Chapter 4 · Section 4.2

4.2 Homogeneous Linear Equations with Constant Coefficients

Section 4.1 told us what the general solution of a homogeneous linear equation looks like: $n$ independent solutions combined with $n$ arbitrary constants. It did not tell us how to find those solutions. For equations with variable coefficients there is no general recipe. But when every coefficient is a constant, the problem reduces to finding the roots of a polynomial. This case is extremely important in practice: mass–spring systems, RLC circuits, vibrating structures and many control systems are all modeled by constant-coefficient equations.

In this section we study

a_{0} y^{(n)} + a_{1} y^{(n - 1)} + \dots + a_{n - 1} y^{'} + a_{n} y = 0,

where $a_{0}, a_{1}, \dots, a_{n}$ are real constants and $a_{0} \neq = 0$ . Since constants are continuous everywhere and $a_{0}$ never vanishes, Theorem 4.1.1 applies on the whole real line: every solution is defined on $(- \infty, \infty)$ .

The key idea: try an exponential

What kind of function could satisfy such an equation? We need a combination of $y, y^{'}, y^{''}, \dots$ with constant weights to cancel exactly. That is easiest if all the derivatives of $y$ look like $y$ itself. The exponential $e^{r x}$ has exactly this property:

\frac{d}{d x} e^{r x} = r e^{r x}, \frac{d ^{2}}{d x ^{2}} e^{r x} = r^{2} e^{r x}, \dots, \frac{d ^{k}}{d x ^{k}} e^{r x} = r^{k} e^{r x} .

Substituting $y = e^{r x}$ into the left-hand side gives

a_{0} r^{n} e^{r x} + a_{1} r^{n - 1} e^{r x} + \dots + a_{n} e^{r x} = (a_{0} r^{n} + a_{1} r^{n - 1} + \dots + a_{n - 1} r + a_{n}) e^{r x} .

Because $e^{r x}$ is never zero, this vanishes for all $x$ exactly when the polynomial in parentheses is zero.

Definition 4.2.1 (Characteristic polynomial and equation)

The polynomial

p (r) = a_{0} r^{n} + a_{1} r^{n - 1} + \dots + a_{n - 1} r + a_{n}

is the characteristic polynomial of the equation, and $p (r) = 0$ is its characteristic equation (sometimes called the auxiliary equation). The function $e^{r x}$ is a solution if and only if $r$ is a root of $p$ .

To write down $p (r)$ , simply replace $y^{(k)}$ by $r^{k}$ (and $y$ by $r^{0} = 1$ ). For example, $3 y^{'''} - y^{'} + 8 y = 0$ has $p (r) = 3 r^{3} - r + 8$ . Do not forget that a missing derivative means a zero coefficient.

In the operator notation of Section 4.1 the equation reads $p (D) y = 0$ : the same polynomial, with $D$ in place of $r$ . Constant-coefficient operators can be multiplied and factored exactly like polynomials, and the order of the factors does not matter. For instance, $(D - 2) (D + 5) y = (D^{2} + 3 D - 10) y$ . We will use this below.

A polynomial of degree $n$ has exactly $n$ roots if we count complex roots and repeated roots with their multiplicity. Three things can happen, and each needs a slightly different treatment:

the roots are real and all different;
some real root is repeated;
some roots are complex (they then come in conjugate pairs, because the coefficients are real).

Case 1: distinct real roots

Theorem 4.2.1 (Distinct real roots)

If the characteristic equation has $n$ distinct real roots $r_{1}, r_{2}, \dots, r_{n}$ , then

e^{r_{1} x}, e^{r_{2} x}, \dots, e^{r_{n} x}

form a fundamental set on $(- \infty, \infty)$ , and the general solution is

y = c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x} + \dots + c_{n} e^{r_{n} x} .

Each $e^{r_{k} x}$ is a solution because $p (r_{k}) = 0$ . They are independent because their Wronskian equals $e^{(r_{1} + \dots + r_{n}) x}$ times a determinant of powers of the $r_{k}$ (a Vandermonde determinant), which equals the product of all differences $r_{j} - r_{i}$ with $i < j$ . When the roots are distinct, none of these differences is zero. You saw a concrete case in Example 4.1.3, where the determinant was $(2 - (- 1)) (5 - (- 1)) (5 - 2) = 54$ .

