Chapter 7 · Section 7.1

7.1 Differential Operators and the Operator Method

Up to now every problem had one unknown function. Real devices rarely behave like that. Two masses joined by a spring pull on each other; the currents in two loops of a network share a branch; salt flows back and forth between connected tanks. In each case the rate of change of one quantity depends on the other quantity too, so the equations are coupled: we cannot solve one without knowing the other.

This section gives the oldest and most direct way to deal with coupled linear equations with constant coefficients: eliminate one unknown, exactly as you would in an algebraic system of two equations. To do that cleanly we need a compact notation for derivatives, the differential operator $D$ , and a few rules for calculating with it.

Throughout this chapter the independent variable is time $t$ , and the unknowns are written $x (t)$ , $y (t)$ (or $x_{1} (t)$ , $x_{2} (t)$ , …). Primes mean derivatives with respect to $t$ .

What a linear system looks like

A system of two first-order linear equations in $x$ and $y$ has the general form

a_{1} (t) x^{'} + a_{2} (t) y^{'} + a_{3} (t) x + a_{4} (t) y b_{1} (t) x^{'} + b_{2} (t) y^{'} + b_{3} (t) x + b_{4} (t) y = F_{1} (t), = F_{2} (t) .

Each equation may contain both derivatives. A solution on an interval $I$ is an ordered pair of functions $x = f (t)$ , $y = g (t)$ that make both equations true for every $t$ in $I$ . Satisfying one equation is not enough.

Systems can also contain higher derivatives. Newton's law for two connected bodies, for instance, gives two second-order equations, one for each body (Section 7.2). Systems with three or more unknowns are built in the same way.

In this section all coefficients are constants. That is the case where the operator method works. A particularly tidy type, the normal form, in which each equation is solved for one derivative,

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

will be studied in detail from Section 7.3 on. The operator method handles it too, but it does not need it.

The differential operator $D$

We write $D$ for "differentiate with respect to $t$ ":

D x = x^{'}, D^{2} x = x^{''}, D^{k} x = x^{(k)} .

You met this notation in Section 4.1 (with $x$ as the variable there). A polynomial in $D$ means "take these derivatives, multiply by these constants, and add". For example

(4 D^{2} - D + 7) x = 4 x^{''} - x^{'} + 7 x .

A lone constant $c$ in such an expression means "multiply by $c$ " (think of it as $c D^{0}$ ).

Definition 7.1.1 (Linear differential operator with constant coefficients)

An expression

L = a_{0} D^{n} + a_{1} D^{n - 1} + \dots + a_{n - 1} D + a_{n},

with real constants $a_{0} \neq = 0, a_{1}, \dots, a_{n}$ , is a linear differential operator of order $n$ with constant coefficients. It acts on an $n$ -times differentiable function $f$ by

L f = a_{0} f^{(n)} + a_{1} f^{(n - 1)} + \dots + a_{n - 1} f^{'} + a_{n} f .

The polynomial $L (r) = a_{0} r^{n} + a_{1} r^{n - 1} + \dots + a_{n}$ , obtained by writing $r$ in place of $D$ , is the polynomial associated with $L$ .

D

is not a number

$D^{2} x$ is the second derivative of $x$ , not " $D$ times $D$ times $x$ " in the sense of multiplying numbers, and $D x$ is not $D \cdot x$ . The power of the notation is that, for constant coefficients, operators may nevertheless be added, multiplied and factored as if they were numbers. The next theorem says exactly what is allowed.

The algebra of operators

Two operators are equal if they give the same result on every (sufficiently differentiable) function. With this meaning:

Theorem 7.1.1 (Rules for constant-coefficient operators)

Let $L$ , $L_{1}$ , $L_{2}$ be linear differential operators with constant coefficients, let $c_{1}, c_{2}$ be constants and $f, g$ functions with enough derivatives.

