7.2 Linear differential equations

The solutions of a homogeneous linear equation of order n form a vector space of dimension n, because "solution to starting values" is an isomorphism. Every solution of the inhomogeneous equation is one particular solution plus a homogeneous one. With constant coefficients, the roots of the characteristic polynomial write the whole solution space down.

In 2.6 the solutions of $Ax = b$ turned out to be one particular solution plus the null space of $A$ . In 2.8 the function $e^{\lambda t}$ turned out to be an eigenvector of the derivative. This chapter puts those two facts together. An equation such as $y'' + 3y' + 2y = 0$ asks for the null space of an operator on functions, and everything we know about null spaces applies: it is a vector space, it has a dimension (here $2$ ), and the full solution of $y'' + 3y' + 2y = e^t$ is one solution plus anything in that null space. For constant coefficients the null space is found by solving a polynomial.

Before you start

You need existence and uniqueness (7.1), dimension and isomorphism, and complex roots of real polynomials. Three quick checks:

Quick checkIf

2 + 3i

is a root of the real polynomial

t^2 - 4t + 13

, what is the other root?

Quick checkA linear map

\mu : V \to \mathbb R^3

is one-to-one and onto. What is

\dim V

Quick checkLet

D

be differentiation. What is the function

e^{3t}

for

D

The picture

Everything below is about the equation $y'' + by' + cy = 0$ (a mass on a spring with friction: $c$ is the spring, $b$ the friction), with the solution that starts at $y(0) = 1$ and rests, $y'(0) = 0$ . The left panel shows the two roots of $r^2 + br + c$ in the complex plane. The right panel shows the solution. Use the presets first, then move the sliders yourself.

Watch three things. Roots on the real axis give pure decay or growth, with no wiggle. A root pair off the axis gives a wiggle, and its imaginary part is the speed of the wiggle. The real part (left or right of the vertical axis) says whether the solution dies away or grows. The rest of the chapter explains why: the formula of the solution is read straight off the roots, and the reason it can be read off is that the solution set is a vector space of dimension $2$ .

Where this came from· Euler and the trick of trying an exponential

Leonhard Euler wanted to solve higher-order linear equations with constant coefficients directly, in closed form, without first lowering their order step by step. He raised the idea in a letter to Johann Bernoulli on 15 September 1739, and gave the full treatment in a paper of 1743 (Miscellanea Berolinensia, volume 7, pages 193 to 242). His method is the one in this chapter: try $y = e^{rt}$ , which turns the differential equation into an ordinary polynomial equation in $r$ ; each linear factor of that polynomial gives a solution. The same paper deals with repeated factors, which is where $te^{rt}$ comes from, and with complex roots, whose solutions come out as sines and cosines. The statement that the solutions form a vector space of dimension $n$ is much younger: it needs the language of linear algebra and, underneath it, the existence and uniqueness theorem of 7.1.

The idea, in words

Why· Linear means superposition, and superposition means linear algebra

An equation is linear when it never multiplies the unknown by itself or by its derivatives, and never feeds it into a function such as $\sin$ . For such an equation, if $y_1$ and $y_2$ are solutions of the version with $0$ on the right, so is $y_1 + y_2$ and so is $5y_1$ . So the solutions are a vector space. The next question is its size. The existence and uniqueness theorem of 7.1 says that a solution of an equation of order $n$ is pinned down by $n$ starting numbers $(y(t_0), \dots, y^{(n-1)}(t_0))$ , and that every choice of those numbers occurs. Therefore “solution $\mapsto$ its starting numbers” is a one-to-one, onto, linear map to $\mathbb R^n$ , and the solution space has dimension $n$ . Writing the general solution means finding $n$ independent solutions. With constant coefficients that is easy: $e^{rt}$ is a solution exactly when $r$ is a root of the characteristic polynomial, and repeated roots and complex roots are handled by two small tricks.

Definitions and theorems

Linear equations and linear operators

Definition T

Linear differential equation

Let $J$ be an interval and let $a_0, \dots, a_{n-1}$ and $R$ be continuous functions on $J$ . The equation

y^{(n)} + a_{n-1}(t)\,y^{(n-1)} + \dots + a_1(t)\,y' + a_0(t)\,y = R(t)

is a linear differential equation of order $n$ . The functions $a_k$ are its coefficients and $R$ is the right-hand side (forcing term). (Technau writes the sum as $\sum p_i\,y^{(n-i)}$ ; it is the same equation with the coefficients numbered from the top.)

In one lineAn equation $y^{(n)} + a_{n-1}(t)y^{(n-1)} + \dots + a_0(t)y = R(t)$ , in which the unknown $y$ and its derivatives each appear to the first power, multiplied only by functions of $t$ .

Why it existsIt is the class of equations where solutions can be added and scaled, so linear algebra applies.

Definition T

Homogeneous equation

The linear equation above is homogeneous if $R(t) = 0$ for all $t \in J$ , and inhomogeneous otherwise. For an inhomogeneous equation $Ly = R$ , the equation $Ly = 0$ is its associated homogeneous equation.

In one lineA linear equation whose right-hand side is zero everywhere: $Ly = 0$ . If $R$ is not zero everywhere the equation is inhomogeneous.

Why it existsThe word 'homogeneous' marks the case where the zero function is a solution and sums of solutions are solutions. Everything else is built from it.

Definition T

Differential operator

For continuous coefficients $a_k$ on $J$ , define $L : C^n(J) \to C(J)$ by

(Ly)(t) = y^{(n)}(t) + a_{n-1}(t)\,y^{(n-1)}(t) + \dots + a_0(t)\,y(t).

Here $C^n(J)$ is the vector space of functions with $n$ continuous derivatives on $J$ and $C(J)$ the space of continuous functions (2.1). The operator $D$ is $Df = f'$ , so $L = D^n + a_{n-1}D^{n-1} + \dots + a_0$ . The equation $Ly = R$ asks: which functions does $L$ send to $R$ ?

In one lineThe map $L$ that sends a function $y$ to the function $y^{(n)} + a_{n-1}y^{(n-1)} + \dots + a_0y$ .

Why it existsIt turns the equation $Ly = R$ into the same shape as $Ax = b$ : a linear map and a target.

Fact T Technau, March 23 (Definition 7.6)

A linear differential operator is a linear map

For $f, g \in C^n(J)$ and a number $c$ : $L(f + g) = Lf + Lg$ and $L(cf) = c\,Lf$ . Hence $L$ is a linear map (2.3) and the set of solutions of $Ly = 0$ is $\ker L = N(L)$ , a subspace of $C^n(J)$ .

In one line $L(f + g) = Lf + Lg$ and $L(cf) = c\,Lf$ , so the solutions of $Ly = 0$ form the null space of a linear map.

Why it is trueDifferentiation is linear, and multiplying by a fixed function is linear.

ProofTechnau, March 23

The idea: Each term of L is a derivative followed by a multiplication by a function of t. Both steps respect sums and scalars, so their sum does.

The

k

-th derivative is linear:

D^k(f + g) = D^kf + D^kg

and

D^k(cf) = cD^kf

(f + g)^{(k)} = f^{(k)} + g^{(k)}

and

(cf)^{(k)} = c f^{(k)}

(rules for derivatives, 1.4).

f \mapsto a_k D^kf

is linear.

Multiplying by the fixed function

a_k(t)

distributes:

a_k(f + g) = a_kf + a_kg

and

a_k(cf) = c\,a_kf

L = D^n + a_{n-1}D^{n-1} + \dots + a_0

is linear.

A sum of linear maps is linear.

\ker L

is a subspace.

\blacksquare

The null space of a linear map is a subspace (2.3). The solutions of

Ly = 0

are by definition the functions sent to

0

The sign of the right-hand side now matters more than anything else. This is the point the 376 final tested in 6(b) and worksheet 19 #4 and #5.

Theorem T Technau, March 23 (Lemma 7.5); Apostol 6.10; worksheet 19 #4 and #5

Homogeneous and inhomogeneous solution sets

Let $L$ be a linear differential operator. (a) $\ker L$ is a vector space. (b) If $L y_p = R$ , then $y$ is a solution of $Ly = R$ if and only if $y - y_p \in \ker L$ . So the solution set is $\{y_p + h : h \in \ker L\}$ . (c) If $R \ne 0$ as a function, the zero function is not a solution, and the solution set is not closed under addition or scaling, so it is not a vector space.

In one lineThe solutions of $Ly = 0$ form a vector space. If $Ly = R$ and $R$ is not the zero function, the solutions do NOT form a vector space: they are $y_p + \ker L$ , where $y_p$ is any one solution.

Why it is trueSubtract two solutions of $Ly = R$ and the right-hand sides cancel. And $y = 0$ would need $0 = R$ .

ProofTechnau, March 23 (Lemma 7.5); worksheet 19

The idea: (a) is the previous fact. (b) Linearity of L turns the difference of two solutions into a homogeneous one. (c) Test the zero vector.

(a)

\ker L

is a subspace of

C^n(J)

, hence a vector space.

The previous theorem.