Example 4.2.1

Find the general solution of $2 y^{''} - 7 y^{'} + 3 y = 0$ .

Solution

Characteristic equation. $2 r^{2} - 7 r + 3 = 0$ .
Roots. By the quadratic formula, $r = \frac{7 \pm 49 - 4 \cdot 2 \cdot 3}{2 \cdot 2} = \frac{7 \pm 25}{4} = \frac{7 \pm 5}{4},$ so $r_{1} = 3$ and $r_{2} = \frac{1}{2}$ . (Equivalently, $2 r^{2} - 7 r + 3 = (2 r - 1) (r - 3)$ .)
General solution. The roots are real and distinct, so $y = c_{1} e^{x /2} + c_{2} e^{3 x} .$

Figure 4.2.1. Basic solutions

e^{r x}

for several real roots

r

, including the roots

3

and

\frac{1}{2}

of Example 4.2.1. Every curve passes through

(0, 1)

. A positive root gives growth, a negative root gives decay, and a larger

∣ r ∣

makes the change faster.

Case 2: repeated real roots

If $r_{1}$ is a double root, the exponential $e^{r_{1} x}$ gives only one solution, but a double root "uses up" two of the $n$ roots. We are one solution short. Where does the missing one come from?

Why $x e^{r x}$ works: first argument (reduction of order)

A second-order equation whose characteristic polynomial is $(r - r_{1})^{2} = r^{2} - 2 r_{1} r + r_{1}^{2}$ is

y^{''} - 2 r_{1} y^{'} + r_{1}^{2} y = 0.

We know $y_{1} = e^{r_{1} x}$ . Apply the reduction-of-order formula (Theorem 4.1.7) with $P = - 2 r_{1}$ :

e^{- \int P d x} \frac{e ^{- \int P d x}}{y _{1}^{2}} v y_{2} = e^{2 r_{1} x}, = \frac{e ^{2 r_{1} x}}{e ^{2 r_{1} x}} = 1, = \int 1 d x = x, = v y_{1} = x e^{r_{1} x} .

So the second solution is $x e^{r_{1} x}$ .

Second argument (works for any multiplicity)

Let $u (x)$ be any differentiable function. Compute $(D - r)$ applied to the product $e^{r x} u$ :

(D - r) [e^{r x} u] = (r e^{r x} u + e^{r x} u^{'}) - r e^{r x} u = e^{r x} u^{'} .

In words: $(D - r)$ acts on $e^{r x} u$ by simply differentiating $u$ . Applying it $m$ times,

(D - r)^{m} [e^{r x} u] = e^{r x} u^{(m)} .

Now choose $u = x^{k}$ with $k \leq m - 1$ . Then $u^{(m)} = 0$ , because differentiating $x^{k}$ more than $k$ times gives zero. Hence

(D - r)^{m} [x^{k} e^{r x}] = 0 for k = 0, 1, \dots, m - 1.

If $r$ is a root of multiplicity $m$ , the characteristic polynomial factors as $p (r) = q (r) (r - r_{1})^{m}$ , and so $p (D) = q (D) (D - r_{1})^{m}$ . Applying $p (D)$ to $x^{k} e^{r_{1} x}$ , the factor $(D - r_{1})^{m}$ acts first and already gives zero. So $x^{k} e^{r_{1} x}$ is a solution for $k = 0, \dots, m - 1$ .

Theorem 4.2.2 (Repeated real roots)

If $r_{1}$ is a real root of the characteristic equation of multiplicity $m$ , then it contributes the $m$ independent solutions

e^{r_{1} x}, x e^{r_{1} x}, x^{2} e^{r_{1} x}, \dots, x^{m - 1} e^{r_{1} x}

to the general solution, that is, the term $(c_{1} + c_{2} x + \dots + c_{m} x^{m - 1}) e^{r_{1} x}$ .

Example 4.2.2

Solve $4 y^{''} + 20 y^{'} + 25 y = 0$ .