Linearity. $L [c_{1} f + c_{2} g] = c_{1} L f + c_{2} Lg$ .
Products. Applying $L_{2}$ first and then $L_{1}$ gives the same result as the single operator whose associated polynomial is the product $L_{1} (r) L_{2} (r)$ .
Commutativity. $L_{1} L_{2} f = L_{2} L_{1} f$ : the order does not matter.
Factoring. If $L (r) = a_{0} (r - r_{1}) (r - r_{2}) \dots (r - r_{n})$ , then $L = a_{0} (D - r_{1}) (D - r_{2}) \dots (D - r_{n})$ .
Exponentials. $L e^{a t} = L (a) e^{a t}$ for every constant $a$ .

Why are these true? Rule 1 holds because derivatives of sums are sums of derivatives. For rule 2, look at a single product, say $(D - p) (D - q)$ :

(D - p) [(D - q) f] = (D - p) [f^{'} - q f] = (f^{''} - q f^{'}) - p (f^{'} - q f) = f^{''} - (p + q) f^{'} + pq f = (D^{2} - (p + q) D + pq) f .

The last operator is exactly the one built from $(r - p) (r - q) = r^{2} - (p + q) r + pq$ . The key step was moving the constant $q$ through a derivative: $D (q f) = q D f$ . Since $p$ and $q$ enter the answer symmetrically, we also get $(D - p) (D - q) = (D - q) (D - p)$ , which is rule 3. Every constant-coefficient operator is built from such factors, so the rules hold in general. Rule 5 follows from $D^{k} e^{a t} = a^{k} e^{a t}$ .

Commutativity needs constant coefficients

If a coefficient depends on $t$ , the order matters. Compare $D (t D)$ and $(t D) D$ acting on $x$ :

D (t x^{'}) = x^{'} + t x^{''}, t D (x^{'}) = t x^{''} .

They differ by $x^{'}$ . This is why the elimination method below is used only for constant coefficients.

Example 7.1.1

Let $L = D^{2} + 4 D - 1$ . Compute $L f$ for (a) $f = t^{3}$ , (b) $f = e^{- 3 t}$ , (c) $f = t^{3} - 2 e^{- 3 t}$ .

Solution

(a) $f^{'} = 3 t^{2}$ and $f^{''} = 6 t$ , so $L t^{3} = 6 t + 4 (3 t^{2}) - t^{3} = - t^{3} + 12 t^{2} + 6 t .$
(b) Use rule 5 with $a = - 3$ : $L (- 3) = 9 - 12 - 1 = - 4$ , so $L e^{- 3 t} = - 4 e^{- 3 t}$ . (Directly: $9 e^{- 3 t} - 12 e^{- 3 t} - e^{- 3 t} = - 4 e^{- 3 t}$ .)
(c) By linearity there is nothing new to compute: $L [t^{3} - 2 e^{- 3 t}] = L t^{3} - 2 L e^{- 3 t} = - t^{3} + 12 t^{2} + 6 t + 8 e^{- 3 t} .$

Example 7.1.2

Let $L_{1} = D - 2$ , $L_{2} = D^{2} + 3$ and $f (t) = t e^{t}$ . Compute $L_{1} L_{2} f$ , $L_{2} L_{1} f$ and $L f$ , where $L$ is the operator with associated polynomial $L_{1} (r) L_{2} (r)$ .

Solution

Derivatives of $f$ . By the product rule, $f^{'} = (t + 1) e^{t}$ , $f^{''} = (t + 2) e^{t}$ , $f^{'''} = (t + 3) e^{t}$ .
$L_{1} L_{2} f$ (apply $L_{2}$ first). $L_{2} f = f^{''} + 3 f = (t + 2 + 3 t) e^{t} = (4 t + 2) e^{t}$ . Then $(D - 2) [(4 t + 2) e^{t}] = (4 + 4 t + 2) e^{t} - 2 (4 t + 2) e^{t} = (2 - 4 t) e^{t} .$
$L_{2} L_{1} f$ (apply $L_{1}$ first). $L_{1} f = f^{'} - 2 f = (t + 1 - 2 t) e^{t} = (1 - t) e^{t}$ . Its derivatives are $(- t) e^{t}$ and $(- t - 1) e^{t}$ , so $(D^{2} + 3) [(1 - t) e^{t}] = (- t - 1) e^{t} + 3 (1 - t) e^{t} = (2 - 4 t) e^{t} .$
The product operator. $L_{1} (r) L_{2} (r) = (r - 2) (r^{2} + 3) = r^{3} - 2 r^{2} + 3 r - 6$ , so $L = D^{3} - 2 D^{2} + 3 D - 6$ and $L f = (t + 3) e^{t} - 2 (t + 2) e^{t} + 3 (t + 1) e^{t} - 6 t e^{t} = (t - 2 t + 3 t - 6 t) e^{t} + (3 - 4 + 3) e^{t} = (2 - 4 t) e^{t} .$
Conclusion. All three agree, as Theorem 7.1.1 promises.

