Chapter 9 · Section 9.3

9.3 Solving Linear Equations with Constant Coefficients

We now have all the pieces: the transform of a derivative (Section 9.1) and the inverse transform by partial fractions (Section 9.2). Put together, they give a complete method for initial-value problems of the form

a_{0} y^{(n)} + a_{1} y^{(n - 1)} + \dots + a_{n - 1} y^{'} + a_{n} y = f (t), y (0) = k_{0}, y^{'} (0) = k_{1}, \dots, y^{(n - 1)} (0) = k_{n - 1},

where $a_{0}, \dots, a_{n}$ are constants with $a_{0} \neq = 0$ , and $f$ is piecewise continuous and of exponential order.

You already know how to solve such problems: Chapter 4 finds the general solution $y = y_{c} + y_{p}$ and then fits the constants to the initial conditions. The Laplace method reaches the same answer by a different road (Figure 9.3.1). Its advantages are that the initial conditions enter at the very start, there are no arbitrary constants to determine at the end, and, as Section 9.4 shows, forcing functions with jumps and impulses cause no extra difficulty.

Figure 9.3.1. The Laplace route: transform, solve an algebraic equation, invert. The direct methods of Chapter 4 go straight down the left side.

Why the method is justified

Two facts make the method legitimate.

By the existence and uniqueness theorem for linear equations (Section 4.1), the IVP has exactly one solution on $[0, \infty)$ .
When $f$ is piecewise continuous and of exponential order, this solution and its first $n - 1$ derivatives are continuous and of exponential order too, so their transforms exist and Theorem 9.1.4 applies. (We accept this fact without proof.)

So we may transform both sides of the equation. Uniqueness of the inverse transform (Theorem 9.2.1) then guarantees that the function we recover is the solution.

The method

Write $Y (s) = L {y (t)}$ and $F (s) = L {f (t)}$ . For a second-order equation $a y^{''} + b y^{'} + cy = f (t)$ with $y (0) = k_{0}$ , $y^{'} (0) = k_{1}$ , the transform of the left side is

a (s^{2} Y - s k_{0} - k_{1}) + b (s Y - k_{0}) + c Y .

Collecting the terms with $Y$ gives

characteristic polynomial (a s^{2} + b s + c) Y = F (s) + from the initial conditions a (s k_{0} + k_{1}) + b k_{0} .

Y (s) = \frac{F ( s )}{a s ^{2} + b s + c} + \frac{a ( s k _{0} + k _{1} ) + b k _{0}}{a s ^{2} + b s + c} .

This formula is worth understanding rather than memorising. The polynomial in front of $Y$ is always the characteristic polynomial $p (s)$ of Chapter 4, with $r$ renamed $s$ . The first fraction is the response to the forcing alone (zero initial conditions); the second is the response to the initial conditions alone (no forcing). The same structure holds for any order $n$ .

Two parts of every solution

Because the formula for $Y (s)$ is a sum, so is the solution: $y = y_{zi} + y_{zs}$ . The zero-input response $y_{zi}$ is what the system does on its own, starting from the given initial values with no forcing. The zero-state response $y_{zs}$ is what the forcing produces in a system that starts at rest. Engineers often compute the two separately: the first depends only on the initial state, the second only on the input, and each can be studied, measured or designed on its own.

Compare this with Chapter 4. There $y = y_{c} + y_{p}$ , and the constants in $y_{c}$ are fitted to the initial values after $y_{p}$ is known, so the two pieces are mixed. In the Laplace solution the split is clean from the start. Neither split is "better"; they simply group the same terms differently. For example, in Example 9.3.5 below the zero-state response is the whole solution, yet it still contains both a decaying part (from the system) and a persistent part (from the input).