Solution

Characteristic equation. $4 r^{2} + 20 r + 25 = 0$ .
Discriminant. $2 0^{2} - 4 \cdot 4 \cdot 25 = 400 - 400 = 0$ , so there is one double root. In fact $4 r^{2} + 20 r + 25 = (2 r + 5)^{2}$ .
Root. $r = - \frac{5}{2}$ , multiplicity $2$ .
General solution. By Theorem 4.2.2, $y = (c_{1} + c_{2} x) e^{- 5 x /2} .$

Figure 4.2.2. Solutions

(c_{1} + c_{2} x) e^{- 5 x /2}

of Example 4.2.2 for four choices of the constants. The factor

x

that comes from the double root lets a solution rise at first or cross zero once, but the exponential always wins and every solution decays to

0

Case 3: complex roots

The equation $y^{''} + y = 0$ has characteristic equation $r^{2} + 1 = 0$ with roots $r = \pm i$ . Yet we know two perfectly real solutions: $cos x$ and $sin x$ . The bridge between " $e^{i x}$ " and " $cos x$ , $sin x$ " is Euler's formula.

Euler's formula, step by step

Start from the power series of the exponential, valid for every number $t$ , real or complex: $e^{t} = 1 + t + \frac{t ^{2}}{2 !} + \frac{t ^{3}}{3 !} + \frac{t ^{4}}{4 !} + \frac{t ^{5}}{5 !} + \dots$
Put $t = i θ$ with $θ$ real, and use $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ , $i^{5} = i$ , and so on: $e^{i θ} = 1 + i θ - \frac{θ ^{2}}{2 !} - i \frac{θ ^{3}}{3 !} + \frac{θ ^{4}}{4 !} + i \frac{θ ^{5}}{5 !} - \dots$
Separate the terms without $i$ from those with $i$ : $e^{i θ} = (1 - \frac{θ ^{2}}{2 !} + \frac{θ ^{4}}{4 !} - \dots) + i (θ - \frac{θ ^{3}}{3 !} + \frac{θ ^{5}}{5 !} - \dots) .$
Recognize the two series. The first bracket is the Maclaurin series of $cos θ$ , the second is that of $sin θ$ . Therefore $e^{i θ} = cos θ + i sin θ (Euler’s formula).$
Apply it to a complex root $r = a + ib$ . Using the law of exponents (which still holds for complex exponents) and Euler's formula with $θ = b x$ : $e^{(a + ib) x} = e^{a x} e^{ib x} = e^{a x} (cos b x + i sin b x) = e^{a x} cos b x + i e^{a x} sin b x .$

One can check directly from the last line, with the product rule, that $\frac{d}{d x} e^{(a + ib) x} = (a + ib) e^{(a + ib) x}$ . So the computation at the start of this section works for complex $r$ too: if $a + ib$ is a root of $p$ , then $e^{(a + ib) x}$ is a (complex-valued) solution.

From a complex solution to two real ones

Let $z (x) = u (x) + i v (x)$ be a complex-valued solution, with $u$ and $v$ real. Because $L$ is linear and its coefficients are real,

0 = L [z] = L [u] + i L [v] .

A complex number is zero only if its real and imaginary parts are both zero. So $L [u] = 0$ and $L [v] = 0$ : the real part and the imaginary part are each real solutions. For $z = e^{(a + ib) x}$ this gives

u = e^{a x} cos b x, v = e^{a x} sin b x .

The conjugate root $a - ib$ produces $e^{a x} cos b x - i e^{a x} sin b x$ , whose real and imaginary parts are the same two functions (up to sign). So a conjugate pair of roots gives exactly two real solutions, which is the right count. Their Wronskian is $b e^{2 a x} \neq = 0$ (since $b \neq = 0$ ), so they are independent.

Theorem 4.2.3 (Complex roots)

If $a \pm ib$ ( $b \neq = 0$ ) is a pair of complex conjugate roots of the characteristic equation, each of multiplicity $m$ , then the pair contributes the $2 m$ independent real solutions

e^{a x} cos b x, x e^{a x} cos b x, \dots, x^{m - 1} e^{a x} cos b x, e^{a x} sin b x, x e^{a x} sin b x, \dots, x^{m - 1} e^{a x} sin b x .

For a simple pair ( $m = 1$ ) this is the term $e^{a x} (c_{1} cos b x + c_{2} sin b x)$ .

Collecting the solutions from all roots as in Theorems 4.2.1–4.2.3 always gives exactly $n$ functions, and they always form a fundamental set on $(- \infty, \infty)$ .