The operator method for two equations

Consider a system with constant coefficients written in operator form,

L_{1} x + L_{2} y L_{3} x + L_{4} y = f_{1} (t), = f_{2} (t), (7.1.1)

where $L_{1}, \dots, L_{4}$ are polynomials in $D$ (possibly constants, possibly zero). For instance, $3 x^{'} + y^{'} - 5 x = e^{t}$ becomes $(3 D - 5) x + D y = e^{t}$ .

Eliminating one unknown

Recall how you eliminate $y$ from $a x + b y = p$ , $c x + d y = q$ : multiply the first equation by $d$ , the second by $b$ , and subtract. We do exactly the same, but "multiply" now means "apply the operator".

Apply $L_{4}$ to the first equation and $L_{2}$ to the second: $L_{4} L_{1} x + L_{4} L_{2} y = L_{4} f_{1}, L_{2} L_{3} x + L_{2} L_{4} y = L_{2} f_{2} .$
Subtract. Because $L_{4} L_{2} y = L_{2} L_{4} y$ (commutativity!), the $y$ -terms cancel: $(L_{1} L_{4} - L_{2} L_{3}) x = L_{4} f_{1} - L_{2} f_{2} . (7.1.2)$
Do the same for $y$ . Applying $L_{3}$ to the first equation, $L_{1}$ to the second, and subtracting the first result from the second gives $(L_{1} L_{4} - L_{2} L_{3}) y = L_{1} f_{2} - L_{3} f_{1} . (7.1.3)$

Both unknowns now satisfy a single constant-coefficient equation with the same operator on the left,

Δ = L_{1} L_{4} - L_{2} L_{3} = L_{1} L_{3} L_{2} L_{4},

the operator determinant of the system. Equations (7.1.2) and (7.1.3) are solved with the methods of Chapter 4: characteristic equation for the homogeneous part, undetermined coefficients (or variation of parameters) for a particular solution. Notice the resemblance to Cramer's rule; the right-hand sides are "determinants" with $f_{1}, f_{2}$ replacing a column, but the operator must always be written in front of the function it acts on.

Too many constants

Suppose $Δ$ has order $N$ . Solving (7.1.2) gives $x$ with $N$ constants $c_{1}, \dots, c_{N}$ , and solving (7.1.3) gives $y$ with $N$ more constants $k_{1}, \dots, k_{N}$ . That is $2 N$ constants. But elimination only produced consequences of the system: every solution of the system satisfies (7.1.2) and (7.1.3), but not every pair satisfying (7.1.2) and (7.1.3) solves the system. The constants are linked, and we find the links by substituting back into the original equations.

Theorem 7.1.2 (Number of arbitrary constants)

Suppose the operator determinant $Δ = L_{1} L_{4} - L_{2} L_{3}$ of system (7.1.1) is not the zero operator, and let $N$ be its order (the degree of the polynomial $Δ (r)$ ). Then the general solution of (7.1.1) contains exactly $N$ arbitrary constants.

The theorem tells us when to stop: once the relations among the $2 N$ constants leave exactly $N$ of them free, we are done. Two special situations deserve a word.

If $Δ$ is a nonzero constant ( $N = 0$ ), then (7.1.2) and (7.1.3) give $x$ and $y$ directly, with no constants at all. The system has exactly one solution and no initial conditions can be imposed. See Example 7.1.6.
If $Δ$ is the zero operator, the two equations are not really independent. The system then has either no solution or infinitely many, with an arbitrary function in the answer (Exercise 12). Such systems are badly posed and we will not consider them further.

