7.4 makes much use of that picture.

For example, $x^{'} = - 3 x + y$ , $y^{'} = 2 x - 4 y$ is homogeneous with constant coefficients, while $x^{'} = t x - y + cos t$ , $y^{'} = e^{t} x + 5$ is nonhomogeneous with variable coefficients.

Every higher-order linear equation is a system

Normal form is more general than it looks. Take a second-order linear equation written with leading coefficient $1$ ,

x^{''} + p (t) x^{'} + q (t) x = F (t) .

Introduce the position and the velocity as two new unknowns: $x_{1} = x$ and $x_{2} = x^{'}$ . Then $x_{1}^{'} = x_{2}$ by definition, and the equation itself says $x_{2}^{'} = x^{''} = - q x_{1} - p x_{2} + F$ . So

x_{1}^{'} x_{2}^{'} = x_{2}, = - q (t) x_{1} - p (t) x_{2} + F (t),

a system in normal form. Conversely, the first component of any solution of this system solves the original equation. An $n$ th-order equation becomes a system of $n$ equations in the same way ( $x_{1} = x$ , $x_{2} = x^{'}$ , …, $x_{n} = x^{(n - 1)}$ ), and two second-order equations, such as those of the coupled springs in Section 7.2, become four first-order equations. This is how computers solve higher-order equations (Section 8.8).

Example 7.3.1

Write the initial-value problem $(t - 2) x^{''} + 3 x^{'} - t x = e^{t}, x (0) = 1, x^{'} (0) = - 1,$ as a system in normal form, and state the initial conditions for the new unknowns.

Solution

Normalize. For $t \neq = 2$ divide by $t - 2$ : $x^{''} = \frac{t}{t - 2} x - \frac{3}{t - 2} x^{'} + \frac{e ^{t}}{t - 2} .$
New unknowns. Put $x_{1} = x$ , $x_{2} = x^{'}$ . Then $x_{1}^{'} x_{2}^{'} = x_{2}, = \frac{t}{t - 2} x_{1} - \frac{3}{t - 2} x_{2} + \frac{e ^{t}}{t - 2} .$ Here $a_{11} = 0$ , $a_{12} = 1$ , $a_{21} = \frac{t}{t - 2}$ , $a_{22} = - \frac{3}{t - 2}$ , $F_{1} = 0$ , $F_{2} = \frac{e ^{t}}{t - 2}$ .
Initial conditions. $x_{1} (0) = x (0) = 1$ and $x_{2} (0) = x^{'} (0) = - 1$ .
Remark. The coefficients are continuous everywhere except at $t = 2$ . We will see in a moment that this guarantees a unique solution on $(- \infty, 2)$ , the largest interval containing $t_{0} = 0$ on which they are continuous.

Existence and uniqueness

For a system we prescribe the value of each unknown at the initial time: two numbers for two unknowns.

Theorem 7.3.1 (Existence and uniqueness)

Suppose $a_{11}, a_{12}, a_{21}, a_{22}, F_{1}, F_{2}$ are continuous on an interval $I$ . Let $t_{0} \in I$ and let $x_{0}, y_{0}$ be any two numbers. Then the system (7.3.1) has exactly one solution $x = f (t)$ , $y = g (t)$ with $f (t_{0}) = x_{0}, g (t_{0}) = y_{0},$ and this solution exists on the whole interval $I$ .

Three remarks.

As in Theorem 4.1.1, a linear system cannot "blow up" inside $I$ : the solution lives on the entire interval of continuity. Nonlinear systems do not have this property.
When all coefficients are constants, $I = (- \infty, \infty)$ .
Uniqueness has a geometric meaning for homogeneous systems with constant coefficients: two different solution curves in the $x y$ -plane can never cross, because at a crossing point both would satisfy the same initial condition. Figure 7.3.2 illustrates this.

A useful special case: the only solution of a homogeneous system with $x (t_{0}) = 0$ , $y (t_{0}) = 0$ is the trivial solution $x \equiv 0$ , $y \equiv 0$ , since the zero pair obviously solves it and there is only one solution.

Homogeneous systems

From now on until the last part of the section we consider the homogeneous system

x^{'} y^{'} = a_{11} (t) x + a_{12} (t) y, = a_{21} (t) x + a_{22} (t) y, (7.3.2)

with coefficients continuous on $I$ .