(b,

\Rightarrow

) If

Ly = R

and

Ly_p = R

, then

y - y_p \in \ker L

Linearity:

L(y - y_p) = Ly - Ly_p = R - R = 0

(b,

\Leftarrow

) If

h \in \ker L

then

y_p + h

solves

Ly = R

Linearity:

L(y_p + h) = Ly_p + Lh = R + 0 = R

(c) The zero function is not a solution, so the set is not a subspace; and the sum of two solutions solves the equation with right-hand side

2R

\blacksquare

L(0) = 0 \ne R

. Also if

Ly_1 = Ly_2 = R

then

L(y_1 + y_2) = 2R \ne R

This is the structure you already know from 2.6: the solutions of $Ax = b$ are $x_p + N(A)$ , a translated copy of the null space. A line that misses the origin is not a subspace, and neither is the solution set of an inhomogeneous equation.

Worked example MATH 376 final 6(b); worksheet 19 #5

Do the solutions of y'' + y = cos t form a vector space of dimension 2?

Solution.

Identify the type. The equation $y'' + y = \cos t$ has right-hand side $\cos t$ , which is not the zero function. So it is inhomogeneous. The first thing to do with any statement about a solution set is to look at the right-hand side.
Test the zero function. $0'' + 0 = 0$ , and $0 \ne \cos t$ . So $y = 0$ is not a solution. Every vector space contains the zero vector, so a set missing it cannot be a vector space.
Find the actual solution set. Try $y_p = \frac t2\sin t$ . Then $y_p' = \frac12\sin t + \frac t2\cos t$ and $y_p'' = \cos t - \frac t2\sin t$ , so $y_p'' + y_p = \cos t$ . The homogeneous equation $y'' + y = 0$ has the solutions $a\cos t + b\sin t$ (two independent solutions, and the dimension is $2$ by the theorem below). So the solution set is $\{\frac t2\sin t + a\cos t + b\sin t\}$ . By the theorem above: one particular solution plus the whole null space.
Describe it. It is a translate of a $2$ -dimensional plane of functions, not a plane through the origin. The sum of two solutions is a solution of $y'' + y = 2\cos t$ . Only the homogeneous solutions have a dimension; the inhomogeneous set is an affine space.

Answer. False. The solution set is not a vector space. (The associated homogeneous equation has a solution space of dimension $2$ .) On the 376 final, a student answered True, solved $y'' + y = 0$ and never used the right-hand side $\cos t$ . They also wrote the solution as $e^t(\cos t + \frac12\sin t)$ with a spurious factor $e^t$ : the roots are $\pm i$ with real part $0$ , so there is no exponential at all.

Quick checkWhich of these is a vector space under the usual addition and scaling of functions?

Existence, uniqueness, and the dimension theorem

To use 7.1 for an equation of order $n$ we turn it into a system, using the trick of the previous chapter.

Definition T

Companion matrix

For $y^{(n)} + a_{n-1}y^{(n-1)} + \dots + a_0y = R$ and $x = (y, y', \dots, y^{(n-1)})$ we have $x_k' = x_{k+1}$ for $k < n$ and $x_n' = -a_0x_1 - a_1x_2 - \dots - a_{n-1}x_n + R$ . In matrix form $x' = A(t)x + g(t)$ with

A(t) = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & & & \ddots & \\ 0 & 0 & 0 & \cdots & 1\\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1}\end{bmatrix}, \qquad g(t) = \begin{bmatrix}0\\0\\\vdots\\0\\R(t)\end{bmatrix}.

$A(t)$ is the companion matrix of the equation. (Technau’s notes of March 23 print the last row with an index mismatch and one dropped sign. With the numbering $y^{(n)} + \sum_{k<n} a_ky^{(k)}$ the last row is $-a_0, \dots, -a_{n-1}$ , as above.)

In one lineThe matrix $A(t)$ with ones directly above the diagonal and the last row $(-a_0, -a_1, \dots, -a_{n-1})$ , for which $x = (y, y', \dots, y^{(n-1)})$ satisfies $x' = A(t)x + (0, \dots, 0, R)$ .

Why it existsIt is what remains of the equation when you remember each derivative as a separate coordinate.

Theorem T Technau, March 23 (Theorem 7.6); Apostol 6.5

Existence and uniqueness for linear equations of order n

Let $a_0, \dots, a_{n-1}, R$ be continuous on an interval $J$ , $t_0 \in J$ and $c_0, \dots, c_{n-1} \in \mathbb R$ . Then there is exactly one function $y \in C^n(J)$ with $Ly = R$ on $J$ and $y^{(k)}(t_0) = c_k$ for $k = 0, \dots, n-1$ .

In one lineFor continuous coefficients and right-hand side on $J$ , any $t_0 \in J$ and any numbers $c_0, \dots, c_{n-1}$ , the equation $Ly = R$ has exactly one solution on all of $J$ with $y^{(k)}(t_0) = c_k$ for $k < n$ .

Why it is trueOrder $n$ is a first-order system for $x = (y, \dots, y^{(n-1)})$ , and 7.1 solves linear systems on the whole interval.

ProofTechnau, March 23

The idea: Translate to the companion system, apply the linear existence-uniqueness theorem of 7.1, and translate back.

The IVP for

y

is equivalent to the IVP

x' = A(t)x + g(t)

x(t_0) = (c_0, \dots, c_{n-1})

The reduction fact of 7.1 (the first coordinate of a solution

x

of the system is a solution

y

, with

x_{k+1} = y^{(k)}

A

and

g

are continuous on

J

The entries of

A

and

g

are

0

1

-a_k

and

R

: continuous on

J

by hypothesis.

So the system has exactly one solution on

J

The theorem 'linear systems have solutions on the whole interval' of 7.1, which rests on the Banach fixed point theorem for a power of the Picard operator.

Hence the original IVP has exactly one solution on

J

\blacksquare

Solutions correspond one to one under

y = x_1

. The first coordinate has

n

continuous derivatives:

x_1^{(k)} = x_{k+1}

for

k < n

, and

x_n' = -\sum a_kx_{k+1} + R

is continuous.

Theorem T Technau, March 25 (Theorem 7.7); Apostol 6.6

Dimension of the solution space of a linear ODE

Let $L = D^n + a_{n-1}D^{n-1} + \dots + a_0$ with continuous coefficients on $J$ . Then $\dim \ker L = n$ . Consequently any $n$ linearly independent solutions $b_1, \dots, b_n$ form a basis, and every solution is $c_1b_1 + \dots + c_nb_n$ for exactly one choice of numbers $c_1, \dots, c_n$ .

In one lineThe solution space $\ker L$ of a homogeneous linear equation of order $n$ has dimension exactly $n$ .

Why it is true'Solution to its starting numbers' is a linear map to $\mathbb R^n$ that is onto by existence and one-to-one by uniqueness, so it is an isomorphism.

Why· The picture behind the proof

In the picture of this chapter, the solution is determined by where it starts: the two numbers $(y(0), y'(0))$ . Every point of the plane $\mathbb R^2$ is the starting point of exactly one solution. The point $(1, 0)$ is the solution drawn on the right of the picture. The point $(0, 1)$ is a different solution, $(\sin t)$ for $y'' + y = 0$ . The map “solution $\mapsto$ starting point” matches the space of solutions with the plane. Since the plane has dimension $2$ , so does the space of solutions. In 7.4 this plane becomes the picture in which solutions are curves.

ProofTechnau, March 25

The idea: Define the map from solutions to starting numbers. It is linear. It is onto because of existence, and one-to-one because of uniqueness. An isomorphism preserves dimension.

Define

\mu : \ker L \to \mathbb R^n

\mu(y) = \big(y(t_0), y'(t_0), \dots, y^{(n-1)}(t_0)\big)

Fix

t_0 \in J

ahead of time.

\ker L

is a vector space (previous theorem), so the map goes between two vector spaces.

\mu

is linear.

Evaluation of a function, and of each derivative, at

t_0

respects sums and scalar multiples.

\mu

is onto.

Given

(c_0, \dots, c_{n-1})

, the existence part of the existence-uniqueness theorem (with

R = 0

) gives a solution

y

with

\mu(y) = (c_0, \dots, c_{n-1})

\mu

is one-to-one.

\mu(y) = O

then

y

and the zero function both solve

Ly = 0

with all starting numbers

0

. The uniqueness part says

y = 0

. So the null space of

\mu

\{0\}

, and a linear map with trivial null space is one-to-one (2.3).

\dim \ker L = \dim \mathbb R^n = n

A one-to-one onto linear map is an isomorphism, and isomorphic spaces have equal dimension (2.4).

The statement about

n

independent solutions follows.

\blacksquare

In an

n

-dimensional space

n

independent vectors form a basis (2.2), and a basis gives unique coordinates.

Proof workshopRebuild the proof that the solution space has dimension nTechnau, March 25

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Quick checkIn the proof of the dimension theorem, which hypothesis of the existence-uniqueness theorem makes the map

\mu

one-to-one?

Corollary T Technau, March 25 (Corollary 7.8)

General solution of a linear equation

Let $b_1, \dots, b_n$ be a basis of $\ker L$ and $Ly_p = R$ . Then the solutions of $Ly = R$ on $J$ are exactly the functions $y = y_p + c_1b_1 + \dots + c_nb_n$ , with $c_1, \dots, c_n$ arbitrary numbers. Given starting numbers $(c_0', \dots, c_{n-1}')$ at $t_0$ the constants $c_i$ are determined: they solve a linear system $\sum_i c_i b_i^{(k)}(t_0) = c_k' - y_p^{(k)}(t_0)$ whose matrix is invertible. (That matrix is the Wronskian matrix of 7.3.)