Engineering reading of the roots

The real part $a$ of a root controls growth or decay ( $e^{a x}$ ), and the imaginary part $b$ controls oscillation ( $cos b x$ , $sin b x$ , angular frequency $b$ ). If every root has negative real part, every solution dies out as $x \to \infty$ ; an engineer would call the system stable. A root with positive real part means some solution grows without bound. Chapter 5 uses exactly this reading for springs and circuits.

Example 4.2.3

Solve the initial-value problem $y^{''} + 6 y^{'} + 34 y = 0$ , $y (0) = 2$ , $y^{'} (0) = - 1$ .

Solution

Characteristic equation. $r^{2} + 6 r + 34 = 0$ .
Roots. $r = \frac{- 6 \pm 36 - 136}{2} = \frac{- 6 \pm - 100}{2} = \frac{- 6 \pm 10 i}{2} = - 3 \pm 5 i .$ So $a = - 3$ , $b = 5$ .
General solution. $y = e^{- 3 x} (c_{1} cos 5 x + c_{2} sin 5 x)$ .
Derivative. By the product rule, $y^{'} = - 3 e^{- 3 x} (c_{1} cos 5 x + c_{2} sin 5 x) + e^{- 3 x} (- 5 c_{1} sin 5 x + 5 c_{2} cos 5 x) = e^{- 3 x} [(- 3 c_{1} + 5 c_{2}) cos 5 x + (- 3 c_{2} - 5 c_{1}) sin 5 x] .$
Apply the initial conditions. At $x = 0$ , $cos 0 = 1$ and $sin 0 = 0$ : $y (0) y^{'} (0) = c_{1} = 2, = - 3 c_{1} + 5 c_{2} = - 1 \Rightarrow - 6 + 5 c_{2} = - 1 \Rightarrow c_{2} = 1.$
Answer. $y = e^{- 3 x} (2 cos 5 x + sin 5 x)$ , an oscillation with angular frequency $5$ whose amplitude decays like $e^{- 3 x}$ .

Figure 4.2.3. The solution of Example 4.2.3. The real part

- 3

of the roots sets the decaying envelope

\pm 5 e^{- 3 x}

(dashed), and the imaginary part

5

sets the angular frequency of the oscillation.

Method in a nutshell

Write the characteristic polynomial $p (r)$ by replacing $y^{(k)}$ with $r^{k}$ .
Find all $n$ roots of $p (r) = 0$ , with multiplicities (see the helper below).
Translate each root into solutions:
- simple real root $r$ : $e^{r x}$ ;
- real root $r$ of multiplicity $m$ : $e^{r x}, x e^{r x}, \dots, x^{m - 1} e^{r x}$ ;
- simple complex pair $a \pm ib$ : $e^{a x} cos b x, e^{a x} sin b x$ ;
- complex pair of multiplicity $m$ : the two functions above multiplied by $1, x, \dots, x^{m - 1}$ .
Check that you have exactly $n$ functions; multiply them by $c_{1}, \dots, c_{n}$ and add.
For an IVP, differentiate the general solution $n - 1$ times, substitute $x_{0}$ , and solve the linear system for $c_{1}, \dots, c_{n}$ .

Try the method on your own equations with the solver below. Enter the coefficients from the highest derivative down to $y$ (order up to 6) and compare its roots and general solution with what the recipe gives you.

Helper: finding the roots of a polynomial

For second-order equations the quadratic formula always works. For higher orders, finding roots is the hardest part of the method. Here are the tools you will use most.

The rational root test

Theorem 4.2.4 (Rational root test)

Let $p (r) = a_{0} r^{n} + a_{1} r^{n - 1} + \dots + a_{n}$ have integer coefficients with $a_{0} \neq = 0$ , $a_{n} \neq = 0$ . If $r = \frac{s}{t}$ (in lowest terms) is a rational root, then $s$ divides the constant term $a_{n}$ and $t$ divides the leading coefficient $a_{0}$ .

In particular, if the leading coefficient is $1$ , every rational root is an integer that divides the constant term. This gives a short, finite list of candidates to test. (If $a_{n} = 0$ , factor out $r$ first; that gives the root $r = 0$ .)