Theorem 7.3.2 (Superposition)

If $(f_{1}, g_{1})$ and $(f_{2}, g_{2})$ are solutions of (7.3.2) on $I$ , then for any constants $c_{1}, c_{2}$ the pair $x = c_{1} f_{1} + c_{2} f_{2}, y = c_{1} g_{1} + c_{2} g_{2}$ is also a solution on $I$ .

The proof is one line per equation. For the first one,

x^{'} = c_{1} f_{1}^{'} + c_{2} f_{2}^{'} = c_{1} (a_{11} f_{1} + a_{12} g_{1}) + c_{2} (a_{11} f_{2} + a_{12} g_{2}) = a_{11} (c_{1} f_{1} + c_{2} f_{2}) + a_{12} (c_{1} g_{1} + c_{2} g_{2}) = a_{11} x + a_{12} y,

and the second is identical. Note that the same constant multiplies both components of a solution.

Linear independence of solutions

Superposition builds new solutions from old ones. To build all solutions we need two that are "really different".

Definition 7.3.2 (Linear dependence and independence)

Two pairs of functions $(f_{1}, g_{1})$ and $(f_{2}, g_{2})$ are linearly dependent on $I$ if there are constants $c_{1}, c_{2}$ , not both zero, such that $c_{1} f_{1} (t) + c_{2} f_{2} (t) = 0 and c_{1} g_{1} (t) + c_{2} g_{2} (t) = 0 for all t \in I .$ Otherwise they are linearly independent on $I$ .

In words: two pairs are dependent exactly when one is a constant multiple of the other, with the same constant for both components. For instance, $(2 e^{- t}, - e^{- t})$ and $(- 6 e^{- t}, 3 e^{- t})$ are dependent (the second is $- 3$ times the first), while $(e^{t}, e^{t})$ and $(e^{t}, 2 e^{t})$ are independent even though their first components coincide.

The Wronskian

Definition 7.3.3 (Wronskian of two solution pairs)

The Wronskian of the pairs $(f_{1}, g_{1})$ and $(f_{2}, g_{2})$ is the determinant whose columns are the two pairs: $W (t) = f_{1} (t) g_{1} (t) f_{2} (t) g_{2} (t) = f_{1} (t) g_{2} (t) - f_{2} (t) g_{1} (t) .$

Compare this with Chapter 4: there the second row contained derivatives $y_{1}^{'}, y_{2}^{'}$ . For the system obtained from $x^{''} + p x^{'} + q x = 0$ by $x_{1} = x$ , $x_{2} = x^{'}$ , the second components are the derivatives, so the two Wronskians coincide.

How does $W$ change in time? Differentiate the determinant row by row (product rule):

W^{'} = (f_{1}^{'} g_{2} - f_{2}^{'} g_{1}) + (f_{1} g_{2}^{'} - f_{2} g_{1}^{'}) .

Now substitute the system. In the first bracket use $f_{k}^{'} = a_{11} f_{k} + a_{12} g_{k}$ :

f_{1}^{'} g_{2} - f_{2}^{'} g_{1} = (a_{11} f_{1} + a_{12} g_{1}) g_{2} - (a_{11} f_{2} + a_{12} g_{2}) g_{1} = a_{11} (f_{1} g_{2} - f_{2} g_{1}) + a_{12} (g_{1} g_{2} - g_{2} g_{1}) = a_{11} W .

In the second bracket use $g_{k}^{'} = a_{21} f_{k} + a_{22} g_{k}$ ; the same cancellation leaves $a_{22} W$ . Therefore

W^{'} = (a_{11} + a_{22}) W,

a first-order linear equation for $W$ (Section 2.3). Solving it gives the following formula.

Theorem 7.3.3 (Abel's formula)

If $(f_{1}, g_{1})$ and $(f_{2}, g_{2})$ are solutions of (7.3.2) on $I$ and $t_{0} \in I$ , then $W (t) = W (t_{0}) exp (\int_{t_{0}}^{t} (a_{11} (s) + a_{22} (s)) d s) for all t \in I .$ In particular, $W$ is either zero everywhere on $I$ or zero nowhere on $I$ .

The sum $a_{11} + a_{22}$ of the diagonal coefficients is called the trace of the system. For constant coefficients Abel's formula reads simply $W (t) = W (0) e^{(a_{11} + a_{22}) t}$ .

Theorem 7.3.4 (Wronskian test)

Two solutions of the homogeneous system (7.3.2) are linearly independent on $I$ if and only if their Wronskian is different from zero at one (and then, by Abel's formula, at every) point of $I$ .