In one lineIf $b_1, \dots, b_n$ is a basis of $\ker L$ and $y_p$ solves $Ly = R$ , then every solution is $y = y_p + c_1b_1 + \dots + c_nb_n$ .

Why it is trueCombine the two structure theorems: the solution set is $y_p$ plus $\ker L$ , and $\ker L$ is spanned by a basis.

Definition T

General solution

The general solution of $Ly = R$ is the expression $y = y_p + c_1b_1 + \dots + c_nb_n$ above. For the homogeneous equation ( $R = 0$ ) it is $y = c_1b_1 + \dots + c_nb_n$ . A particular solution is any one function $y_p$ with $Ly_p = R$ ; it does not depend on any constants.

In one lineThe formula $y = y_p + c_1b_1 + \dots + c_nb_n$ with arbitrary constants, which lists every solution of $Ly = R$ .

Why it existsIt is the answer to 'find all solutions', the way a parametric description is the answer to 'find all solutions of Ax = b'.

Definition T

Particular solution

A particular solution of $Ly = R$ is any one function $y_p$ that satisfies the equation. Finding one is the job of 7.3 (variation of parameters, undetermined coefficients).

In one lineAny single function $y_p$ with $Ly_p = R$ .

Why it existsYou need only one. The homogeneous solutions supply all the rest.

Worked example MATH 376 worksheet 19 #6; worksheet 21 #9

Why are cos t and sin t every solution of y'' + y = 0? And when do two solutions form a basis?

Solution.

Count the dimension. The equation $y'' + y = 0$ is homogeneous of order $2$ , so its solution space has dimension $2$ . The dimension theorem.
Check that $\cos t$ and $\sin t$ are solutions. $(\cos t)'' = -\cos t$ and $(\sin t)'' = -\sin t$ , so both satisfy $y'' + y = 0$ . Direct substitution.
Check that they are independent. Their starting numbers at $t_0 = 0$ are $\mu(\cos) = (1, 0)$ and $\mu(\sin) = (0, 1)$ . The map $\mu$ is an isomorphism, so independent starting vectors mean independent solutions.
Conclude. Two independent solutions in a $2$ -dimensional space form a basis, so every solution is $c_1\cos t + c_2\sin t$ . “Any solution” is what the dimension theorem gives. Linearity alone gives only that each combination IS a solution.
A basis test from starting values (worksheet 21 #9). Suppose $y_1, y_2$ solve a second-order homogeneous equation with $y_1(0) = 1$ , $y_1'(0) = 2$ , $y_2(0) = 0$ , $y_2'(0) = 3$ . Then $\mu(y_1) = (1, 2)$ and $\mu(y_2) = (0, 3)$ are independent in $\mathbb R^2$ (the determinant is $1 \cdot 3 - 0 \cdot 2 = 3 \ne 0$ ), so $y_1, y_2$ are independent solutions and a basis. The determinant $3$ is the Wronskian at $0$ , which 7.3 makes into a general test.

Answer. $y = c_1\cos t + c_2\sin t$ is the general solution; and $y_1, y_2$ in part 5 are a basis.

Quick checkTwo solutions of a second-order homogeneous equation have

y_1(0) = 1

y_1'(0) = 2

and

y_2(0) = 2

y_2'(0) = 4

. Could

y_1, y_2

be a basis of the solution space?

Equations with constant coefficients

From here on the coefficients are numbers, and the whole solution space can be written down.

Definition T

Constant-coefficient operator

A constant-coefficient operator is $A = D^n + a_{n-1}D^{n-1} + \dots + a_1D + a_0$ with real or complex numbers $a_k$ , acting on infinitely differentiable functions $C^\infty(J)$ (the solutions of $Ay = 0$ turn out to be infinitely differentiable automatically, because $y^{(n)}$ is a combination of lower derivatives). The polynomial $P_A(t) = t^n + a_{n-1}t^{n-1} + \dots + a_0$ is its characteristic polynomial; we write $A = P_A(D)$ . Do not confuse this with the characteristic polynomial $\det(\lambda I - M)$ of a matrix in 2.8; the two are linked in the example on the companion matrix below.

In one lineAn operator $A = D^n + a_{n-1}D^{n-1} + \dots + a_0$ with numbers $a_k$ as coefficients. It equals $P_A(D)$ for the polynomial $P_A(t) = t^n + a_{n-1}t^{n-1} + \dots + a_0$ .

Why it existsWriting $A$ as a polynomial in $D$ turns the differential equation $Ay = 0$ into algebra with polynomials.

Definition T

Characteristic polynomial of an operator

For a constant-coefficient operator $A = P_A(D)$ the characteristic polynomial of the operator is $P_A$ . The equation $P_A(r) = 0$ is the characteristic equation of $Ay = 0$ . The key computation: $A(e^{rt}) = P_A(r)\,e^{rt}$ , because $D^k e^{rt} = r^ke^{rt}$ .

In one lineFor $A = D^n + a_{n-1}D^{n-1} + \dots + a_0$ , the polynomial $P_A(t) = t^n + a_{n-1}t^{n-1} + \dots + a_0$ . Its roots decide the solutions.

Why it existsTrying $y = e^{rt}$ in $Ay = 0$ gives $P_A(r)e^{rt} = 0$ , so the solutions of the form $e^{rt}$ are exactly the roots.

Theorem T Technau, March 25; Apostol 6.7

Constant-coefficient operators multiply like polynomials

For constant-coefficient operators $A$ and $B$ acting on $C^\infty(J)$ : the composition $AB$ (apply $B$ first) is the constant-coefficient operator with characteristic polynomial $P_AP_B$ . Hence $AB = BA$ .

In one lineIf $A = P_A(D)$ and $B = P_B(D)$ then $AB = (P_AP_B)(D)$ . In particular $AB = BA$ .

Why it is trueComposing two operators multiplies out like composing polynomials in $D$ , because $D^iD^j = D^{i+j}$ and the coefficients are numbers that pass through $D$ .

ProofTechnau, March 25

The idea: Expand the composition term by term with the distributive law and collect the powers of D.

AB = \sum_{i,j}a_iD^i(b_jD^j)

Write

A = \sum_i a_iD^i

and

B = \sum_j b_jD^j

(with

a_n = b_m = 1

). Composition of linear maps distributes over sums (2.4).

= \sum_{i,j}a_ib_jD^{i+j}

A number

b_j

passes through

D^i

D^i(b_jf) = b_jD^if

. And

D^iD^j = D^{i+j}

, since differentiating

j

times and then

i

times is differentiating

i + j

times.

= (P_AP_B)(D)

This is the sum we get on multiplying the polynomials

P_A(t) = \sum a_it^i

and

P_B(t) = \sum b_jt^j

, with

t

replaced by

D

AB = BA

\blacksquare

Polynomials commute:

P_AP_B = P_BP_A

. So

AB

and

BA

have the same characteristic polynomial and are the same operator. (A different thing for variable coefficients:

D \circ t \ne t \circ D

Worked example Technau, March 25 (Example 7.10)

Factor the operator A = D² − 5D + 6.

Solution.

Write the characteristic polynomial. $P_A(t) = t^2 - 5t + 6$ . Replace $D^k$ by $t^k$ .
Factor it. $t^2 - 5t + 6 = (t - 2)(t - 3)$ , since $2 + 3 = 5$ and $2 \cdot 3 = 6$ . Roots of a polynomial give its factors (0.6).
Translate back to operators. $A = (D - 2)(D - 3)$ . By the theorem, multiplication of polynomials in $D$ is composition of operators.
Check on a function. $(D - 2)(D - 3)f = (D - 2)(f' - 3f) = f'' - 3f' - 2f' + 6f = f'' - 5f' + 6f$ . Both factors are applied; the order does not matter.

Answer. $A = (D - 2)(D - 3) = (D - 3)(D - 2)$ .

Fact T Apostol 6.7; Euler's formula of 0.6

Derivative of a complex exponential

Define $e^{rt} = e^{\alpha t}(\cos\beta t + i\sin\beta t)$ for $r = \alpha + i\beta$ and real $t$ (Euler’s formula). Then $(e^{rt})' = re^{rt}$ .

In one lineFor a complex number $r = \alpha + i\beta$ , $e^{rt} = e^{\alpha t}(\cos\beta t + i\sin\beta t)$ satisfies $D(e^{rt}) = re^{rt}$ .

Why it is trueDifferentiate the real and imaginary parts and use $i \cdot i = -1$ .

ProofApostol 6.7

The idea: Product rule on the real part e^{αt} times the unit-size part cos + i sin.

(e^{rt})' = \alpha e^{\alpha t}(\cos\beta t + i\sin\beta t) + e^{\alpha t}(-\beta\sin\beta t + i\beta\cos\beta t)

Product rule on each of the real and imaginary parts, with

(e^{\alpha t})' = \alpha e^{\alpha t}

(\cos\beta t)' = -\beta\sin\beta t

and

(\sin\beta t)' = \beta\cos\beta t

The second bracket equals

i\beta(\cos\beta t + i\sin\beta t)

i\beta(\cos\beta t + i\sin\beta t) = i\beta\cos\beta t - \beta\sin\beta t

, since

i^2 = -1

(e^{rt})' = (\alpha + i\beta)e^{\alpha t}(\cos\beta t + i\sin\beta t) = re^{rt}

\blacksquare

Factor out the common part.