Synthetic division

Once a root $r_{1}$ is found, divide $p (r)$ by $(r - r_{1})$ to get a polynomial of lower degree. Synthetic division does this quickly using only the coefficients. Let us divide $p (r) = r^{3} - 2 r^{2} - 11 r + 12$ by $(r - 1)$ :

111 - 2 1 - 1 - 11 - 1 - 12 12 - 12 0

How to read it: write the root $1$ on the left and the coefficients on the top row. Bring down the first coefficient ( $1$ ). Multiply it by the root and write the result under the next coefficient ( $1 \cdot 1 = 1$ ); add ( $- 2 + 1 = - 1$ ). Repeat: $(- 1) \cdot 1 = - 1$ , and $- 11 + (- 1) = - 12$ ; then $(- 12) \cdot 1 = - 12$ , and $12 + (- 12) = 0$ . The last number is the remainder; it is $0$ because $1$ is a root. The other numbers on the bottom row are the coefficients of the quotient:

r^{3} - 2 r^{2} - 11 r + 12 = (r - 1) (r^{2} - r - 12) .

Other useful tricks

Grouping: $r^{3} + 4 r^{2} + 9 r + 36 = r^{2} (r + 4) + 9 (r + 4) = (r + 4) (r^{2} + 9)$ .
Equations in $r^{2}$ only: for $r^{4} + 5 r^{2} - 36$ , substitute $s = r^{2}$ to get $s^{2} + 5 s - 36 = (s + 9) (s - 4)$ , so $r = \pm 2$ and $r = \pm 3 i$ .
Pure powers: $r^{3} - c^{3} = (r - c) (r^{2} + cr + c^{2})$ and $r^{3} + c^{3} = (r + c) (r^{2} - cr + c^{2})$ .
Sign check: if all coefficients of $p$ are positive, $p$ has no positive real roots, so only negative candidates need testing. Similarly, if $p (- r)$ has only positive coefficients, $p$ has no negative real roots.
Perfect squares: a quartic whose first and last coefficients are squares may be the square of a quadratic; see Example 4.2.6.
If none of this works, the roots must be found numerically (for example with a computer algebra system). Exam and textbook problems are normally designed so that the tricks above suffice.

Example 4.2.4

Solve $y^{'''} - 2 y^{''} - 11 y^{'} + 12 y = 0$ with $y (0) = 2$ , $y^{'} (0) = - 9$ , $y^{''} (0) = 3$ .

Solution

Characteristic equation. $r^{3} - 2 r^{2} - 11 r + 12 = 0$ .
Candidates. The leading coefficient is $1$ , so rational roots are divisors of $12$ : $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ .
Test. $p (1) = 1 - 2 - 11 + 12 = 0$ . So $r = 1$ is a root.
Divide. The synthetic division above gives $p (r) = (r - 1) (r^{2} - r - 12)$ .
Factor the quadratic. $r^{2} - r - 12 = (r - 4) (r + 3)$ . The roots are $1, 4, - 3$ : real and distinct.
General solution. $y = c_{1} e^{x} + c_{2} e^{- 3 x} + c_{3} e^{4 x}$ .
Derivatives. $y^{'} y^{''} = c_{1} e^{x} - 3 c_{2} e^{- 3 x} + 4 c_{3} e^{4 x}, = c_{1} e^{x} + 9 c_{2} e^{- 3 x} + 16 c_{3} e^{4 x} .$
Initial conditions at $x = 0$ . $c_{1} + c_{2} + c_{3} c_{1} - 3 c_{2} + 4 c_{3} c_{1} + 9 c_{2} + 16 c_{3} = 2, = - 9, = 3. (1) (2) (3)$
Eliminate $c_{1}$ . $(2) - (1)$ : $- 4 c_{2} + 3 c_{3} = - 11$ . $(3) - (1)$ : $8 c_{2} + 15 c_{3} = 1$ .
Eliminate $c_{2}$ . Doubling the first gives $- 8 c_{2} + 6 c_{3} = - 22$ ; adding the second: $21 c_{3} = - 21$ , so $c_{3} = - 1$ . Then $- 4 c_{2} - 3 = - 11$ gives $c_{2} = 2$ , and (1) gives $c_{1} = 2 - 2 + 1 = 1$ .
Answer. $y = e^{x} + 2 e^{- 3 x} - e^{4 x}$ .

Example 4.2.5

Find the general solution of $y^{(4)} - 4 y^{'''} + 13 y^{''} - 36 y^{'} + 36 y = 0$ .