Why should the order of $Δ$ matter? Count derivatives in a simple case. In a normal-form system $x^{'} = \dots$ , $y^{'} = \dots$ we expect to choose $x (0)$ and $y (0)$ freely, two constants, and indeed $Δ = (D - a_{11}) (D - a_{22}) - a_{12} a_{21}$ has order 2. In two second-order equations for two masses we choose two positions and two velocities, four constants, and $Δ$ has order 4 (Section 7.2).

Figure 7.1.1. The operator method at a glance. The shortcut on the right (find

y

algebraically) avoids the second differential equation and the matching of constants.

Method in a nutshell

Write the system in operator form $L_{1} x + L_{2} y = f_{1}$ , $L_{3} x + L_{4} y = f_{2}$ .
Compute $Δ = L_{1} L_{4} - L_{2} L_{3}$ . Its order $N$ is the number of arbitrary constants in the answer.
Solve $Δ x = L_{4} f_{1} - L_{2} f_{2}$ for $x$ (constants $c_{1}, \dots, c_{N}$ ).
Solve $Δ y = L_{1} f_{2} - L_{3} f_{1}$ for $y$ (new constants $k_{1}, \dots, k_{N}$ ). The homogeneous parts use the same basic functions as for $x$ .
Substitute $x$ and $y$ into the original equations and equal the coefficients of each basic function to zero. Solve these relations for $k_{1}, \dots, k_{N}$ in terms of $c_{1}, \dots, c_{N}$ .
Shortcut (often available for first-order systems): instead of steps 4–5, combine the two original equations so that $y^{'}$ (and any higher derivatives of $y$ ) cancel. The result gives $y$ algebraically in terms of $x$ , $x^{'}$ , … . Substitute the $x$ found in step 3.

Common mistakes

Writing down $x$ with $c$ 's and $y$ with independent $k$ 's and stopping there. That pair has $2 N$ constants and is not the general solution.
Substituting into only one equation when that equation gives no information about some term. Check that $N$ constants remain; if more remain, use the other equation as well (Example 7.1.3).
Computing $L_{4} f_{1}$ as " $L_{4}$ times $f_{1}$ ". It means: apply $L_{4}$ to $f_{1}$ , that is, differentiate.

Example 7.1.3

Find the general solution of $x^{'} + y^{'} - 2 x + y 2 x^{'} + y^{'} - 2 x - y = 0, = 0.$

Solution

Operator form. $(D - 2) x + (D + 1) y = 0$ and $(2 D - 2) x + (D - 1) y = 0$ .
Operator determinant. $Δ = (D - 2) (D - 1) - (D + 1) (2 D - 2) = (D^{2} - 3 D + 2) - (2 D^{2} - 2) = - D^{2} - 3 D + 4 = - (D - 1) (D + 4) .$ It has order $N = 2$ , so the answer will contain two constants.
Equations for $x$ and $y$ . The right-hand sides are zero, so $Δ x = 0$ and $Δ y = 0$ . The characteristic roots are $1$ and $- 4$ : $x = c_{1} e^{t} + c_{2} e^{- 4 t}, y = k_{1} e^{t} + k_{2} e^{- 4 t} .$
Substitute into the first equation. Using rule 5, $(D - 2) e^{t} = - e^{t}$ , $(D - 2) e^{- 4 t} = - 6 e^{- 4 t}$ , $(D + 1) e^{t} = 2 e^{t}$ , $(D + 1) e^{- 4 t} = - 3 e^{- 4 t}$ . Hence $(- c_{1} + 2 k_{1}) e^{t} + (- 6 c_{2} - 3 k_{2}) e^{- 4 t} = 0 for all t .$ Since $e^{t}$ and $e^{- 4 t}$ are linearly independent, both coefficients vanish: $k_{1} = \frac{1}{2} c_{1}$ and $k_{2} = - 2 c_{2}$ .
Check the second equation. $(2 D - 2) e^{t} = 0$ and $(D - 1) e^{t} = 0$ , so the $e^{t}$ -terms give $0 = 0$ : this equation says nothing about $k_{1}$ . For $e^{- 4 t}$ : $- 10 c_{2} - 5 k_{2} = - 10 c_{2} + 10 c_{2} = 0$ . Consistent. (Had we used only the second equation, $k_{1}$ would have stayed free and we would have had three constants, one too many.)
General solution. $x = c_{1} e^{t} + c_{2} e^{- 4 t}, y = \frac{1}{2} c_{1} e^{t} - 2 c_{2} e^{- 4 t} .$ If you dislike fractions, rename $c_{1} = 2 a$ : $x = 2 a e^{t} + c_{2} e^{- 4 t}$ , $y = a e^{t} - 2 c_{2} e^{- 4 t}$ . Both forms describe the same family.