So for every complex number $r$ , $D(e^{rt}) = re^{rt}$ , and $e^{rt}$ is an eigenfunction of $D$ with eigenvalue $r$ . Every statement of this chapter holds verbatim for complex-valued functions (a complex-valued function is a pair of real ones, and the solution space is then a complex vector space of dimension $n$ ). We use them, and return to real functions at the end.

Fact T Technau, March 27 (Proposition 7.13); Apostol 6.8

Shifting an exponential

For a polynomial $p(t)$ and a complex number $r$ : $(D - r)\big(p(t)e^{rt}\big) = p'(t)e^{rt}$ . Consequently $(D - r)^k\big(t^je^{rt}\big) = j(j-1)\cdots(j-k+1)\,t^{j-k}e^{rt}$ for $k \le j$ and $(D - r)^{j+1}\big(t^je^{rt}\big) = 0$ .

In one lineFor a polynomial $p$ and a number $r$ : $(D - r)\big(p(t)e^{rt}\big) = p'(t)\,e^{rt}$ .

Why it is trueThe product rule gives $p'e^{rt} + p\,re^{rt}$ , and subtracting $r\,pe^{rt}$ cancels the second term. So $D - r$ lowers the degree of the polynomial by one.

ProofTechnau, March 27

The idea: Product rule, then one term cancels. Induct for the powers.

D(pe^{rt}) = p'e^{rt} + p\,re^{rt}

Product rule, and

D(e^{rt}) = re^{rt}

from the previous fact.

(D - r)(pe^{rt}) = p'e^{rt} + rpe^{rt} - rpe^{rt} = p'e^{rt}

Subtract

r

times the function.

(D - r)^k(t^je^{rt}) = \big(\tfrac{d^k}{dt^k}t^j\big)e^{rt} = j(j-1)\cdots(j-k+1)\,t^{j-k}e^{rt}

Apply the formula

k

times to

p = t^j

: each application differentiates the polynomial once, and

\frac{d}{dt}t^i = it^{i-1}

For

k = j + 1

the result is

0

\blacksquare

After

j

differentiations

t^j

becomes the constant

j!

; one more gives

0

Worked example MATH 376 practice final 4(b)

Show directly that b_j(t) = t^j e^{rt} satisfies (D − r)^100 y = 0 for j = 0, 1, ..., 99.

Solution.

Compute one application. $(D - r)(t^je^{rt}) = jt^{j-1}e^{rt} + rt^je^{rt} - rt^je^{rt} = jt^{j-1}e^{rt}$ . Product rule; the two $r\,t^je^{rt}$ terms cancel.
State the pattern. $(D - r)^k b_j = j(j-1)\cdots(j-k+1)\,t^{j-k}e^{rt}$ for $0 \le k \le j$ . Induction on $k$ : the case $k = 1$ is step 1, and applying step 1 to $t^{j-k}e^{rt}$ multiplies by $j - k$ and lowers the power by one.
Reach zero. At $k = j$ we get $j!\,e^{rt}$ , and one more application gives $(D - r)(j!\,e^{rt}) = 0$ . So $(D - r)^{j+1}b_j = 0$ . $e^{rt}$ has derivative $re^{rt}$ , so $(D - r)e^{rt} = 0$ .
Use the size of $j$ . For $j \le 99$ we have $j + 1 \le 100$ , and applying $(D - r)$ more times to $0$ still gives $0$ . So $(D - r)^{100}b_j = 0$ . The equation is of order $100$ and each power $t^j$ with $j < 100$ is killed by it.

Answer. All of $e^{rt}, te^{rt}, \dots, t^{99}e^{rt}$ solve $(D - r)^{100}y = 0$ . (By the dimension theorem the solution space has dimension $100$ , so these $100$ functions, once shown independent, are a basis.)

Theorem T Technau, March 27 (Proposition 7.13); Apostol 6.8; worksheet 20

Basis for repeated roots

Let $r$ be a complex number and $m \ge 1$ . The functions $b_j(t) = t^je^{rt}$ , $j = 0, 1, \dots, m-1$ , are a basis of $\ker(D - r)^m$ .

In one lineThe $m$ functions $e^{rt}, te^{rt}, \dots, t^{m-1}e^{rt}$ are a basis of the solutions of $(D - r)^my = 0$ .

Why it is trueThey all solve it by the shifting fact, they are independent because a polynomial with too many zeros is zero, and the dimension theorem says $m$ independent solutions are a basis.

ProofTechnau, March 27

The idea: Each b_j is a solution, by the shifting fact. They are independent since dividing by e^{rt} leaves a polynomial. The operator has order m, so the dimension theorem finishes.

Each

b_j

lies in

\ker(D - r)^m

The shifting fact:

(D - r)^{j+1}b_j = 0

, and

j + 1 \le m

Suppose

\sum_{j<m}c_jt^je^{rt} = 0

J

. Then

\sum_{j<m}c_jt^j = 0

for all

t \in J

Divide by

e^{rt}

, which is never zero.

So all

c_j = 0

: the

b_j

are independent.

A non-zero polynomial of degree less than

m

has fewer than

m

roots (0.6), and

J

contains infinitely many points.

So they are a basis.

\blacksquare

(D - r)^m

is a constant-coefficient operator of order

m

, so its kernel has dimension

m

(dimension theorem), and

m

independent vectors in an

m

-dimensional space are a basis.

Proof workshopRebuild: t^j e^(rt), j below m, are a basis of the solutions of (D − r)^m y = 0Technau, March 27

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$(D - r)\big(p(t)e^{rt}\big) = p'(t)e^{rt}$ for every polynomial $p$ .
$(D - r)^{j+1}\big(t^je^{rt}\big) = 0$ , so $t^je^{rt}$ solves $(D - r)^my = 0$ whenever $j < m$ .
If $\sum_{j<m}c_jt^je^{rt} = 0$ then $\sum c_jt^j = 0$ for infinitely many $t$ , so all $c_j = 0$ .
So the $m$ functions are independent solutions of an equation of order $m$ , hence a basis.

Quick checkThe characteristic roots of a third-order equation are

1, 1, 2

. Which is the general solution?

Fact T Technau, March 25 (Proposition 7.11)

Kernel of a factor lies in the kernel of the product

If $A = A_1\cdots A_k$ with each $A_i$ a constant-coefficient operator, then every solution of $A_jy = 0$ is a solution of $Ay = 0$ .

In one lineIf $A = A_1A_2\cdots A_k$ are constant-coefficient operators, then $\ker A_j \subseteq \ker A$ for every $j$ .

Why it is trueThe operators commute, so the factor $A_j$ can be moved to act first, and it sends the function to zero.

ProofTechnau, March 25

The idea: Commute A_j to the right end, so it acts first.

A = \big(\prod_{i \ne j}A_i\big)A_j

The constant-coefficient operators commute (previous theorem).

y \in \ker A

\blacksquare

A_jy = 0

, then

Ay = \big(\prod_{i \ne j}A_i\big)(A_jy) = \big(\prod_{i \ne j}A_i\big)0 = 0

, since linear maps send

0

0

Theorem T Technau, April 13 and 29 (Proposition 9.18); Apostol 6.8; homework 10 #1

Basis of solutions from the roots

Let $A = P_A(D)$ be a constant-coefficient operator of order $n$ with $P_A(t) = \prod_{i=1}^{k}(t - r_i)^{m_i}$ , the $r_i$ distinct complex numbers. Then the functions $t^je^{r_it}$ with $1 \le i \le k$ and $0 \le j < m_i$ form a basis of $\ker A$ (over $\mathbb C$ ).

In one lineIf $P_A(t) = \prod_{i=1}^{k}(t - r_i)^{m_i}$ with distinct roots $r_i$ , the $n = \sum m_i$ functions $t^je^{r_it}$ ( $i \le k$ , $0 \le j < m_i$ ) are a basis of the solutions of $Ay = 0$ .

Why it is trueEach is killed by its own factor of $A$ , they are independent by a degree-lowering argument, and there are $n$ of them in an $n$ -dimensional space.

ProofTechnau, April 13; homework 10 #1

The idea: Membership by the factor fact, independence by induction on the number of distinct roots, applying the factor that kills one root's block and keeps the others alive, and then the count n.

Every function

t^je^{r_it}

(

j < m_i

) is in

\ker A

A = \prod(D - r_i)^{m_i}

by the multiplication theorem; each factor

(D - r_i)^{m_i}

kills

t^je^{r_it}

for

j < m_i

(basis for repeated roots), and a factor's kernel lies in the kernel of the product.

Independence: we show the claim for

k = 1

first.

Claim to prove by induction on

k

: if

r_1, \dots, r_k

are distinct and

p_1, \dots, p_k

are polynomials with

\sum_ip_i(t)e^{r_it} = 0

on an interval, then all

p_i = 0

. For

k = 1

p_1e^{r_1t} = 0

gives

p_1(t) = 0

at infinitely many

t

, so

p_1 = 0

For the step from

k - 1

k

, apply the operator

(D - r_1)^N

\sum_{i \le k}p_ie^{r_it} = 0

. The first term disappears.