Why. If the solutions are dependent, one is a multiple of the other, so the columns of $W$ are proportional and $W \equiv 0$ . Conversely, suppose $W (t_{0}) = 0$ at some $t_{0}$ . Then the algebraic system $c_{1} f_{1} (t_{0}) + c_{2} f_{2} (t_{0}) = 0$ , $c_{1} g_{1} (t_{0}) + c_{2} g_{2} (t_{0}) = 0$ has a solution $(c_{1}, c_{2}) \neq = (0, 0)$ , because its determinant is zero. With these constants, $x = c_{1} f_{1} + c_{2} f_{2}$ , $y = c_{1} g_{1} + c_{2} g_{2}$ is a solution (superposition) that vanishes at $t_{0}$ . By uniqueness it is the trivial solution, so $c_{1} f_{1} + c_{2} f_{2} \equiv 0$ and $c_{1} g_{1} + c_{2} g_{2} \equiv 0$ : the solutions are dependent.

The test is for solutions only

For pairs of functions that are not solutions of a common system with continuous coefficients, a Wronskian that vanishes (even identically) does not imply dependence. Example 7.3.5 shows two independent pairs with $W \equiv 0$ .

Fundamental sets and the general solution

Definition 7.3.4 (Fundamental set)

Two linearly independent solutions of the homogeneous system (7.3.2) on $I$ form a fundamental set of solutions on $I$ .

Theorem 7.3.5 (General solution of the homogeneous system)

The homogeneous system (7.3.2) with coefficients continuous on $I$ has a fundamental set on $I$ . If $(f_{1}, g_{1})$ , $(f_{2}, g_{2})$ is a fundamental set, then every solution on $I$ can be written as $x = c_{1} f_{1} (t) + c_{2} f_{2} (t), y = c_{1} g_{1} (t) + c_{2} g_{2} (t)$ for suitable constants $c_{1}, c_{2}$ . This expression is called the general solution.

Why. Existence: pick $t_{0} \in I$ and let $(f_{1}, g_{1})$ be the solution with initial values $(1, 0)$ and $(f_{2}, g_{2})$ the solution with initial values $(0, 1)$ (Theorem 7.3.1). Their Wronskian at $t_{0}$ is $1 \cdot 1 - 0 \cdot 0 = 1 \neq = 0$ . Every solution is a combination: given a solution $(x, y)$ , solve $c_{1} f_{1} (t_{0}) + c_{2} f_{2} (t_{0}) = x (t_{0}), c_{1} g_{1} (t_{0}) + c_{2} g_{2} (t_{0}) = y (t_{0})$ for $c_{1}, c_{2}$ ; this is possible because the determinant is $W (t_{0}) \neq = 0$ . Then $(x, y)$ and $c_{1} (f_{1}, g_{1}) + c_{2} (f_{2}, g_{2})$ are two solutions with the same initial values, so by uniqueness they are equal.

The argument is also the practical recipe for initial-value problems: once a fundamental set is known, an IVP is just a $2 \times 2$ linear system for $c_{1}$ and $c_{2}$ .

Nonhomogeneous systems

Theorem 7.3.6 (General solution of the nonhomogeneous system)

Let $(x_{p}, y_{p})$ be any one solution of the nonhomogeneous system (7.3.1) on $I$ , and let $(f_{1}, g_{1})$ , $(f_{2}, g_{2})$ be a fundamental set of the corresponding homogeneous system (7.3.2). Then every solution of (7.3.1) on $I$ has the form $x = c_{1} f_{1} + c_{2} f_{2} + x_{p}, y = c_{1} g_{1} + c_{2} g_{2} + y_{p} .$

The reason: if $(x, y)$ is any solution of (7.3.1), the difference $(x - x_{p}, y - y_{p})$ solves the homogeneous system, because the forcing terms cancel when we subtract. By Theorem 7.3.5 that difference is $c_{1} (f_{1}, g_{1}) + c_{2} (f_{2}, g_{2})$ . This is the same structure "complementary + particular" that you know from Chapter 4.

Method in a nutshell: using the theory

Check that the coefficients and forcing are continuous on an interval $I$ containing $t_{0}$ ; the solution then exists and is unique on all of $I$ .
Find (or verify) two solutions of the homogeneous system. Compute their Wronskian at one convenient point. If it is not zero, they form a fundamental set. Abel's formula is a free check of your algebra.
For a nonhomogeneous system add one particular solution.
For an IVP substitute $t = t_{0}$ and solve the $2 \times 2$ linear system for $c_{1}, c_{2}$ .