Example 7.1.4

Solve $2 x^{'} + y^{'} + 3 x x^{'} + y^{'} - 2 y = 6 e^{t}, = 6 t,$ first in general and then with $x (0) = 3$ , $y (0) = - 1$ . Use the shortcut to find $y$ .

Solution

Operator form. $(2 D + 3) x + D y = 6 e^{t}$ and $D x + (D - 2) y = 6 t$ . So $L_{1} = 2 D + 3$ , $L_{2} = D$ , $L_{3} = D$ , $L_{4} = D - 2$ .
Determinant. $Δ = (2 D + 3) (D - 2) - D \cdot D = 2 D^{2} - D - 6 - D^{2} = D^{2} - D - 6 = (D - 3) (D + 2) .$ Order $2$ : two constants.
Right-hand side for $x$ . $L_{4} f_{1} - L_{2} f_{2} = (D - 2) [6 e^{t}] - D [6 t] = (6 e^{t} - 12 e^{t}) - 6 = - 6 e^{t} - 6.$ So $x^{''} - x^{'} - 6 x = - 6 e^{t} - 6$ .
Solve for $x$ . Homogeneous part: roots $3, - 2$ . Particular part: try $x_{p} = A e^{t} + B$ . Then $x_{p}^{''} - x_{p}^{'} - 6 x_{p} = (1 - 1 - 6) A e^{t} - 6 B$ , so $- 6 A = - 6$ and $- 6 B = - 6$ , i.e. $A = B = 1$ : $x = c_{1} e^{3 t} + c_{2} e^{- 2 t} + e^{t} + 1.$
Shortcut for $y$ . Subtract the second original equation from the first: the $y^{'}$ terms cancel and $x^{'} + 3 x + 2 y = 6 e^{t} - 6 t ⟹ y = \frac{1}{2} (6 e^{t} - 6 t - x^{'} - 3 x) .$
Substitute. $x^{'} = 3 c_{1} e^{3 t} - 2 c_{2} e^{- 2 t} + e^{t}$ and $3 x = 3 c_{1} e^{3 t} + 3 c_{2} e^{- 2 t} + 3 e^{t} + 3$ . So $x^{'} + 3 x = 6 c_{1} e^{3 t} + c_{2} e^{- 2 t} + 4 e^{t} + 3$ and $y = - 3 c_{1} e^{3 t} - \frac{1}{2} c_{2} e^{- 2 t} + e^{t} - 3 t - \frac{3}{2} .$ No constants had to be matched: the shortcut produces the correct number automatically.
General solution. $x y = c_{1} e^{3 t} + c_{2} e^{- 2 t} + e^{t} + 1, = - 3 c_{1} e^{3 t} - \frac{1}{2} c_{2} e^{- 2 t} + e^{t} - 3 t - \frac{3}{2} .$
Initial conditions. At $t = 0$ : $x (0) = c_{1} + c_{2} + 2 = 3$ and $y (0) = - 3 c_{1} - \frac{1}{2} c_{2} + 1 - \frac{3}{2} = - 1$ . So $c_{1} + c_{2} = 1$ and $6 c_{1} + c_{2} = 1$ . Subtracting, $5 c_{1} = 0$ : $c_{1} = 0$ , $c_{2} = 1$ .
Answer. $x = e^{- 2 t} + e^{t} + 1, y = - \frac{1}{2} e^{- 2 t} + e^{t} - 3 t - \frac{3}{2} .$ Figure 7.1.2 shows both functions.

Figure 7.1.2. The solution of Example 7.1.4. Both unknowns are driven by the forcing:

e^{t}

appears in

x

and

y

, while the

- 3 t

y

comes from the forcing

6 t

. The

e^{- 2 t}

terms die out quickly.