Let

N = \deg p_1 + 1

(take

N = 1

p_1 = 0

). Apply

(D - r_1)^N

. The shifting fact: it kills

p_1e^{r_1t}

because

p_1

has degree less than

N

Each other term becomes

q_ie^{r_it}

where

q_i

p_i

transformed

N

times by

p \mapsto p' + (r_i - r_1)p

q_i = 0

if and only if

p_i = 0

For

i \ne 1

(D - r_1)(pe^{r_it}) = (p' + (r_i - r_1)p)e^{r_it}

(product rule). The polynomial

p' + (r_i - r_1)p

has the same degree as

p

and leading coefficient multiplied by

r_i - r_1 \ne 0

. So the map

p \mapsto p' + (r_i - r_1)p

sends a non-zero polynomial to a non-zero one.

\sum_{i \ge 2}q_ie^{r_it} = 0

forces all

q_i = 0

, hence

p_2 = \dots = p_k = 0

Induction hypothesis for the

k - 1

distinct roots

r_2, \dots, r_k

So all

p_i = 0

Only the first term remains:

p_1e^{r_1t} = 0

, so

p_1 = 0

by the case

k = 1

So the

n

functions are independent.

A combination

\sum c_{ij}t^je^{r_it}

is of the form

\sum p_ie^{r_it}

with

p_i = \sum_jc_{ij}t^j

. All

p_i = 0

means all

c_{ij} = 0

They are a basis.

\blacksquare

The dimension theorem:

\dim\ker A = n

over

\mathbb C

, and

n

independent vectors are a basis.

Worked example MATH 376 homework 8 #2; worksheet 20 #4

Distinct real roots: solve y''' − 6y'' + 11y' − 6y = 0 with y(0) = 2, y'(0) = 1, y''(0) = 0.

Solution.

Write the characteristic polynomial. $P(t) = t^3 - 6t^2 + 11t - 6$ . Replace $y^{(k)}$ by $t^k$ .
Factor it. $P(t) = (t - 1)(t - 2)(t - 3)$ . Try $t = 1$ : $1 - 6 + 11 - 6 = 0$ , so $t - 1$ is a factor; divide to get $t^2 - 5t + 6$ .
Write the general solution. $y = C_1e^t + C_2e^{2t} + C_3e^{3t}$ . Three distinct roots give three exponentials, and order $3$ means three is the dimension.
Differentiate for the starting numbers. $y' = C_1e^t + 2C_2e^{2t} + 3C_3e^{3t}$ and $y'' = C_1e^t + 4C_2e^{2t} + 9C_3e^{3t}$ . Order $3$ has three conditions, on $y$ , $y'$ , $y''$ .
Solve for the constants. At $t = 0$ : $C_1 + C_2 + C_3 = 2$ , $C_1 + 2C_2 + 3C_3 = 1$ , $C_1 + 4C_2 + 9C_3 = 0$ . Subtracting consecutive equations gives $C_2 + 2C_3 = -1$ and $2C_2 + 6C_3 = -1$ , so $C_3 = \frac12$ , $C_2 = -2$ , $C_1 = \frac72$ . Elimination as in 2.6.
Check. $y'(0) = \frac72 - 4 + \frac32 = 1$ and $y''(0) = \frac72 - 8 + \frac92 = 0$ . Test the two conditions you did not use to write the answer.

Answer. $y = \frac72e^t - 2e^{2t} + \frac12e^{3t}$ .

Your answerThe solution of

y'' - 3y' + 2y = 0

with

y(0) = 2

y'(0) = 1

y = 3e^t - e^{2t}

(worksheet 28 #1). What is

y(1)

? (Four decimals.)

Worked example MATH 376 worksheet 20 #5, #6, #7, #9; homework 9 #1

Repeated roots: solve y'' − 2y' + y = 0 with y(0) = 3, y'(0) = 1; solve x'' + 6x' + 9x = 0 with x(0) = 2, x'(0) = 1; give the general solution of y''' − 3y'' + 3y' − y = 0, of y''' − y'' − y' + y = 0 and of y⁗ − 2y''' − y'' + 2y' = 0.

Solution.

First equation. $P(t) = t^2 - 2t + 1 = (t - 1)^2$ : a double root $1$ , so $y = (c_1 + c_2t)e^t$ . A double root contributes $e^t$ and $te^t$ .
Use the starting values. $y(0) = c_1 = 3$ . And $y' = c_2e^t + (c_1 + c_2t)e^t$ , so $y'(0) = c_2 + c_1 = 1$ and $c_2 = -2$ . Differentiate with the product rule BEFORE substituting $t = 0$ .
Second equation. $t^2 + 6t + 9 = (t + 3)^2$ , so $x = (C_1 + C_2t)e^{-3t}$ . And $x' = (C_2 - 3C_1 - 3C_2t)e^{-3t}$ , so $x(0) = C_1 = 2$ and $x'(0) = C_2 - 3C_1 = 1$ gives $C_2 = 7$ . The same two moves.
Triple root. $t^3 - 3t^2 + 3t - 1 = (t - 1)^3$ , so $y = (c_1 + c_2t + c_3t^2)e^t$ . A root of multiplicity $m$ gives the powers $1, t, \dots, t^{m-1}$ .
Mixed. $t^3 - t^2 - t + 1 = t^2(t - 1) - (t - 1) = (t - 1)(t^2 - 1) = (t - 1)^2(t + 1)$ , so $y = c_1e^t + c_2te^t + c_3e^{-t}$ . Factor by grouping, then read multiplicities.
Fourth order. $t^4 - 2t^3 - t^2 + 2t = t(t - 2)(t - 1)(t + 1)$ , so $y = c_1 + c_2e^{2t} + c_3e^t + c_4e^{-t}$ . The root $0$ gives $e^{0t} = 1$ , a constant.

Answer. $y = (3 - 2t)e^t$ ; $x = (2 + 7t)e^{-3t}$ ; $(c_1 + c_2t + c_3t^2)e^t$ ; $c_1e^t + c_2te^t + c_3e^{-t}$ ; $c_1 + c_2e^{2t} + c_3e^t + c_4e^{-t}$ .

Your answerThe solution of

x'' + 6x' + 9x = 0

with

x(0) = 2

x'(0) = 1

x = (2 + 7t)e^{-3t}

. What is

x(1)

? (Three decimals.)

Complex roots, and real solutions

The roots of a real polynomial can be complex, but the equation is real and we want real solutions. Euler’s formula does it.

Theorem T Technau, April 6 (Lemma 7.20); Apostol 6.8; worksheets 21 and 23

Real solutions from complex roots

Let $A$ have real coefficients and let $r = \alpha + i\beta$ , $\beta \ne 0$ , be a root of $P_A$ of multiplicity $m$ . Then $\bar r = \alpha - i\beta$ is a root of the same multiplicity, and in a basis of $\ker A$ the $2m$ functions $t^je^{rt}$ , $t^je^{\bar rt}$ may be replaced by the $2m$ real functions $u_j = t^je^{\alpha t}\cos\beta t$ and $v_j = t^je^{\alpha t}\sin\beta t$ , $j = 0, \dots, m-1$ . Hence the real solutions of $Ay = 0$ are the real linear combinations of the real basis functions made this way from all roots.

In one lineIf $A$ has real coefficients and $r = \alpha + i\beta$ ( $\beta \ne 0$ ) is a root of multiplicity $m$ , then so is $\bar r$ , and the $2m$ functions $t^je^{\alpha t}\cos\beta t$ and $t^je^{\alpha t}\sin\beta t$ ( $0 \le j < m$ ) replace the $2m$ complex solutions $t^je^{rt}$ and $t^je^{\bar rt}$ .

Why it is trueThe real and imaginary parts of $t^je^{rt}$ are real solutions, and the change from $\{f, \bar f\}$ to $\{\operatorname{Re} f, \operatorname{Im} f\}$ is an invertible linear change of basis.

ProofTechnau, April 6 and 10; Apostol 6.8

The idea: Conjugating the factorization shows the multiplicities agree. Then u and v are linear combinations of f and its conjugate with an invertible 2x2 coefficient matrix, so the new family has the same span, is still independent and consists of real functions.

\bar r

is a root of the same multiplicity

m

Write

P_A(t) = (t - r)^mQ(t)

with

Q(r) \ne 0

. Since the coefficients of

P_A

are real, conjugating all coefficients does not change

P_A

(this is the argument of 0.6):

P_A(t) = (t - \bar r)^m\bar Q(t)

with

\bar Q(\bar r) = \overline{Q(r)} \ne 0

Put

f_j = t^je^{rt}

and

\bar f_j = t^je^{\bar rt}

Euler's formula (0.6):

f = t^je^{rt} = t^je^{\alpha t}(\cos\beta t + i\sin\beta t)

and

\bar f = t^je^{\alpha t}(\cos\beta t - i\sin\beta t)

u_j = \tfrac12(f_j + \bar f_j)

and

v_j = \tfrac1{2i}(f_j - \bar f_j)

Add and subtract the two lines above.

u_j, v_j

are real solutions of

Ay = 0

Combinations of solutions are solutions. And

u_j, v_j

are real-valued.

Replacing each pair

f_j, \bar f_j

u_j, v_j

in the basis of the previous theorem gives another basis of the complex solution space.

The matrix

\begin{bmatrix}\frac12 & \frac12\\ \frac1{2i} & -\frac1{2i}\end{bmatrix}

has determinant

-\frac1{2i} \ne 0

, so

f_j, \bar f_j

are combinations of

u_j, v_j

as well. Replacing basis vectors by an invertible combination gives a basis (over

\mathbb C

So every real solution is a real combination of the real basis functions.