Solution

Characteristic equation. $p (r) = r^{4} - 4 r^{3} + 13 r^{2} - 36 r + 36 = 0$ .
Narrow the candidates. Divisors of $36$ are candidates. For $r < 0$ every term of $p (r)$ is positive ( $r^{4} > 0$ , $- 4 r^{3} > 0$ , $13 r^{2} > 0$ , $- 36 r > 0$ , $36 > 0$ ), so there are no negative roots. Test positive divisors: $p (1) = 10$ , $p (2) = 16 - 32 + 52 - 72 + 36 = 0$ .
First division by $(r - 2)$ . $211 - 4 2 - 2 13 - 4 9 - 36 18 - 18 36 - 36 0$ So $p (r) = (r - 2) (r^{3} - 2 r^{2} + 9 r - 18)$ .
Test $2$ again on the cubic: $8 - 8 + 18 - 18 = 0$ . It is a root again.
Second division. $211 - 2 20 909 - 18 180$ So $p (r) = (r - 2)^{2} (r^{2} + 9)$ . (Grouping also works: $r^{3} - 2 r^{2} + 9 r - 18 = r^{2} (r - 2) + 9 (r - 2)$ .)
Roots. $r = 2$ (double) and $r = \pm 3 i$ (so $a = 0$ , $b = 3$ ).
General solution. $y = (c_{1} + c_{2} x) e^{2 x} + c_{3} cos 3 x + c_{4} sin 3 x .$

Check Example 4.2.5 with the solver, then change a coefficient and see how the roots and the solution change.

Example 4.2.6 (Repeated complex roots)

Solve $y^{(4)} + 4 y^{'''} + 24 y^{''} + 40 y^{'} + 100 y = 0$ .

Solution

Characteristic equation. $p (r) = r^{4} + 4 r^{3} + 24 r^{2} + 40 r + 100$ . All coefficients are positive, so there are no positive real roots; testing the negative divisors of $100$ finds no root either. So we look for another structure.
Guess a perfect square. The first term is $(r^{2})^{2}$ and the last is $1 0^{2}$ . Try $(r^{2} + β r + 10)^{2}$ : $(r^{2} + β r + 10)^{2} = r^{4} + 2 β r^{3} + (β^{2} + 20) r^{2} + 20 β r + 100.$
Match coefficients. $2 β = 4$ gives $β = 2$ . Check the rest: $β^{2} + 20 = 24$ ✓ and $20 β = 40$ ✓. So $p (r) = (r^{2} + 2 r + 10)^{2}$ .
Roots of the quadratic. $r = \frac{- 2 \pm 4 - 40}{2} = - 1 \pm 3 i$ . Because of the square, each root has multiplicity $2$ .
General solution. By Theorem 4.2.3 with $a = - 1$ , $b = 3$ , $m = 2$ : $y = e^{- x} [(c_{1} + c_{2} x) cos 3 x + (c_{3} + c_{4} x) sin 3 x] .$

Example 4.2.7 (Reading off a solution)

The characteristic polynomial of a sixth-order homogeneous equation is

p (r) = (r - 3) (r + 1)^{3} (r^{2} + 4 r + 13) .

Write the general solution without expanding $p$ .

Solution

Factor $r - 3$ : simple real root $3$ , giving $e^{3 x}$ .
Factor $(r + 1)^{3}$ : real root $- 1$ of multiplicity $3$ , giving $e^{- x}, x e^{- x}, x^{2} e^{- x}$ .
Factor $r^{2} + 4 r + 13$ : $r = \frac{- 4 \pm 16 - 52}{2} = - 2 \pm 3 i$ , giving $e^{- 2 x} cos 3 x$ and $e^{- 2 x} sin 3 x$ .
Count. $1 + 3 + 2 = 6$ functions, matching the order.
General solution. $y = c_{1} e^{3 x} + (c_{2} + c_{3} x + c_{4} x^{2}) e^{- x} + e^{- 2 x} (c_{5} cos 3 x + c_{6} sin 3 x) .$

Figure 4.2.4. The roots of

p (r) = (r - 3) (r + 1)^{3} (r^{2} + 4 r + 13)

from Example 4.2.7 in the complex plane, each labelled with the solutions it contributes. Roots in the shaded left half-plane give decaying solutions; the root

3

gives the growing solution

e^{3 x}

. The distance of a root from the real axis is the angular frequency of the oscillation it produces.

Common mistakes

Writing $e^{(a + ib) x}$ in the final answer. The answer should be real: use $e^{a x} cos b x$ and $e^{a x} sin b x$ .
For roots $\pm ib$ (zero real part) the factor is $e^{0 x} = 1$ : the solution is $c_{1} cos b x + c_{2} sin b x$ , not $e^{b x} (\dots)$ .
Using $b$ with the wrong value: for $r = - 3 \pm 5 i$ the frequency is $5$ , not $- 5$ or $\pm 5 i$ .
For a repeated root writing $c_{1} e^{r x} + c_{2} e^{r x}$ . That is only one independent function; the second must be $x e^{r x}$ .
Losing a root after synthetic division, or forgetting $r = 0$ when the equation has no $y$ term (for instance $y^{'''} - 9 y^{'} = 0$ has roots $0, \pm 3$ , and $e^{0 x} = 1$ is a solution).

Exercises

In Exercises 1–8 find the general solution.

$y^{''} - y^{'} - 20 y = 0$
Answer
$y = c_{1} e^{5 x} + c_{2} e^{- 4 x}$
$3 y^{''} + 5 y^{'} - 2 y = 0$
Answer
$y = c_{1} e^{x /3} + c_{2} e^{- 2 x}$
$y^{''} + 14 y^{'} + 49 y = 0$
Answer
$y = (c_{1} + c_{2} x) e^{- 7 x}$
$y^{''} + 4 y^{'} + 40 y = 0$
Answer
$y = e^{- 2 x} (c_{1} cos 6 x + c_{2} sin 6 x)$
$4 y^{''} + 49 y = 0$
Answer
$y = c_{1} cos \frac{7 x}{2} + c_{2} sin \frac{7 x}{2}$
$y^{'''} + 5 y^{''} + 3 y^{'} - 9 y = 0$
Answer
$p (r) = (r - 1) (r + 3)^{2}$ ; $y = c_{1} e^{x} + (c_{2} + c_{3} x) e^{- 3 x}$
$y^{'''} - 27 y = 0$
Answer
$p (r) = (r - 3) (r^{2} + 3 r + 9)$ ; $y = c_{1} e^{3 x} + e^{- 3 x /2} (c_{2} cos \frac{3 3}{2} x + c_{3} sin \frac{3 3}{2} x)$
$y^{(4)} + 18 y^{''} + 81 y = 0$
Answer
$p (r) = (r^{2} + 9)^{2}$ ; $y = (c_{1} + c_{2} x) cos 3 x + (c_{3} + c_{4} x) sin 3 x$
Solve $y^{''} + 2 y^{'} - 15 y = 0$ , $y (0) = 1$ , $y^{'} (0) = 11$ .
Answer
$y = 2 e^{3 x} - e^{- 5 x}$
Solve $9 y^{''} + 30 y^{'} + 25 y = 0$ , $y (0) = 3$ , $y^{'} (0) = - 3$ .
Answer
$y = (3 + 2 x) e^{- 5 x /3}$
Solve $y^{'''} - 2 y^{''} + 9 y^{'} - 18 y = 0$ , $y (0) = 0$ , $y^{'} (0) = 8$ , $y^{''} (0) = 13$ .
Answer
$p (r) = (r - 2) (r^{2} + 9)$ ; $y = e^{2 x} - cos 3 x + 2 sin 3 x$
Find a homogeneous linear equation with constant coefficients, of the lowest possible order, that has $y = 3 x e^{- x} + 5 e^{2 x} sin x$ as a solution.
Answer
The roots must include $- 1$ (double) and $2 \pm i$ , so $p (r) = (r + 1)^{2} (r^{2} - 4 r + 5) = r^{4} - 2 r^{3} - 2 r^{2} + 6 r + 5$ : $y^{(4)} - 2 y^{'''} - 2 y^{''} + 6 y^{'} + 5 y = 0$ .

The key idea: try an exponential

Case 1: distinct real roots

Case 2: repeated real roots

Why xerx works: first argument (reduction of order)

Second argument (works for any multiplicity)

Case 3: complex roots

Euler's formula, step by step

From a complex solution to two real ones

Helper: finding the roots of a polynomial

The rational root test

Synthetic division

Other useful tricks

Exercises

Why $x e^{r x}$ works: first argument (reduction of order)