Example 7.1.5 (A system with a second derivative)

Find the general solution of $x^{''} - x^{'} - 3 x + y^{'} x^{'} - x + y^{'} - y = 0, = 0.$

Solution

Operator form. $(D^{2} - D - 3) x + D y = 0$ and $(D - 1) x + (D - 1) y = 0$ .
Determinant. $Δ = (D^{2} - D - 3) (D - 1) - D (D - 1) = (D - 1) (D^{2} - D - 3 - D) = (D - 1) (D^{2} - 2 D - 3) = (D - 1) (D - 3) (D + 1) .$ Order $3$ , so the general solution has three constants, not four, even though the first equation is of second order.
Solve for $x$ . $Δ x = 0$ with roots $1, 3, - 1$ : $x = c_{1} e^{3 t} + c_{2} e^{t} + c_{3} e^{- t} .$
Shortcut for $y$ . Subtract the second equation from the first. The $y^{'}$ terms cancel: $x^{''} - 2 x^{'} - 2 x + y = 0 ⟹ y = - x^{''} + 2 x^{'} + 2 x .$
Substitute term by term. For $e^{a t}$ the operator $- D^{2} + 2 D + 2$ gives the factor $- a^{2} + 2 a + 2$ : $a = 3$ : $- 9 + 6 + 2 = - 1$ ; $a = 1$ : $- 1 + 2 + 2 = 3$ ; $a = - 1$ : $- 1 - 2 + 2 = - 1$ .
General solution. $x = c_{1} e^{3 t} + c_{2} e^{t} + c_{3} e^{- t}, y = - c_{1} e^{3 t} + 3 c_{2} e^{t} - c_{3} e^{- t} .$ You can check that this pair satisfies both equations for every choice of $c_{1}, c_{2}, c_{3}$ .

Example 7.1.6 (A determinant that is a constant)

Solve $x^{'} + y^{'} + y x^{'} - x + y^{'} = e^{2 t}, = t .$

Solution

Operator form. $D x + (D + 1) y = e^{2 t}$ and $(D - 1) x + D y = t$ .
Determinant. $Δ = D \cdot D - (D + 1) (D - 1) = D^{2} - (D^{2} - 1) = 1.$ A nonzero constant: by Theorem 7.1.2 there are no arbitrary constants.
Find $x$ . Equation (7.1.2) reads $1 \cdot x = L_{4} f_{1} - L_{2} f_{2}$ : $x = D [e^{2 t}] - (D + 1) [t] = 2 e^{2 t} - 1 - t .$
Find $y$ . Equation (7.1.3) reads $y = L_{1} f_{2} - L_{3} f_{1}$ : $y = D [t] - (D - 1) [e^{2 t}] = 1 - (2 e^{2 t} - e^{2 t}) = 1 - e^{2 t} .$
Check. $x^{'} = 4 e^{2 t} - 1$ , $y^{'} = - 2 e^{2 t}$ . First equation: $4 e^{2 t} - 1 - 2 e^{2 t} + 1 - e^{2 t} = e^{2 t}$ . Second: $4 e^{2 t} - 1 - 2 e^{2 t} + 1 + t - 2 e^{2 t} = t$ . Both hold.
Comment. The system has the single solution $x = 2 e^{2 t} - 1 - t$ , $y = 1 - e^{2 t}$ . Its value at $t = 0$ is forced to be $(1, 0)$ : an initial condition such as $x (0) = 5$ cannot be met. Systems that are not in normal form can behave like this, which is one reason why the normal form of Section 7.3 is preferred for theory.

Exercises

Let $L = 2 D^{2} - 3 D + 4$ . Compute (a) $L t^{3}$ , (b) $L e^{2 t}$ , (c) $L sin t$ .
Answer
(a) $4 t^{3} - 9 t^{2} + 12 t$ ; (b) $6 e^{2 t}$ ; (c) $2 sin t - 3 cos t$ .
Write each operator in factored form: (a) $D^{2} + D - 12$ , (b) $2 D^{2} - 5 D - 3$ , (c) $D^{3} - 7 D + 6$ .
Answer
(a) $(D + 4) (D - 3)$ ; (b) $(2 D + 1) (D - 3)$ ; (c) $(D - 1) (D - 2) (D + 3)$ .
Let $f (t) = t sin t$ . Compute $(D - 1) (D + 2) f$ and $(D + 2) (D - 1) f$ and show that both equal $(D^{2} + D - 2) f$ .
Answer
All three equal $(1 - 3 t) sin t + (2 + t) cos t$ .

In Exercises 4–9 find the general solution by the operator method.

$2 x^{'} + x + y^{'} - y = 0$ , $x^{'} + 3 x + y^{'} - y = 0$ .
Answer
$Δ = (D - 1) (D - 2)$ . $x = c_{1} e^{2 t}$ , $y = - 5 c_{1} e^{2 t} + c_{2} e^{t}$ .
$x^{'} - x + y^{'} - y = 0$ , $x^{'} + 2 y^{'} - 3 y = 0$ .
Answer
$Δ = (D - 1) (D - 3)$ . $x = c_{1} e^{t} + c_{2} e^{3 t}$ , $y = c_{1} e^{t} - c_{2} e^{3 t}$ .
$x^{'} + 2 x + y^{'} + y = 3 e^{t}$ , $2 x^{'} - 2 x + y^{'} + y = 0$ .
Answer
$Δ = - (D - 4) (D + 1)$ . $x = 5 c_{1} e^{4 t} + e^{t}$ , $y = - 6 c_{1} e^{4 t} + c_{2} e^{- t}$ .
$x^{'} - x + y^{'} - y = 0$ , $x^{'} + 2 y^{'} - 3 y = 4 e^{2 t}$ .
Answer
$x = c_{1} e^{t} + c_{2} e^{3 t} + 4 e^{2 t}$ , $y = c_{1} e^{t} - c_{2} e^{3 t} - 4 e^{2 t}$ .
$x^{'} = 2 x - 3 y$ , $y^{'} = x - 2 y + 2 e^{2 t}$ . (Write it as $(D - 2) x + 3 y = 0$ , $- x + (D + 2) y = 2 e^{2 t}$ .)
Answer
$Δ = D^{2} - 1$ . $x = c_{1} e^{- t} + 3 c_{2} e^{t} - 2 e^{2 t}$ , $y = c_{1} e^{- t} + c_{2} e^{t}$ .
$x^{''} - 2 x^{'} + y^{'} = 0$ , $x^{'} - 2 x + 2 y^{'} + 3 y = 0$ .
Answer
$Δ = 2 D (D - 2) (D + 1)$ , three constants. $x = 3 c_{1} + c_{2} e^{- t} + c_{3} e^{2 t}$ , $y = 2 c_{1} + 3 c_{2} e^{- t}$ .
Solve the system of Exercise 5 with $x (0) = 4$ , $y (0) = 0$ .
Answer
$x = 2 e^{t} + 2 e^{3 t}$ , $y = 2 e^{t} - 2 e^{3 t}$ .
Without solving, find the number of arbitrary constants in the general solution of $(D^{2} + 1) x + D y = 0$ , $(D - 1) x + (D^{2} + 2) y = t$ . Then do the same for $D x + (D + 2) y = e^{t}$ , $(D - 3) x + (D - 1) y = 0$ , and solve that second system.
Answer
First system: $Δ = D^{4} + 2 D^{2} + D + 2$ , four constants. Second: $Δ = 6$ , no constants; the only solution is $x = 0$ , $y = \frac{1}{3} e^{t}$ .
Show that $x^{'} + y^{'} = t$ , $2 x^{'} + 2 y^{'} = 1$ has no solution. What is $Δ$ ? What happens if the second right-hand side is changed to $2 t$ ?
Answer
$Δ = D \cdot 2 D - D \cdot 2 D = 0$ . Doubling the first equation gives $2 x^{'} + 2 y^{'} = 2 t \neq = 1$ , a contradiction. With $2 t$ the equations coincide: $x$ can be any differentiable function and $y = \frac{1}{2} t^{2} - x + c$ .

What a linear system looks like

The differential operator D

The algebra of operators

The operator method for two equations

Eliminating one unknown

Too many constants

Exercises

The differential operator $D$