\blacksquare

y

is a real solution,

y = \sum c_k\varphi_k

with complex

c_k

and real

\varphi_k

. Taking real parts,

y = \sum(\operatorname{Re}c_k)\varphi_k

, because

y

and

\varphi_k

are real.

Worked example MATH 376 worksheet 21 #4, #5, #6; worksheet 23 #2

Complex roots: solve y'' + 4y' + 13y = 0 with y(0) = 2, y'(0) = 0; give the general solutions of y'' − 2y' + 5y = 0 and y''' − y'' + y' − y = 0.

Solution.

Characteristic roots. $t^2 + 4t + 13 = 0$ gives $t = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i$ . The quadratic formula; the discriminant $-36$ is negative.
Real general solution. With $\alpha = -2$ , $\beta = 3$ : $y = e^{-2t}(C_1\cos3t + C_2\sin3t)$ . Real part of the root in the exponent, imaginary part as the frequency.
First starting value. $y(0) = C_1 = 2$ . At $t = 0$ , $e^0 = 1$ , $\cos0 = 1$ , $\sin0 = 0$ .
Second starting value. $y' = e^{-2t}\big[-2(C_1\cos3t + C_2\sin3t) - 3C_1\sin3t + 3C_2\cos3t\big]$ , so $y'(0) = -2C_1 + 3C_2 = 0$ and $C_2 = \frac43$ . Product rule, then substitute $t = 0$ .
The second equation. $t^2 - 2t + 5 = 0$ gives $1 \pm 2i$ , so $y = e^t(C_1\cos2t + C_2\sin2t)$ . $\alpha = 1$ , $\beta = 2$ .
The third equation. $t^3 - t^2 + t - 1 = (t - 1)(t^2 + 1)$ , with roots $1$ and $\pm i$ , so $y = C_1e^t + C_2\cos t + C_3\sin t$ . The root pair $\pm i$ has $\alpha = 0$ , so there is no exponential factor. This is the fix for the extra $e^t$ in your 376 final 6(b).

Answer. $y = e^{-2t}\big(2\cos3t + \frac43\sin3t\big)$ ; $e^t(C_1\cos2t + C_2\sin2t)$ ; $C_1e^t + C_2\cos t + C_3\sin t$ .

Your answerThe solution of

y'' + 4y' + 13y = 0

y(0) = 2

y'(0) = 0

y = e^{-2t}(2\cos3t + \frac43\sin3t)

. What is

y(\pi/3)

? (Three decimals.)

Your answerA suspended mass obeys

y'' + 2y' + 26y = 0

. What is the angular frequency of its oscillation, in rad/s?

Worked example Technau, April 6 (Lemma 7.20); new numbers

A repeated complex pair: find a real basis for y⁗ + 2y'' + y = 0.

Solution.

Factor the characteristic polynomial. $t^4 + 2t^2 + 1 = (t^2 + 1)^2 = (t - i)^2(t + i)^2$ . The roots $\pm i$ each have multiplicity $m = 2$ .
List the complex basis. $e^{it}, te^{it}, e^{-it}, te^{-it}$ . Each double root contributes $e^{rt}$ and $te^{rt}$ .
Pass to real functions. With $\alpha = 0$ and $\beta = 1$ : $\cos t$ , $\sin t$ , $t\cos t$ , $t\sin t$ . Real and imaginary parts of $e^{it}$ and $te^{it}$ .
Check one. $(t\sin t)' = \sin t + t\cos t$ , $(t\sin t)'' = 2\cos t - t\sin t$ , $(t\sin t)''' = -3\sin t - t\cos t$ , $(t\sin t)^{(4)} = -4\cos t + t\sin t$ . Then $y^{(4)} + 2y'' + y = (-4\cos t + t\sin t) + (4\cos t - 2t\sin t) + t\sin t = 0$ . The terms in $t\sin t$ and in $\cos t$ cancel separately.

Answer. $y = (c_1 + c_2t)\cos t + (c_3 + c_4t)\sin t$ . The factor $t$ makes the oscillation grow without bound, although no root has positive real part.

Quick checkThe characteristic polynomial of a real constant-coefficient equation is

(r^2 - 2r + 5)^2

. Which list is a basis of the real solutions?

Why· The recipe for a constant-coefficient equation

For $y^{(n)} + a_{n-1}y^{(n-1)} + \dots + a_0y = 0$ with real numbers $a_k$ : (1) Write $P(t) = t^n + a_{n-1}t^{n-1} + \dots + a_0$ and factor it completely, over the complex numbers. (2) For each real root $r$ of multiplicity $m$ write $e^{rt}, te^{rt}, \dots, t^{m-1}e^{rt}$ . (3) For each pair $\alpha \pm i\beta$ of multiplicity $m$ write $t^je^{\alpha t}\cos\beta t$ and $t^je^{\alpha t}\sin\beta t$ , $0 \le j < m$ . (4) You should now have exactly $n$ functions. (5) The general solution is an arbitrary real combination. (6) With starting values, differentiate with the product rule, put $t = t_0$ , and solve the $n \times n$ linear system. Step (4) is the check that catches a missing power of $t$ .

The first-order case and the companion matrix

Worked example MATH 376 homework 9 #4; new

First order with a variable coefficient: solve y' − P(t)y = 0 for continuous P, and solve y' + y = t with y(0) = 0.

Solution.

Choose an integrating factor. Let $F(t) = \int_0^tP(s)\,ds$ , so $F' = P$ (the continuity of $P$ makes this work). Multiply the equation by $e^{-F}$ . We want the left side to become a derivative of a product.
Recognise the derivative. $\big(e^{-F}y\big)' = e^{-F}y' - Pe^{-F}y = e^{-F}(y' - Py) = 0$ . Product and chain rule; the equation says the bracket is $0$ .
Integrate. $e^{-F}y = C$ , so $y = Ce^{F(t)} = C\exp\int_0^tP$ . A function with zero derivative on an interval is constant (1.4). One constant, as the dimension theorem predicts for order $1$ .
The second problem. For $y' + y = t$ take $F = t$ and multiply by $e^t$ : $(e^ty)' = te^t$ . Integrate by parts: $e^ty = (t - 1)e^t + C$ . $\int te^t\,dt = te^t - e^t$ ; differentiate to confirm.
Fix the constant. $y(0) = -1 + C = 0$ , so $C = 1$ and $y = t - 1 + e^{-t}$ . Starting value.
Check. $y' + y = (1 - e^{-t}) + (t - 1 + e^{-t}) = t$ and $y(0) = -1 + 1 = 0$ . Equation and starting value both checked.

Answer. $y = C\exp\int_0^tP$ ; and $y = t - 1 + e^{-t}$ .

Worked example Technau, March 23 and April 15; new

The companion matrix of y''' − 6y'' + 11y' − 6y = 0, and its eigenvalues.

Solution.

Write the companion matrix. Here $a_0 = -6$ , $a_1 = 11$ , $a_2 = -6$ , so the last row is $(6, -11, 6)$ : $A = \begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\6 & -11 & 6\end{bmatrix}$ . The last row is $(-a_0, -a_1, -a_2)$ .
Compute $\det(\lambda I - A)$ . Expanding gives $\lambda^3 - 6\lambda^2 + 11\lambda - 6$ . The same polynomial as the characteristic polynomial of the equation.
Factor it. $(\lambda - 1)(\lambda - 2)(\lambda - 3)$ , so the eigenvalues of $A$ are $1, 2, 3$ . The roots of the equation’s polynomial. The solutions $e^{t}, e^{2t}, e^{3t}$ of the scalar equation are the first coordinates of $e^{\lambda t}v$ for the eigenvectors $v$ of $A$ .

Answer. The characteristic polynomial of the companion matrix equals the characteristic polynomial of the equation. In general the two coincide (stretch problem below), and 7.4 uses the matrix side to solve systems.

Where it is used

Used inside maths· Inside maths: the same structure everywhere linear algebra goes

The pattern “particular solution plus null space, and the null space has a basis” is the pattern of $Ax = b$ (2.6), of integrals (" $+\,C$ " is the null space of $D$ : the constants, 2.3), and of linear recurrences such as $F_{k+2} = F_{k+1} + F_k$ (the characteristic equation $\lambda^2 = \lambda + 1$ gives Binet’s formula, 2.8). The recurrence is the discrete twin of the ODE $y'' = y' + y$ and has the same proof structure. Solving partial differential equations (the heat equation, the vibrating string) starts by separating variables, which produces exactly the equations of this chapter, and then uses eigenfunctions of $D^2$ as a basis, which is the idea of Fourier series.

Used in the real world· Mechanical engineering: a car's suspension

One wheel of a car carries a mass $m = 400$ kg on a spring of stiffness $k = 16000$ N/m and a shock absorber with damping $c = 1500$ N s/m. The displacement $y$ from rest satisfies $my'' + cy' + ky = 0$ , that is $y'' + 3.75y' + 40y = 0$ . The roots are $r = -1.875 \pm 6.040i$ . So the car bounces at angular frequency $6.04$ rad/s (period $1.04$ s, about once a second) and each bounce is smaller than the one before by the factor $e^{-1.875 \times 1.04} = 0.142$ . After three bounces the amplitude is $0.142^3 = 0.0029$ of the first. Compare the critical value $c_{\text{crit}} = 2\sqrt{km} = 5060$ N s/m. With this $c$ the damping ratio is $c/c_{\text{crit}} = 0.30$ , a deliberately underdamped setting: at critical damping (the double root, the second preset of the picture) the car would sink back without bouncing, which designers avoid because the ride becomes stiff. Make $c$ very large and you reach the overdamped case, where the car returns slowly. All three cases are the three cases of the characteristic polynomial.