How do we find the two homogeneous solutions? For constant coefficients this is the subject of Section 7.4. You can already experiment with the solver below: enter the coefficient matrix row by row ( $a_{11}, a_{12}; a_{21}, a_{22}$ ) and an initial point, and it returns a fundamental set, the solution of the IVP and its phase-plane picture. The preset is the system of Example 7.3.2.

Example 7.3.2

Consider $x^{'} = - 3 x + y$ , $y^{'} = 2 x - 4 y$ .

Verify that $(e^{- 2 t}, e^{- 2 t})$ and $(e^{- 5 t}, - 2 e^{- 5 t})$ are solutions.
Show that they form a fundamental set and check Abel's formula.
Write the general solution and solve the IVP $x (0) = 3$ , $y (0) = 0$ .

Solution

First pair. $x = y = e^{- 2 t}$ give $x^{'} = - 2 e^{- 2 t}$ and $- 3 x + y = - 2 e^{- 2 t}$ ; $y^{'} = - 2 e^{- 2 t}$ and $2 x - 4 y = - 2 e^{- 2 t}$ . Both equations hold.
Second pair. $x = e^{- 5 t}$ , $y = - 2 e^{- 5 t}$ : $x^{'} = - 5 e^{- 5 t}$ and $- 3 x + y = (- 3 - 2) e^{- 5 t} = - 5 e^{- 5 t}$ ; $y^{'} = 10 e^{- 5 t}$ and $2 x - 4 y = (2 + 8) e^{- 5 t} = 10 e^{- 5 t}$ . Both hold.
Wronskian. $W (t) = e^{- 2 t} e^{- 2 t} e^{- 5 t} - 2 e^{- 5 t} = - 2 e^{- 7 t} - e^{- 7 t} = - 3 e^{- 7 t} \neq = 0.$ So the solutions are independent on $(- \infty, \infty)$ and form a fundamental set.
Abel's check. The trace is $a_{11} + a_{22} = - 3 + (- 4) = - 7$ , so Abel predicts $W (t) = W (0) e^{- 7 t} = - 3 e^{- 7 t}$ . It matches.
General solution. $x = c_{1} e^{- 2 t} + c_{2} e^{- 5 t}, y = c_{1} e^{- 2 t} - 2 c_{2} e^{- 5 t} .$
IVP. At $t = 0$ : $c_{1} + c_{2} = 3$ and $c_{1} - 2 c_{2} = 0$ . Subtracting, $3 c_{2} = 3$ , so $c_{2} = 1$ and $c_{1} = 2$ : $x = 2 e^{- 2 t} + e^{- 5 t}, y = 2 e^{- 2 t} - 2 e^{- 5 t} .$ Figure 7.3.1 shows the two components; Figure 7.3.2 shows the same solution as a curve in the $x y$ -plane.

Figure 7.3.1. The solution of Example 7.3.2 as two functions of time. Salt-in-two-tanks behaviour again:

y

starts at zero, is "fed" by

x

, peaks, and then both decay.

Figure 7.3.2. The same system drawn in the

x y

-plane. Each arrow shows the direction

(x^{'}, y^{'})

at that point; each curve is one solution, traced as

t

increases. The highlighted curve starts at

(3, 0)

(Example 7.3.2). By uniqueness, no two curves cross.

Example 7.3.3 (Variable coefficients)

Show that $(t^{2}, t)$ and $(t^{3}, 2 t^{2})$ form a fundamental set of $x^{'} = \frac{1}{t} x + y, y^{'} = - \frac{2}{t ^{2}} x + \frac{3}{t} y (t > 0),$ and solve the IVP $x (1) = 1$ , $y (1) = 0$ . On which interval is the solution guaranteed?

Solution

First pair. $x = t^{2}$ , $y = t$ : $x^{'} = 2 t$ and $\frac{1}{t} t^{2} + t = 2 t$ . ✓. $y^{'} = 1$ and $- \frac{2}{t ^{2}} t^{2} + \frac{3}{t} t = - 2 + 3 = 1$ . ✓
Second pair. $x = t^{3}$ , $y = 2 t^{2}$ : $x^{'} = 3 t^{2}$ and $t^{2} + 2 t^{2} = 3 t^{2}$ . ✓. $y^{'} = 4 t$ and $- 2 t + 6 t = 4 t$ . ✓
Wronskian. $W (t) = t^{2} t t^{3} 2 t^{2} = 2 t^{4} - t^{4} = t^{4} > 0 on (0, \infty) .$ So the pairs form a fundamental set on $(0, \infty)$ .
Abel's check. The trace is $\frac{1}{t} + \frac{3}{t} = \frac{4}{t}$ , so $W (t) = W (1) exp (\int_{1}^{t} \frac{4}{s} d s) = 1 \cdot e^{4 l n t} = t^{4}$ . ✓
IVP. $x = c_{1} t^{2} + c_{2} t^{3}$ , $y = c_{1} t + 2 c_{2} t^{2}$ . At $t = 1$ : $c_{1} + c_{2} = 1$ and $c_{1} + 2 c_{2} = 0$ , so $c_{2} = - 1$ , $c_{1} = 2$ : $x = 2 t^{2} - t^{3}, y = 2 t - 2 t^{2} .$
Interval. The coefficients $\frac{1}{t}$ , $- \frac{2}{t ^{2}}$ , $\frac{3}{t}$ are continuous on $(0, \infty)$ , which contains $t_{0} = 1$ . Theorem 7.3.1 guarantees exactly one solution there. (The formulas happen to make sense for all $t$ , but the theorem promises nothing across $t = 0$ , where the coefficients are undefined.)

Example 7.3.4 (A nonhomogeneous system)

Show that $x_{p} = t$ , $y_{p} = 2 t$ is a solution of $x^{'} = - 3 x + y + t + 1, y^{'} = 2 x - 4 y + 6 t + 2,$ and find the solution with $x (0) = 2$ , $y (0) = - 1$ .

Solution

Check the particular solution. $x_{p}^{'} = 1$ and $- 3 t + 2 t + t + 1 = 1$ . ✓. $y_{p}^{'} = 2$ and $2 t - 8 t + 6 t + 2 = 2$ . ✓
Homogeneous part. The homogeneous system is the one of Example 7.3.2, so by Theorem 7.3.6 $x = c_{1} e^{- 2 t} + c_{2} e^{- 5 t} + t, y = c_{1} e^{- 2 t} - 2 c_{2} e^{- 5 t} + 2 t .$
Initial conditions. $c_{1} + c_{2} = 2$ and $c_{1} - 2 c_{2} = - 1$ . Subtracting, $3 c_{2} = 3$ : $c_{2} = 1$ , $c_{1} = 1$ .
Answer. $x = e^{- 2 t} + e^{- 5 t} + t$ , $y = e^{- 2 t} - 2 e^{- 5 t} + 2 t$ . For large $t$ the solution follows the particular solution $(t, 2 t)$ .

Example 7.3.5 (A vanishing Wronskian that proves nothing)

Let $P_{1} = (t, 1)$ and $P_{2} = (t^{2}, t)$ . Show that $P_{1}$ and $P_{2}$ are linearly independent on $(- \infty, \infty)$ although their Wronskian is identically zero. What does this say about systems they could solve?

Solution

Wronskian. $W = t \cdot t - t^{2} \cdot 1 = 0$ for every $t$ .
Independence. Suppose $c_{1} t + c_{2} t^{2} = 0$ and $c_{1} + c_{2} t = 0$ for all $t$ . Put $t = 0$ in the second: $c_{1} = 0$ . Then $c_{2} t = 0$ for all $t$ , so $c_{2} = 0$ . Hence the pairs are independent.
Conclusion. If $P_{1}$ and $P_{2}$ were solutions of one homogeneous system (7.3.2) with coefficients continuous on an interval $I$ , Theorem 7.3.4 would force $W \neq = 0$ on $I$ . So no such system exists on any interval. The Wronskian test works for solutions, not for arbitrary pairs of functions.

Exercises

Consider $x^{'} = 3 x + y$ , $y^{'} = 2 x + 2 y$ . (a) Verify that $(e^{4 t}, e^{4 t})$ and $(e^{t}, - 2 e^{t})$ are solutions. (b) Compute their Wronskian and check it with Abel's formula. (c) Solve the IVP $x (0) = 0$ , $y (0) = 6$ .
Answer
(b) $W = - 3 e^{5 t}$ , trace $5$ . (c) $x = 2 e^{4 t} - 2 e^{t}$ , $y = 2 e^{4 t} + 4 e^{t}$ .
Write $x^{''} - 4 t x^{'} + 3 x = cos t$ as a system in normal form.
Answer
$x_{1}^{'} = x_{2}$ , $x_{2}^{'} = - 3 x_{1} + 4 t x_{2} + cos t$ (with $x_{1} = x$ , $x_{2} = x^{'}$ ).
Write $x^{'''} + 2 x^{''} - x^{'} + 5 x = 0$ as a system of three first-order equations in normal form.
Answer
$x_{1}^{'} = x_{2}$ , $x_{2}^{'} = x_{3}$ , $x_{3}^{'} = - 5 x_{1} + x_{2} - 2 x_{3}$ .
Find the largest interval on which Theorem 7.3.1 guarantees a unique solution of $x^{'} = \frac{x}{t - 4} + (ln t) y$ , $y^{'} = x + \frac{y}{t + 1}$ , $x (1) = 2$ , $y (1) = 0$ .
Answer
$(0, 4)$ : $ln t$ needs $t > 0$ , and $\frac{1}{t - 4}$ is discontinuous at $4$ .
Without solving, find the Wronskian of any two solutions of $x^{'} = 5 x - 7 y$ , $y^{'} = 3 x - 2 y$ , up to a constant factor.
Answer
$W (t) = C e^{3 t}$ .
Without solving, find the Wronskian of any two solutions of $x^{'} = \frac{2}{t} x - y$ , $y^{'} = t^{3} x + \frac{1}{t} y$ on $t > 0$ , up to a constant factor.
Answer
$W (t) = C t^{3}$ .
Verify that $(t, 1)$ and $(t^{- 1}, - t^{- 2})$ are solutions of $x^{'} = y$ , $y^{'} = \frac{x}{t ^{2}} - \frac{y}{t}$ on $t > 0$ . Compute $W$ , check Abel's formula, and solve the IVP $x (1) = 2$ , $y (1) = 0$ .
Answer
$W = - \frac{2}{t}$ (trace $- \frac{1}{t}$ ). $x = t + \frac{1}{t}$ , $y = 1 - \frac{1}{t ^{2}}$ .
Find a constant solution of $x^{'} = 3 x + y - 5$ , $y^{'} = 2 x + 2 y - 6$ and then the general solution (use Exercise 1).
Answer
$(x_{p}, y_{p}) = (1, 2)$ . $x = c_{1} e^{4 t} + c_{2} e^{t} + 1$ , $y = c_{1} e^{4 t} - 2 c_{2} e^{t} + 2$ .
Can two solutions of a homogeneous system with coefficients continuous on $(- \infty, \infty)$ have Wronskian $W (t) = t - 1$ ? Explain.
Answer
No. By Abel's formula $W$ is either never zero or identically zero, but $t - 1$ vanishes only at $t = 1$ .
Show that the pairs $(2 e^{- t}, - e^{- t})$ and $(- 6 e^{- t}, 3 e^{- t})$ have $W \equiv 0$ , and find constants, not both zero, that show they are dependent.
Answer
$W = 6 e^{- 2 t} - 6 e^{- 2 t} = 0$ ; $3 (2 e^{- t}, - e^{- t}) + 1 \cdot (- 6 e^{- t}, 3 e^{- t}) = (0, 0)$ .
Let $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ be solutions of the nonhomogeneous system (7.3.1). Show that $(x_{1} - x_{2}, y_{1} - y_{2})$ solves the homogeneous system, but $(x_{1} + x_{2}, y_{1} + y_{2})$ in general does not solve (7.3.1).
Answer
Subtracting, the forcing terms cancel. Adding, the forcing appears twice: the sum solves the system with forcing $2 F_{1}, 2 F_{2}$ .
For the system of Exercise 1, find the solutions with initial values $(1, 0)$ and $(0, 1)$ at $t = 0$ , and check that their Wronskian at $t = 0$ is $1$ .
Answer
$(\frac{2}{3} e^{4 t} + \frac{1}{3} e^{t}, \frac{2}{3} e^{4 t} - \frac{2}{3} e^{t})$ and $(\frac{1}{3} e^{4 t} - \frac{1}{3} e^{t}, \frac{1}{3} e^{4 t} + \frac{2}{3} e^{t})$ ; $W (t) = e^{5 t}$ , so $W (0) = 1$ .