Method in a nutshell

Transform both sides of the equation. Replace $L {y^{(k)}}$ by $s^{k} Y - s^{k - 1} y (0) - \dots - y^{(k - 1)} (0)$ and insert the initial values immediately.
Solve the resulting linear equation for $Y (s)$ : collect the $Y$ terms on the left; the coefficient of $Y$ is the characteristic polynomial $p (s)$ .
Decompose $Y (s)$ into partial fractions (Section 9.2), completing squares for irreducible quadratics.
Invert term by term with Table 9.1.1 to get $y (t)$ .
Check (recommended): $y (0)$ and $y^{'} (0)$ should match the data; substituting back into the equation is the full check.

First examples

Example 9.3.1 (A first-order problem)

Solve $y^{'} + 3 y = 10 e^{2 t}$ , $y (0) = 5$ .

Solution

Transform. $L {y^{'}} = s Y - y (0) = s Y - 5$ and $L {10 e^{2 t}} = \frac{10}{s - 2}$ . So $(s Y - 5) + 3 Y = \frac{10}{s - 2} .$
Solve for $Y$ . $(s + 3) Y = 5 + \frac{10}{s - 2}$ , hence $Y = \frac{5}{s + 3} + \frac{10}{( s - 2 ) ( s + 3 )} .$
Partial fractions for the second term. Cover-up: at $s = 2$ , $\frac{10}{2 + 3} = 2$ ; at $s = - 3$ , $\frac{10}{- 3 - 2} = - 2$ . So $\frac{10}{( s - 2 ) ( s + 3 )} = \frac{2}{s - 2} - \frac{2}{s + 3}$ and $Y = \frac{5}{s + 3} + \frac{2}{s - 2} - \frac{2}{s + 3} = \frac{3}{s + 3} + \frac{2}{s - 2} .$
Invert. $y (t) = 3 e^{- 3 t} + 2 e^{2 t}$ .
Check. $y (0) = 3 + 2 = 5$ . And $y^{'} + 3 y = (- 9 e^{- 3 t} + 4 e^{2 t}) + (9 e^{- 3 t} + 6 e^{2 t}) = 10 e^{2 t}$ .

Example 9.3.2 (A homogeneous second-order problem)

Solve $y^{''} + 7 y^{'} + 10 y = 0$ , $y (0) = 1$ , $y^{'} (0) = 4$ .

Solution

Transform each term. $L {y^{''}} L {y^{'}} = s^{2} Y - s \cdot 1 - 4 = s^{2} Y - s - 4, = s Y - 1.$ The equation becomes $(s^{2} Y - s - 4) + 7 (s Y - 1) + 10 Y = 0$ .
Solve for $Y$ . $(s^{2} + 7 s + 10) Y = s + 4 + 7 = s + 11$ . The characteristic polynomial factors as $(s + 2) (s + 5)$ : $Y = \frac{s + 11}{( s + 2 ) ( s + 5 )} .$
Partial fractions. At $s = - 2$ : $\frac{- 2 + 11}{- 2 + 5} = 3$ . At $s = - 5$ : $\frac{- 5 + 11}{- 5 + 2} = - 2$ . So $Y = \frac{3}{s + 2} - \frac{2}{s + 5}$ .
Invert. $y (t) = 3 e^{- 2 t} - 2 e^{- 5 t}$ .
Check. $y (0) = 3 - 2 = 1$ ; $y^{'} (t) = - 6 e^{- 2 t} + 10 e^{- 5 t}$ , so $y^{'} (0) = 4$ .

Example 9.3.3 (Repeated root and a forcing of the same type)

Solve $y^{''} + 4 y^{'} + 4 y = 2 e^{- 2 t}$ , $y (0) = 1$ , $y^{'} (0) = - 1$ .

Solution

With undetermined coefficients this problem needs the modification rule with $k = 2$ ( $- 2$ is a double root of the characteristic polynomial, so the trial would be $A t^{2} e^{- 2 t}$ ). The Laplace method does not need any special rule.

Transform. $(s^{2} Y - s + 1) + 4 (s Y - 1) + 4 Y = \frac{2}{s + 2} .$
Solve for $Y$ . The constant terms move to the right: $- s + 1 - 4 = - (s + 3)$ . So $(s^{2} + 4 s + 4) Y = s + 3 + \frac{2}{s + 2}$ , and since $s^{2} + 4 s + 4 = (s + 2)^{2}$ , $Y = \frac{s + 3}{( s + 2 ) ^{2}} + \frac{2}{( s + 2 ) ^{3}} .$
Partial fractions without effort. Write $s + 3 = (s + 2) + 1$ : $\frac{s + 3}{( s + 2 ) ^{2}} = \frac{1}{s + 2} + \frac{1}{( s + 2 ) ^{2}} .$ Hence $Y = \frac{1}{s + 2} + \frac{1}{( s + 2 ) ^{2}} + \frac{2}{( s + 2 ) ^{3}}$ .
Invert with row 8 ( $n = 0, 1, 2$ ; note $2/ (s + 2)^{3}$ is exactly $2! / (s + 2)^{3}$ ): $y (t) = e^{- 2 t} + t e^{- 2 t} + t^{2} e^{- 2 t} = e^{- 2 t} (1 + t + t^{2}) .$
Check. $y (0) = 1$ . $y^{'} = e^{- 2 t} (- 2 (1 + t + t^{2}) + (1 + 2 t)) = e^{- 2 t} (- 1 - 2 t^{2})$ , so $y^{'} (0) = - 1$ .

Step inputs, sinusoidal inputs and resonance

The next example is the classic "step response". A spring–mass–damper system or an RLC circuit at rest is suddenly subjected to a constant force or voltage. In normalised form,

y^{''} + 2 ζ ω_{n} y^{'} + ω_{n}^{2} y = ω_{n}^{2}, y (0) = y^{'} (0) = 0,

so that the steady value is $y = 1$ . We take $ω_{n}^{2} = 8$ and $2 ζ ω_{n} = 4$ , an underdamped case.

Example 9.3.4 (Step response of an underdamped system)

Solve $y^{''} + 4 y^{'} + 8 y = 8$ , $y (0) = 0$ , $y^{'} (0) = 0$ .

Solution

Transform. With zero initial values, $L {y^{''}} = s^{2} Y$ and $L {y^{'}} = s Y$ . Also $L {8} = 8/ s$ . So $(s^{2} + 4 s + 8) Y = \frac{8}{s}, Y = \frac{8}{s ( s ^{2} + 4 s + 8 )} .$
Factor type. The quadratic has discriminant $16 - 32 = - 16 < 0$ : irreducible. The decomposition is $\frac{8}{s ( s ^{2} + 4 s + 8 )} = \frac{A}{s} + \frac{B s + C}{s ^{2} + 4 s + 8} .$
Constants. Cover-up at $s = 0$ : $A = 8/8 = 1$ . Clearing denominators, $8 = (s^{2} + 4 s + 8) + (B s + C) s$ . Comparing $s^{2}$ : $0 = 1 + B$ , so $B = - 1$ . Comparing $s$ : $0 = 4 + C$ , so $C = - 4$ . Hence $Y = \frac{1}{s} - \frac{s + 4}{s ^{2} + 4 s + 8} .$
Complete the square. $s^{2} + 4 s + 8 = (s + 2)^{2} + 2^{2}$ and $s + 4 = (s + 2) + 2$ : $\frac{s + 4}{( s + 2 ) ^{2} + 4} = \frac{s + 2}{( s + 2 ) ^{2} + 2 ^{2}} + \frac{2}{( s + 2 ) ^{2} + 2 ^{2}} .$
Invert. $y (t) = 1 - e^{- 2 t} (cos 2 t + sin 2 t) .$
Interpretation. $y \to 1$ as $t \to \infty$ , as expected. The derivative is $y^{'} = 4 e^{- 2 t} sin 2 t$ , which first vanishes at $t = π /2$ . There $y (π /2) = 1 + e^{- π} \approx 1.043$ : the response overshoots its final value by about $4.3%$ (Figure 9.3.2).

Figure 9.3.2. Step response of

y^{''} + 4 y^{'} + 8 y = 8

from rest (Example 9.3.4). The complex poles

- 2 \pm 2 i

cause a small overshoot; the real part

- 2

sets how fast it dies out.

Next comes the most common case in practice: a damped system driven by a sinusoidal input. The partial fractions now involve two quadratic factors, one from the system and one from the forcing, so this is the place to be careful.

Example 9.3.5 (Sinusoidal forcing: transient and steady state)

Solve $y^{''} + 4 y^{'} + 5 y = 8 cos t$ , $y (0) = 0$ , $y^{'} (0) = 0$ .

Solution

Transform. $(s^{2} + 4 s + 5) Y = \frac{8 s}{s ^{2} + 1}$ , so $Y = \frac{8 s}{( s ^{2} + 1 ) ( s ^{2} + 4 s + 5 )} .$ Both quadratics are irreducible (discriminants $- 4$ and $16 - 20 = - 4$ ).
Form of the decomposition. Each quadratic gets a linear numerator: $\frac{8 s}{( s ^{2} + 1 ) ( s ^{2} + 4 s + 5 )} = \frac{A s + B}{s ^{2} + 1} + \frac{C s + D}{s ^{2} + 4 s + 5} .$
Clear denominators and expand. $8 s = (A s + B) (s^{2} + 4 s + 5) + (C s + D) (s^{2} + 1) = (A + C) s^{3} + (4 A + B + D) s^{2} + (5 A + 4 B + C) s + (5 B + D) .$
Compare coefficients. $s^{3} : s^{2} : s^{1} : s^{0} : A + C = 0, 4 A + B + D = 0, 5 A + 4 B + C = 8, 5 B + D = 0.$
Solve step by step. From $s^{3}$ : $C = - A$ . Put it into $s^{1}$ : $4 A + 4 B = 8$ , so $B = 2 - A$ . From $s^{0}$ : $D = - 5 B = 5 A - 10$ . Put $B$ and $D$ into $s^{2}$ : $4 A + (2 - A) + (5 A - 10) = 0$ , so $8 A = 8$ and $A = 1$ . Hence $A = 1, B = 1, C = - 1, D = - 5.$
Rewrite. $Y = \frac{s + 1}{s ^{2} + 1} - \frac{s + 5}{( s + 2 ) ^{2} + 1}, s + 5 = (s + 2) + 3.$
Invert. $y (t) = steady state cos t + sin t - transient e^{- 2 t} (cos t + 3 sin t) .$
Check. $y (0) = 1 - 1 = 0$ . $y^{'} (t) = - sin t + cos t - e^{- 2 t} (- 2 cos t - 6 sin t - sin t + 3 cos t)$ , so $y^{'} (0) = 1 - (- 2 + 3) = 0$ .
Interpretation. The terms from the factor $s^{2} + 1$ (the input) form the steady-state response $cos t + sin t = 2 cos (t - π /4)$ : an oscillation at the input frequency, with amplitude $2$ , lagging the input $8 cos t$ by $π /4$ . The terms from $(s + 2)^{2} + 1$ (the system) form the transient, which dies out like $e^{- 2 t}$ (Figure 9.3.3). After a couple of time units only the steady state is visible.

Figure 9.3.3. Example 9.3.5. The poles of the input (

\pm i

) give the steady state; the poles of the system (

- 2 \pm i

) give a transient that decays like

e^{- 2 t}

. Their sum starts from rest as required.

This splitting is general for a stable system (all characteristic roots with negative real part): the partial fractions coming from the input's poles give the steady state, those coming from the system's poles give the transient. If the input's poles coincide with the system's poles, the two cannot be separated, and we get resonance. For instance, for $y^{''} + 4 y = 4 cos 2 t$ with $y (0) = y^{'} (0) = 0$ , the transform is

(s^{2} + 4) Y = \frac{4 s}{s ^{2} + 4}, Y = \frac{4 s}{( s ^{2} + 4 ) ^{2}} .

By row 11 of Table 9.1.1 with $b = 2$ , $L {t sin 2 t} = \frac{4 s}{( s ^{2} + 4 ) ^{2}}$ , so $y (t) = t sin 2 t$ . The doubled pair of poles $\pm 2 i$ , one pair from the system and one from the input, produces the factor $t$ : the amplitude grows without bound (Figure 9.3.4).

Figure 9.3.4. Pure resonance: the response

t sin 2 t

y^{''} + 4 y = 4 cos 2 t

from rest oscillates inside the growing envelope

\pm t

Try the solver below on your own problems. Enter the coefficients from the highest derivative down, a forcing made of polynomials, exponentials, sines and cosines, and the initial values $y (0), y^{'} (0), \dots$ . It shows the transformed equation, $Y (s)$ , the partial fractions and $y (t)$ . The first preset is Example 9.3.4. Change the damping coefficient from $4$ to $1$ and watch the overshoot grow; change it to $6$ and it disappears (why? look at the poles).

A third-order problem

Nothing changes for higher order, except that more initial values enter and the partial fractions get longer. Keep the bookkeeping tidy: write $L {y^{'''}}$ out in full before substituting numbers.

Example 9.3.6 (Third order with nonzero initial values)

Solve $y^{'''} - y^{'} = 6 e^{2 t}$ , $y (0) = 0$ , $y^{'} (0) = 2$ , $y^{''} (0) = 0$ .

Solution

Transform the derivatives. $L {y^{'''}} L {y^{'}} = s^{3} Y - s^{2} y (0) - s y^{'} (0) - y^{''} (0) = s^{3} Y - 2 s, = s Y - y (0) = s Y .$
Transformed equation. $(s^{3} Y - 2 s) - s Y = \frac{6}{s - 2}$ , so $(s^{3} - s) Y = 2 s + \frac{6}{s - 2} .$
Solve for $Y$ . $s^{3} - s = s (s - 1) (s + 1)$ . Dividing, $Y = \frac{2 s}{s ( s - 1 ) ( s + 1 )} + \frac{6}{s ( s - 1 ) ( s + 1 ) ( s - 2 )} = \frac{2}{( s - 1 ) ( s + 1 )} + \frac{6}{s ( s - 1 ) ( s + 1 ) ( s - 2 )} .$
First fraction. Cover-up at $s = 1$ gives $2/2 = 1$ , at $s = - 1$ gives $2/ (- 2) = - 1$ : $\frac{1}{s - 1} - \frac{1}{s + 1}$ .
Second fraction. Four simple factors, so four cover-ups: $s = 0 : s = - 1 : \frac{6}{( - 1 ) ( 1 ) ( - 2 )} = 3, \frac{6}{( - 1 ) ( - 2 ) ( - 3 )} = - 1, s = 1 : s = 2 : \frac{6}{( 1 ) ( 2 ) ( - 1 )} = - 3, \frac{6}{( 2 ) ( 1 ) ( 3 )} = 1.$
Add the pieces. $Y = \frac{3}{s} + \frac{1 - 3}{s - 1} + \frac{- 1 - 1}{s + 1} + \frac{1}{s - 2} = \frac{3}{s} - \frac{2}{s - 1} - \frac{2}{s + 1} + \frac{1}{s - 2} .$
Invert. $y (t) = 3 - 2 e^{t} - 2 e^{- t} + e^{2 t}$ .
Check the initial values. $y (0) = 3 - 2 - 2 + 1 = 0$ ; $y^{'} (0) = - 2 + 2 + 2 = 2$ ; $y^{''} (0) = - 2 - 2 + 4 = 0$ . All correct.

The second preset loads Example 9.3.6 into the solver. Change the initial values and see that only the "initial-condition part" of $Y (s)$ changes.

Initial conditions at

t_{0} \neq = 0

The transform uses values at $t = 0$ . If the data are given at $t = t_{0}$ , introduce the shifted time $τ = t - t_{0}$ and the function $w (τ) = y (τ + t_{0})$ . Then $w^{'} = y^{'}$ , $w^{''} = y^{''}$ , the equation keeps its constant coefficients (the forcing becomes $f (τ + t_{0})$ ), and the conditions become $w (0) = y (t_{0})$ , $w^{'} (0) = y^{'} (t_{0})$ . Solve for $w$ by the usual method and substitute $τ = t - t_{0}$ back.

Common mistakes

Forgetting the initial-value terms, for example writing $L {y^{''}} = s^{2} Y$ when $y (0) \neq = 0$ .
Sign errors when moving the initial-value terms to the right-hand side. Write the transformed equation completely before rearranging.
Stopping at $Y (s)$ or at the partial fractions. The answer is the function $y (t)$ .

Exercises

Solve each initial-value problem with the Laplace transform.

$y^{'} - 4 y = 8$ , $y (0) = 1$
Answer
$y = 3 e^{4 t} - 2$
$y^{'} + 2 y = 10 cos t$ , $y (0) = 0$
Answer
$y = 4 cos t + 2 sin t - 4 e^{- 2 t}$
$y^{''} - 2 y^{'} - 15 y = 0$ , $y (0) = 0$ , $y^{'} (0) = 8$
Answer
$y = e^{5 t} - e^{- 3 t}$
$y^{''} + 16 y = 0$ , $y (0) = 2$ , $y^{'} (0) = - 4$
Answer
$y = 2 cos 4 t - sin 4 t$
$y^{''} + 10 y^{'} + 25 y = 0$ , $y (0) = 2$ , $y^{'} (0) = - 7$
Answer
$Y = \frac{2 s + 13}{( s + 5 ) ^{2}}$ , $y = (2 + 3 t) e^{- 5 t}$
$y^{''} - 4 y^{'} + 20 y = 0$ , $y (0) = 1$ , $y^{'} (0) = 10$
Answer
$y = e^{2 t} (cos 4 t + 2 sin 4 t)$
$y^{''} + 5 y^{'} + 6 y = 20 e^{2 t}$ , $y (0) = 0$ , $y^{'} (0) = 0$
Answer
$y = e^{2 t} - 5 e^{- 2 t} + 4 e^{- 3 t}$
$y^{''} + y = 6 cos 2 t$ , $y (0) = 1$ , $y^{'} (0) = 0$
Answer
$y = 3 cos t - 2 cos 2 t$
$y^{''} + 2 y^{'} + y = 4 t e^{- t}$ , $y (0) = 0$ , $y^{'} (0) = 0$
Answer
$Y = \frac{4}{( s + 1 ) ^{4}}$ , $y = \frac{2}{3} t^{3} e^{- t}$
$y^{''} + 9 y = 6 sin 3 t$ , $y (0) = 0$ , $y^{'} (0) = 1$
Answer
$y = \frac{2}{3} sin 3 t - t cos 3 t$ (use row 13 of Table 9.1.1)
$y^{'''} + 4 y^{'} = 8$ , $y (0) = y^{'} (0) = y^{''} (0) = 0$
Answer
$Y = \frac{8}{s ^{2} ( s ^{2} + 4 )} = \frac{2}{s ^{2}} - \frac{2}{s ^{2} + 4}$ , $y = 2 t - sin 2 t$
$y^{'} + 2 y = 4$ , $y (1) = 5$ (use the shift $τ = t - 1$ )
Answer
$w (τ) = 2 + 3 e^{- 2 τ}$ , so $y = 2 + 3 e^{- 2 (t - 1)}$