Used in the real world· Electrical engineering: an RLC circuit

A coil of inductance $L = 0.1$ H, a resistor $R = 20\ \Omega$ and a capacitor $C = 100\ \mu$ F are in series. The charge $q$ on the capacitor satisfies $Lq'' + Rq' + q/C = 0$ , that is $q'' + 200q' + 10^5q = 0$ . The characteristic roots are $-100 \pm 300i$ (check: $(-100)^2 - 10^5 = -90000$ , whose square root is $300i$ ). So $q(t) = e^{-100t}(A\cos300t + B\sin300t)$ . The charge rings at $300/(2\pi) = 47.7$ Hz and dies away with a time constant $\frac{1}{100} = 10$ ms: after $23$ ms the amplitude has fallen to a tenth. The roles are the mass-spring ones with $L \leftrightarrow m$ , $R \leftrightarrow c$ , $1/C \leftrightarrow k$ . This is why one equation serves for suspensions, circuits, and the vibrations of buildings.

Traps

Watch out· Only the homogeneous solution set is a vector space

On the 376 final (6(b)) you answered True to “the solutions of $y'' + y = \cos t$ form a vector space of dimension 2”. The solution set of an inhomogeneous equation misses the zero function and is not closed under addition. It is a translate $y_p + \ker L$ . The dimension $2$ belongs to $\ker L$ , the associated homogeneous equation. Always look at the right-hand side first.

Watch out· Purely imaginary roots have no exponential factor

On the same question you wrote the solution with roots $\pm i$ as $e^t(\cos t + \frac12\sin t)$ . A root $\alpha \pm i\beta$ gives $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$ . For $\pm i$ , $\alpha = 0$ , so the solutions are $\cos t$ and $\sin t$ with no exponential. Also, the two functions are separate basis elements, not one function with a fixed ratio $\frac12$ between its parts.

Watch out· A repeated root gives t times the exponential, not the exponential twice

For roots $1, 1, 2$ the solution $c_1e^t + c_2e^t + c_3e^{2t}$ has only two independent functions. The count of functions must equal the order. Worksheet 20 #8(b) tests exactly this.

Watch out· Differentiate with the product rule before you substitute the starting values

For $y = (c_1 + c_2t)e^t$ the derivative is $c_2e^t + (c_1 + c_2t)e^t$ , not $c_2e^t$ . Substituting $t = 0$ in $y'$ then needs both terms. It is also easy to write $y'(0) = c_2$ only. The check that catches it is to substitute your answer back.

Watch out· Operators commute only for constant coefficients

$(D - 2)(D - 3) = (D - 3)(D - 2)$ because the coefficients are numbers. With a variable coefficient, $D(ty) = y + ty'$ but $t\,Dy = ty'$ , so $D \circ t \ne t \circ D$ . The factoring method and the ‘kernel of a factor’ fact are only for constant coefficients.

Watch out· The order, not the number of distinct roots, is the dimension

A cubic with roots $1, 1, 1$ has three solutions $e^t, te^t, t^2e^t$ even though there is one root. The multiplicities add to $n$ .

Problems

checkMATH 376 worksheet 20 #1 to #4

Find the characteristic roots and the general solution of: (a) $y'' - 3y' + 2y = 0$ ; (b) $y'' + y' - 6y = 0$ ; (c) $y'' - 4y' + 4y = 0$ ; (d) $y'' - 7y' + 10y = 0$ .

Solution

(a) $t^2 - 3t + 2 = (t - 1)(t - 2)$ : $y = c_1e^t + c_2e^{2t}$ . (b) $t^2 + t - 6 = (t + 3)(t - 2)$ : $y = c_1e^{-3t} + c_2e^{2t}$ . (c) $(t - 2)^2$ : $y = (c_1 + c_2t)e^{2t}$ . (d) $(t - 5)(t - 2)$ : $y = C_1e^{5t} + C_2e^{2t}$ .

checkMATH 376 worksheet 19 #5; final 6(b)

For each, say whether the solution set is a vector space and, if yes, its dimension: (a) $y'' - y = 0$ ; (b) $y'' - y = e^t$ ; (c) $y''' + y' = 0$ ; (d) $y' = y^2$ .

Solution

(a) Yes, dimension $2$ . (b) No: the zero function is not a solution ( $0 \ne e^t$ ); it is $y_p +$ (a $2$ -dimensional space), with $y_p = \frac t2e^t$ . (c) Yes, dimension $3$ . (d) No. The equation is not linear. $y = \frac{1}{1-t}$ is a solution but $2y = \frac{2}{1-t}$ is not: $(2y)' = \frac{2}{(1-t)^2}$ while $(2y)^2 = \frac{4}{(1-t)^2}$ . So the set is not closed under scaling.

checkMATH 376 worksheet 20 #8

True or false: (a) If $2$ is a double root, $c_1e^{2x} + c_2xe^{2x}$ is the general solution of the associated part. (b) If the roots are $1, 1, 2$ the general solution is $c_1e^x + c_2e^x + c_3e^{2x}$ . (c) If the roots are $-1, 2, 3$ the general solution is $c_1e^{-x} + c_2e^{2x} + c_3e^{3x}$ .

Solution

(a) True. (b) False: it is $c_1e^x + c_2xe^x + c_3e^{2x}$ . (c) True.

checkMATH 376 worksheet 21 #9

The solutions $y_1, y_2$ of a second-order homogeneous equation satisfy $y_1(0) = 1$ , $y_1'(0) = 2$ , $y_2(0) = 0$ , $y_2'(0) = 3$ . Are they a basis of the solution space?

Solution

Yes. Their starting vectors $(1, 2)$ and $(0, 3)$ are independent in $\mathbb R^2$ (determinant $3 \ne 0$ ); $\mu$ is an isomorphism, so $y_1, y_2$ are independent, and two independent solutions of an order- $2$ equation form a basis.

useMATH 376 worksheet 28 #1; homework 8 #3

Solve $y'' - 3y' + 2y = 0$ with $y(0) = 2$ , $y'(0) = 1$ ; and give the general solution of $y'' - 7y' + 10y = 0$ .

Solution

$y = c_1e^t + c_2e^{2t}$ ; $c_1 + c_2 = 2$ , $c_1 + 2c_2 = 1$ gives $c_2 = -1$ , $c_1 = 3$ . So $y = 3e^t - e^{2t}$ . Check: $y' = 3e^t - 2e^{2t}$ , $y'(0) = 1$ . Second: $y = C_1e^{5t} + C_2e^{2t}$ .

useMATH 376 worksheet 21 #4; worksheet 28 #2

Solve $y'' - 2y' + 5y = 0$ with $y(0) = 1$ , $y'(0) = 0$ . Then give the general solution of $y'' - 4y' + 13y = 0$ .

Solution

Roots $1 \pm 2i$ . $y = e^t(C_1\cos2t + C_2\sin2t)$ . $y(0) = C_1 = 1$ . $y' = e^t[(C_1 + 2C_2)\cos2t + (C_2 - 2C_1)\sin2t]$ , so $y'(0) = C_1 + 2C_2 = 0$ gives $C_2 = -\frac12$ . $y = e^t(\cos2t - \frac12\sin2t)$ . For the second equation, $t^2 - 4t + 13 = 0$ gives $t = 2 \pm 3i$ , so $y = e^{2t}(C_1\cos3t + C_2\sin3t)$ (worksheet 23 #2).

use

Find a real basis for the solutions of $y^{(5)} - 2y^{(4)} + y''' = 0$ and for $y''' + 4y' = 0$ .

Solution

First: $t^5 - 2t^4 + t^3 = t^3(t - 1)^2$ : roots $0$ (triple) and $1$ (double). Basis: $1, t, t^2, e^t, te^t$ . Second: $t^3 + 4t = t(t^2 + 4)$ : roots $0, \pm2i$ . Basis: $1, \cos2t, \sin2t$ .

use

A mass-spring system obeys $2x'' + 12x' + 50x = 0$ . Find the roots, the frequency of the oscillation and the time for the amplitude to halve.

Solution

Divide by $2$ : $x'' + 6x' + 25x = 0$ , roots $-3 \pm 4i$ . Angular frequency $4$ rad/s ( $0.64$ Hz). The amplitude is multiplied by $e^{-3t}$ ; halving when $e^{-3t} = \frac12$ , $t = \frac{\ln2}{3} = 0.231$ s.

proveMATH 376 worksheet 19 #4

Prove that $S = \{y \in C^2(I) : y'' + p(t)y' + q(t)y = 0\}$ is a vector space.

Solution

Plan. Show $S$ is a subspace of $C^2(I)$ : contains zero, closed under sum and scalar multiple.

$0'' + p0' + q0 = 0$ , so $0 \in S$ . If $y_1, y_2 \in S$ then $(y_1 + y_2)'' + p(y_1 + y_2)' + q(y_1 + y_2) = (y_1'' + py_1' + qy_1) + (y_2'' + py_2' + qy_2) = 0$ . If $y \in S$ and $c$ is a number, $(cy)'' + p(cy)' + q(cy) = c(y'' + py' + qy) = 0$ . By the subspace test, $S$ is a subspace of the vector space $C^2(I)$ , hence a vector space. $\blacksquare$

proveMATH 376 worksheet 19 #5

Let $g \not\equiv 0$ be continuous. Prove that $S_g = \{y : y'' + p(t)y' + q(t)y = g(t)\}$ is not a vector space, and describe it using a particular solution.

Solution

Plan. Show the zero function is missing; then write $S_g = y_p + S$ .

$L[0] = 0 \ne g$ since $g \not\equiv 0$ , so $0 \notin S_g$ and $S_g$ is not a subspace. (Also if $y_1, y_2 \in S_g$ then $L[y_1 + y_2] = 2g \ne g$ .) If $y_p \in S_g$ , then for $y \in S_g$ we have $L[y - y_p] = g - g = 0$ , so $y - y_p \in S$ ; conversely $L[y_p + h] = g$ for $h \in S$ . Hence $S_g = y_p + S$ . $\blacksquare$

proveMATH 376 homework 8 #1(3)

Prove that $e^{t_1t}, \dots, e^{t_rt}$ are linearly independent on any interval when $t_1, \dots, t_r$ are distinct. (Use the independence theorem of this chapter.)

Solution

Plan. Apply the key claim of the proof of the basis theorem with all polynomials constant.

Suppose $\sum_jc_je^{t_jt} = 0$ on an interval. This is the case of the claim ” $\sum p_je^{r_jt} = 0$ with $r_j$ distinct implies all $p_j = 0$ ” in which each $p_j = c_j$ is a constant polynomial. The claim, proved by induction on the number of exponentials, says $c_j = 0$ for all $j$ . (Alternative: the Wronskian of the functions is $e^{(t_1 + \dots + t_r)t}$ times a Vandermonde determinant $\prod_{i<j}(t_j - t_i) \ne 0$ ; see 7.3.) $\blacksquare$

prove

Prove that $A$ and $B$ commute when $A = D^2 + 3D + 2$ and $B = D - 5$ by computing $AB$ and $BA$ on a function $f$ .

Solution

Plan. Expand both orders and compare.

$ABf = (D^2 + 3D + 2)(f' - 5f) = f''' - 5f'' + 3f'' - 15f' + 2f' - 10f = f''' - 2f'' - 13f' - 10f$ . $BAf = (D - 5)(f'' + 3f' + 2f) = f''' + 3f'' + 2f' - 5f'' - 15f' - 10f = f''' - 2f'' - 13f' - 10f$ . They agree, and both equal $(D^3 - 2D^2 - 13D - 10)f$ , with characteristic polynomial $(t^2 + 3t + 2)(t - 5) = t^3 - 2t^2 - 13t - 10$ . $\blacksquare$

proveTechnau, March 25

Prove that the characteristic polynomial $\det(\lambda I - A)$ of the companion matrix of $y'' + by' + cy = 0$ equals $\lambda^2 + b\lambda + c$ . Then explain why $e^{rt}$ is a solution of the equation exactly when $r$ is an eigenvalue of the companion matrix.

Solution

Plan. Compute the $2\times 2$ determinant; then find the eigenvectors.

$A = \begin{bmatrix}0 & 1\\-c & -b\end{bmatrix}$ , so $\lambda I - A = \begin{bmatrix}\lambda & -1\\c & \lambda + b\end{bmatrix}$ and $\det = \lambda(\lambda + b) + c = \lambda^2 + b\lambda + c$ . For the second part: if $e^{rt}$ solves the equation then $x = (y, y')= e^{rt}(1, r)$ solves $x' = Ax$ , so $A(1, r) = r(1, r)$ , and $r$ is an eigenvalue with eigenvector $(1, r)$ . Conversely, if $Av = rv$ then $e^{rt}v$ solves $x' = Ax$ and its first coordinate solves the equation. (In general the characteristic polynomial of the companion matrix of an order- $n$ equation is $\lambda^n + a_{n-1}\lambda^{n-1} + \dots + a_0$ , by induction on $n$ expanding along the first column.) $\blacksquare$

proveMATH 376 practice final 4(b); new generalisation

Show that for $r \in \mathbb R$ and $j \in \{0, \dots, m-1\}$ , $t^je^{rt}$ solves $(D - r)^my = 0$ , and that these $m$ functions are independent.

Solution

Plan. Use the shifting fact; then divide by the exponential.

$(D - r)(pe^{rt}) = p'e^{rt}$ for a polynomial $p$ (product rule). Applying this $j + 1$ times to $p = t^j$ gives $\frac{d^{j+1}}{dt^{j+1}}t^j \cdot e^{rt} = 0$ , and $m \ge j + 1$ gives $(D - r)^mt^je^{rt} = 0$ . If $\sum_{j<m}c_jt^je^{rt} = 0$ , divide by $e^{rt} > 0$ : the polynomial $\sum c_jt^j$ has infinitely many roots, so it is zero and all $c_j = 0$ . $\blacksquare$

stretch

The equation $y'' + p(t)y' + q(t)y = 0$ on $(0, \infty)$ with $p = 1/t$ , $q = -1/t^2$ is $t^2y'' + ty' - y = 0$ (an Euler equation), whose coefficients are not constant. Try $y = t^s$ and find two independent solutions. Why does the dimension theorem say that is all of them?

Solution

$y = t^s$ gives $t^2s(s-1)t^{s-2} + t\,st^{s-1} - t^s = (s^2 - 1)t^s$ , which is zero when $s = \pm1$ . So $t$ and $t^{-1}$ are solutions. They are independent (neither is a multiple of the other) on $(0, \infty)$ . The coefficients $p = 1/t$ and $q = -1/t^2$ are continuous on $(0, \infty)$ , so the solution space has dimension $2$ by the dimension theorem; two independent solutions are a basis, and the general solution is $c_1t + c_2/t$ . (Rewrite with $t = e^s$ to reduce an Euler equation to constant coefficients.)

stretch

Show that every real solution of $y'' + \omega^2y = 0$ ( $\omega > 0$ ) can be written $y = A\cos(\omega t - \varphi)$ with $A \ge 0$ , and find $A$ and $\varphi$ for $y = 3\cos2t + 4\sin2t$ .

Solution

The general real solution is $y = c_1\cos\omega t + c_2\sin\omega t$ . Put $A = \sqrt{c_1^2 + c_2^2}$ and choose $\varphi$ with $\cos\varphi = c_1/A$ , $\sin\varphi = c_2/A$ . Then $A\cos(\omega t - \varphi) = A\cos\omega t\cos\varphi + A\sin\omega t\sin\varphi = c_1\cos\omega t + c_2\sin\omega t$ . For $3\cos2t + 4\sin2t$ : $A = 5$ , $\cos\varphi = \frac35$ , $\sin\varphi = \frac45$ , so $\varphi = \arctan\frac43 = 0.9273$ radians. So $y = 5\cos(2t - 0.9273)$ : an oscillation of amplitude $5$ and frequency $2$ rad/s. $\blacksquare$

Summary

A linear differential equation of order $n$ is $Ly = R$ with $L = D^n + a_{n-1}D^{n-1} + \dots + a_0$ . $L$ is a linear map, so the solutions of the homogeneous equation $Ly = 0$ are its null space, a vector space.
The solutions of an inhomogeneous equation ( $R \ne 0$ ) are NOT a vector space: they are $y_p + \ker L$ , a translate of the null space. Subtract two solutions and the right-hand sides cancel.
The map “solution $\mapsto (y(t_0), \dots, y^{(n-1)}(t_0))$ ” is linear, onto by existence and one-to-one by uniqueness (7.1, via the companion matrix). So $\dim\ker L = n$ , and the general solution is $y = y_p + c_1b_1 + \dots + c_nb_n$ .
With constant coefficients $L = P(D)$ and operators multiply like polynomials, so they commute. A real root $r$ of multiplicity $m$ gives $e^{rt}, te^{rt}, \dots, t^{m-1}e^{rt}$ , because $(D - r)$ lowers the degree of a polynomial factor by one.
A complex pair $\alpha \pm i\beta$ gives $t^je^{\alpha t}\cos\beta t$ and $t^je^{\alpha t}\sin\beta t$ (real and imaginary parts of $t^je^{rt}$ ). No exponential factor appears when $\alpha = 0$ .
The recipe: factor $P$ , list $n$ functions, take an arbitrary real combination; with starting values, differentiate with the product rule and solve an $n \times n$ system. The count of functions must equal the order.
Your 376 traps: look at the right-hand side before talking about vector spaces (final 6(b)), the root pair $\pm i$ has no exponential, and a repeated root needs the factor $t$ .

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

The picture

The idea, in words

Definitions and theorems

Linear equations and linear operators

Linear differential equation

Homogeneous equation

Differential operator

A linear differential operator is a linear map

Homogeneous and inhomogeneous solution sets

Existence, uniqueness, and the dimension theorem

Companion matrix

Existence and uniqueness for linear equations of order n

Dimension of the solution space of a linear ODE

General solution of a linear equation

General solution

Particular solution

Equations with constant coefficients

Constant-coefficient operator

Characteristic polynomial of an operator

Constant-coefficient operators multiply like polynomials

Derivative of a complex exponential

Shifting an exponential

Basis for repeated roots

Kernel of a factor lies in the kernel of the product

Basis of solutions from the roots

Complex roots, and real solutions

Real solutions from complex roots

The first-order case and the companion matrix

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere