Convolution.Pdf

“Convolution” is an operation involving two functions that turns out to be rather useful in many applications. We have two reasons for introducing it here. First of all, convolution will give us a way to deal with inverse transforms of fairly arbitrary products of functions. Secondly, it will be a major element in some relatively simple formulas for solving a number of differential equations. Let us start with just seeing what “convolution” is. After that, we’ll discuss using it with the Laplace transform and in solving differential equations.

27.1 Convolution, the Basics Deﬁnition and Notation Let f (t) and g(t) be two functions. The convolution of f and g , denoted by f g , is the ∗ function on t 0 given by ≥ t f g(t) f (x)g(t x) dx . ∗ = x 0 − Z = !◮Example 27.1: Let

f (t) e3t and g(t) e7t . = = Since we will use f (x) and g(t x) in computing the convolution, let us note that − 3x 7(t x) f (x) e and g(t x) e − . = − = So, t f g(t) f (x)g(t x) dx ∗ = x 0 − Z = t 3x 7(t x) e e − dx = x 0 Z = t 3x 7t 7x e e e− dx = x 0 Z =

539 540 Convolution and Laplace Transforms

t 7t 4x e e− dx = x 0 Z = t 7t 1 4x e − e− = · 4 x 0 = 1 7t 4t 1 7t 4 0 1 3t 1 7t − e e− − e e− · − e e . = 4 − 4 = 4 + 4 Simplifying this slightly, we have 1 f g(t) e7t e3t when f (t) e3t and g(t) e7t . ∗ = 4 − = = It is common practice to also denote the convolution f g(t) by f (t) g(t) where, here, ∗ ∗ f (t) and g(t) denote the formulas for f and g . Thus, instead of writing 1 f g(t) e7t e3t when f (t) e3t and g(t) e7t , ∗ = 4 − = = we may just write 1 e3t e7t e7t e3t . ∗ = 4 − This simplifies notation a little, but be careful — t is being used for two different things in this equation: On the left side, t is used to describe f and g ; on the right side, t is the variable in the formula for the convolution. By convention, if we assign t a value, say, t 2, then we are = setting t 2 in the final formula for the convolution. That is, = e3t e7t with t 2 ∗ = means compute the convolution and replace the t in the resulting formula with 2, which, by the above computations, is 1 7 2 3 2 1 14 6 e · e · e e . 4 − = 4 − It does NOT mean to compute 3 2 7 2 e · e · , ∗ which would give you a completely different result, namely, t e6 e14 e6e14 dt e20t . ∗ = x 0 = Z = !◮Example 27.2: Let us find 1 t2 when t 4 . √t ∗ = Here, 1 f (t) and g(t) t2 . = √t = So 1 f (x) and g(t x) (t x)2 , = √x − = − and 1 t 1 t2 f g(t) (t x)2 √t ∗ = ∗ = x 0 √x − Z = t 1/ 2 2 x− 2 t 2tx x dx = x 0 − + Z = Convolution, the Basics 541

t 2 1/ 1/ 3/ t x− 2 2tx 2 x 2 dx = x 0 − + Z = 1 2 3 2 5 t t22x /2 2t x /2 x /2 = − 3 + 5 x 0 = 1 4 3 2 5 2t2 t /2 t t /2 t /2 . = · − 3 · + 5 After a little algebra and arithmetic, this reduces to

1 16 5 t2 t /2 . (27.1) √t ∗ = 15 Thus, to compute 1 t2 when t 4 , √t ∗ = we actually compute 16 5 t /2 with t 4 , 15 = obtaining 16 5 16 512 4 /2 25 . 15 = 15 = 15

Basic Identities Let us quickly note a few easily veriﬁed identities that can simplify the computation of some convolutions. The ﬁrst identity is trivial to derive. Let α be a constant, and let f and g be two functions. Then, of course, t t t α f (x) g(t x) dx f (x) αg(t x) dx α f (x)g(t x) dx , x 0[ ] − = x 0 [ − ] = x 0 − Z = Z = Z = which we can rewrite as α f g f αg α f g . [ ]∗ = ∗[ ] = [ ∗ ] In other words, we can “factor out constants”. A more substantial identity comes from looking at how switching the roles of f and g changes the convolution. That is, how does the result of computing t g f (t) g(x) f (t x) dx ∗ = x 0 − Z = compare to what we get by computing t f g(t) f (x)g(t x) dx ? ∗ = x 0 − Z = Well, in the last integral, let’s use the substitution y t x . Then x t y , dx dy and t = − = − =− f g(t) f (x)g(t x) dx ∗ = x 0 − Z = t t − f (t y)g(y)( 1) dy = y t 0 − − Z = − 0 t g(y) f (t y) dy g(y) f (t y) dy . = − y t − = y 0 − Z = Z = 542 Convolution and Laplace Transforms

Thelastintegralisexactlythesameastheintegralforcomputing g f (t) , exceptforthecosmetic ∗ change of denoting the variable of integration by y instead of x . So that integral is the formula for formula for g f (t) , and our computations just above reduce to ∗ f g(t) g f (t) . (27.2) ∗ = ∗ Thus we see that convolution is “commutative”.

!◮Example 27.3: Let’s consider the convolution

1 t2 . ∗ √t Since we just showed that convolution is commutative, we know that

1 1 t2 t2 . ∗ √t = √t ∗ What an incredible stroke of luck! We’ve already computed the convolution on the right in example 27.2. Checking back to equation (27.1), we ﬁnd

1 16 5 t2 t /2 . √t ∗ = 15 Hence, 1 1 16 5 t2 t2 t /2 . ∗ √t = √t ∗ = 15

In addition to being commutative, convolution is “distributive” and “associative”. That is, given three functions f , g and h ,

f g h f h g h , (27.3) [ + ]∗ = [ ∗ ]+[ ∗ ] f g h f g f h (27.4) ∗[ + ]=[ ∗ ]+[ ∗ ] and f g h f g h . (27.5) ∗[ ∗ ]=[ ∗ ]∗ The first and second equations are that “addition distributes over convolution”. They are easily confirmed using the basic definition of convolution. For the first:

t f g h(t) f (x) g(x) h(t x) dx [ + ]∗ = x 0[ + ] − Z = t f (x)h(t x) g(x)h(t x) dx = x 0[ − + − ] Z = t t f (x)h(t x) dx g(x)h(t x) dx = x 0 − + x 0 − ] Z = Z = f g g h . = [ ∗ ]+[ ∗ ] The second, equation (27.4) follows in a similar manner or by combining (27.3) with the commu- tativity of the convolution. The last equation in the list, equation (27.5), states that convolution Convolution and Products of Transforms 543

is “associative”,that is, when convolving three functions together, it does not matter which two you convolve first. It’s verification requires showing that the two double integrals defining

f g h and f g h ∗[ ∗ ] [ ∗ ]∗ are equivalent. This is a relatively straightforward exercise in substitution, and will be left as a challenge for the interested student (exercise 27.3 on page 552). Finally, just for fun, let’s make a few more simple observations:

t 0 g(t) g 0(t) 0 g(t x) dx 0 . ∗ = ∗ = x 0 · − = Z = t t f 1(t) 1 f (t) f (s) 1 dx f (s) dx . ∗ = ∗ = x 0 · = x 0 Z = Z = 0 f g(0) f (x)g(0 x) dx 0 . ∗ = x 0 − = Z =

Observations on the Existence of the Convolution The observant reader will have noted that, if f and g are at least piecewise continuous on (0, ) , then, for any positive value t , the product f (x)g(t x) is a piecewise continuous ∞ − function of x on (0, t) . It then follows that the integral in

t f g(t) f (x)g(t x) dx ∗ = x 0 − Z = is well defined and finite for every positive value of t . In other words, f g is a well-defined ∗ function on (0, ) , at least whenever f and g are both piecewise continuous on (0, ) . (In ∞ ∞ fact, it can then even be shown that f g(t) is a continuous function on 0, ) .) ∗ [ ∞1/2 But now observe that one of the functions in example 27.2, namely t− , ‘blows up’ at t 0 and, thus, is not piecewise continuous on (0, ) . So that example also demonstrates that, = ∞ sometimes, f g is well defined on (0, ) even though f or g is not piecewise continuous. ∗ ∞

27.2 Convolution and Products of Transforms To see one reason convolution is important in the study of Laplace transforms, let us examine the Laplace transform of the convolution of two functions f (t) and g(t) . Our goal is a surprisingly simple formula of the corresponding transforms,

∞ st F(s) L[ f (t)] f (t)e− dt = |s = Z0 and ∞ st G(s) L[g(t)] g(t)e− dt . = |s = Z0 544 Convolution and Laplace Transforms

(The impatient can turn to theorem 27.1 on page 545 for that formula.) Keep in mind that we can rename the variable of integration in each of the above integrals. In particular, note (for future reference) that

∞ sx ∞ sy F(s) e− f (x) dx and G(s) e− g(y) dy . = = Z0 Z0 Now, simply writing out the integral formulas for the Laplace transform and for the convolution yields L ∞ st [ f g(t)] s e− f g(t) dt ∗ | = t 0 ∗ Z = t ∞ st e− f (x)g(t x) dxdt = t 0 x 0 − Z = Z = t ∞ st e− f (x)g(t x) dxdt . = t 0 x 0 − Z = Z = Combined with the observation that

st st sx sx s(t x) sx e− e− + − e− − e− , = = the above sequence becomes

t L ∞ sx s(t x) [ f g(t)] s e− f (x) e− − g(t x) dxdt ∗ | = t 0 x 0 − Z = Z = t (27.6) ∞ K (x, t) dxdt = t 0 x 0 Z = Z = where, simply to simplify expressions in the next few lines, we’ve set

sx s(t x) K (x, t) e− f (x) e− − g(t x) . = − It is now convenient to switch the order of integration in the last double integral. According to the limits in that double integral, we are integrating over the region R in the XT –plane consisting of all (x, t) for which 0 < t < ∞ and, for each of these values of t , 0 < x < t .

As illustrated in figure 27.1, region R is the portion of the first quadrant of the XT –plane to the left of the line t x . Equivalently, as can also be seen in this figure, R is the portion of = the first quadrant above the line t x . So R can also be described as the set of all (x, t) for = which 0 < x < ∞ and, for each of these values of x , x < t < . ∞ Thus,

t ∞ K (x, t) dxdt K (x, t) d A ∞ ∞ K (x, t) dtdx . t 0 x 0 = R = x 0 t x Z = Z = ZZ Z = Z = Convolution and Products of Transforms 545

T R x t = (x0, t0) t t0 (t , t ) = 0 0

t x0 (x , x ) = 0 0

x x0 x t0 = = X Figure 27.1: The region R for the transform of the convolution. Note that the coordinates of any point (x0, t0) in R must satisfy 0 < x0 < t0 < . ∞

Combining this with equation (27.6), and then continuing, we have

L ∞ ∞ [ f g(t)] s K (x, t) dtdx ∗ | = x 0 t x Z = Z = ∞ ∞ sx s(t x) e− f (x) e− − g(t x) dtdx = x 0 t x − Z = Z =

∞ sx ∞ s(t x) e− f (x) e− − g(t x) dt dx . = x 0 t x − Z = Z = Let us simplify the inner integral with the substitution y t x (remembering that t is the = − variable of integration in this integral):

x ∞ s(t x) ∞− sy ∞ sy e− − g(t x) dt e− g(y) dy e− g(y) dy G(s) ! t x − = y x x = y 0 = Z = Z = − Z = Combining this with our last formula for L[ f g] then yields ∗ L ∞ sx [ f g(t)] s f (x)e− G(s) dx ∗ | = x 0 Z = ∞ sx e− f (x) dx G(s) F(s) G(s) ! = x 0 · = · Z = Thus, L[ f g(t)] F(s)G(s) . ∗ |s = Equivalently, 1 f g(t) L− [F(s)G(s)] . ∗ = |t If we had been a little more complete in our computations, we would have kept track of the exponential order of all the functions involved (see exercise 27.9), and obtained all of the following theorem.

Theorem 27.1 (Laplace convolution identities) Assume f (t) and g(t) are two functions of exponential order s0 , and with Laplace transforms F(s) L[ f (t)] and G(s) L[g(t)] . = |s = |s 546 Convolution and Laplace Transforms

Then the convolution f g(t) is of exponential order s1 for any s1 > s0 . Moreover, ∗

L[ f g(t)] F(s)G(s) for s > s0 (27.7) ∗ |s = and 1 L− [F(s)G(s)] f g(t) . (27.8) |t = ∗ Do remember that identities (27.7) and (27.8) are equivalent. It is also worthwhile to rewrite these identities as L[ f g(t)] L[ f ] L[g(t)] (27.7 ′) ∗ |s = |s · |s and 1 1 1 L− [F(s)G(s)] L− [F(s)] L− [G(s)] , (27.8 ′) |t = |t ∗ |t respectively. These forms, especially the latter,are sometimes a little more convenient in practice.

!◮Example 27.4: Consider ﬁnding the inverse Laplace transform of

1 . s2 10s 21 − + Factoring the denominator and applying the above, we get

1 1 1 1 L− L− s2 10s 21 = (s 3)(s 7) − + t − − t

1 1 1 L− = s 3 · s 7 − − t

1 1 1 1 3t 7t L− L− e e . = s 3 ∗ s 7 = ∗ − t − t

As luck would have it, this convolution was computed in example 27.1 on page 539),

1 e3t e7t e7t e3t . ∗ = 4 − Thus, 1 1 3t 7t 1 7t 3t L− e e e e . s2 10s 21 = ∗ = 4 − − + t

The inverse transform in the last example could also have been computed using partial fractions. Indeed, many of the inverse transforms we computed using partial fractions can also be computed using convolution. Whether one approach or the other is preferred depends on the opinion of the person doing the computing. However, as the next example shows, there are cases where convolution can be applied, but not partial fractions. We will also use this example to demonstrate how convolution naturally arises when solving differential equations.

!◮Example 27.5: Consider solving the initial-value problem

1 y′′ 9y with y(0) 0 and y′(0) 0 . + = √t = = Convolution and Products of Transforms 547

Taking the Laplace transform of both sides:

1 L y′′ 9y L + s = √t s √π [L y′′ 9L[y ֒ → s + |s = √s 2 √π (s Y (s) sy(0) y′(0) 9Y (s ֒ → − − + = √s 0 0 |{z} |{z} √π (s2 9 Y (s ֒ → + = √s √π . (Y (s ֒ = √s s2 9 → + Thus, y(t) is the inverse Laplace transform of

√π . √s s2 9 + Because the denominator does not factor into two polynomial s, we cannot use partial fractions — we must use convolution,

1 √π 1 √π 1 1 √π 1 1 L− L− L− L− . √s s2 9 = √s · s2 9 = √s ∗ s2 9 t + t t + t + Reversing the transform made on the right side of the above equations, we have

1 √π 1 L− . √s = √t t

Using our tables, we ﬁnd that

1 1 1 1 3 1 L− L− sin(3t) . s2 9 = 3 s2 32 = 3 + t + t

Combining the above and recalling that “constants factor out”,we then obtain

1 √π 1 √π 1 1 y(t) L− L− L− = √s s2 9 = √s ∗ s2 9 t t + t + 1 1 sin(3t) = √t ∗ 3 1 1 sin(3t) . = 3 √t ∗ That is, 1 t 1 y(t) sin(3 t x ) dx . = 3 x 0 √x [ − ] Z = Admittedly, this last integral is not easily evaluated by hand. But it is something that can be accurately approximated for any speciﬁc (nonnegative) value of t using routines found in many computer math packages. So it is still a usable formula. 548 Convolution and Laplace Transforms

27.3 Convolution and Differential Equations (Duhamel’s Principle) As illustrated in our last example, convolution has a natural role in solving differential equations when using the Laplace transform. However, if we look a little more carefully at the process of solving differential equations using the Laplace transform, we will ﬁnd that convolution can play an even more signiﬁcant role.

!◮Example 27.6: Let’s consider solving the nonhomogeneous initial-value problem

y′′ 10y′ 21y f (t) with y(0) 0 and y′(0) 0 − + = = = where f f (t) is any Laplace transformable function. Naturally, we will use the Laplace = transform. So let

Y (s) L[y(t)] and F(s) L[ f (t)] . = |s = |s Because of our initial conditions, the “transform of the derivatives” identities simplify con- siderably: 2 2 L y′′ s Y (s) sy(0) y′(0) s Y (s) s = − − = 0 0 and |{z} |{z} L y′ sY (s) y(0) sY (s) . s = − = 0

Consequently, |{z}

L y′′ 10y′ 21y L[ f (t)] − + s = |s (L y′′ 10L y′ 21L[y] F(s ֒ → s − s + |s = (s2 Y (s) 10sY (s) 21Y (s) F(s ֒ → − + = . (s2 10s 21 Y (s) F(s ֒ → − + = Dividing through by the polynomial, we get

1 Y (s) H(s)F(s) where H(s) . = = s2 10s 21 − + Thus, 1 1 y(t) L− [Y (s)] L− [H(s)F(s)] . = |t = |t Applying the convolution identity then yields

y(t) h f (t) (27.9a) = ∗ where 1 1 1 h(x) L− [H(s)] L− . (27.9b) = |x = s2 10s 21 − + t

Convolution and Differential Equations (Duhamel’s Principle) 549

The convolution h f can be computed using either ∗ t t h(x) f (t x) dx or h(t x) f (x) dx . − − Z0 Z0 For no particular reason, we will choose the ﬁrst integral formula. To compute h(x) , we can use partial fractions or convolution. Or we can glance at example 27.4 a few pages ago, discover that we’ve already computed h(t) , and just replace the t in that formula with x , 1 h(x) e7x e3x . = 4 − With this and our chosen integral formula for h f , formula (27.9a), the solution to our ∗ initial-value problem, becomes

t 1 y(t) e7x e3x f (t x) dx . (27.10) = 4 − − Z0 Formula (27.10) is a convenient way to describe the solutions to our initial-value problem, especially if we want to solve this problem for a number of different choices of f (t) . Using it, we can quickly write out a relatively simple integral formula for the solution corresponding to each f (t) . For example: 4t 4(t x) If f (t) e , then f (t x) e − and formula (27.10) yields = − = t 1 7x 3x 4(t x) y(t) e e e − dx . = 4 − Z0 If f (t) 1 , then f (t x) 1 and and formula (27.10) yields = − = t 1 y(t) e7x e3x 1 dx . = 4 − · Z0 And ﬁnally, if f (t) √3 t , then f (t x) √3 t x and formula (27.10) yields = − = − t 1 y(t) e7x e3x √3 t xdx . = 4 − − Z0 The ﬁrst two integrals are easily evaluated, giving us

1 1 1 1 1 1 y(t) e7t e3t e4t and y(t) e7t e3t , = 2 + 4 − 3 = 28 − 12 − 21 respectively. The last integral,

t 1 y(t) e7x e3x √3 t xdx , = 4 − − Z0 is not easily evaluated by hand, but can be accurately approximated for any value of t using routines found in our favorite computer math package.

?◮Exercise 27.1: Using the formula (27.10), ﬁnd the solution to

3t y′′ 10y′ 21y e with y(0) 0 and y′(0) 0 . − + = = = 550 Convolution and Laplace Transforms

Generalizing what we just derived in the last example is easy. Suppose we have any second- order initial-value problem of the form

ay′′ by′ cy f (t) with y(0) 0 and y′(0) 0 + + = = = where a , b and c are constants, and f is any Laplace transformable function. Then, taking the Laplace transform of both sides of the differential equation, letting

Y (s) L[y(t)] and F(s) L[ f (t)] , = |s = |s and noting that, because of our initial conditions, the “transform of the derivatives” identities simplify to 2 2 L y′′ s Y (s) sy(0) y′(0) s Y (s) s = − − = 0 0 and |{z} |{z} L y′ sY (s) y(0) sY (s) , s = − = 0 we see that |{z} L ay′′ by′ cy L[ f (t)] + + s = |s (aL y′′ bL y′ cL[y] F(s ֒ → s + s + |s = (as2 Y (s) bsY (s) cY (s) F(s ֒ → + + = . (as2 bs c Y (s) F(s ֒ → + + = Dividing through by the polynomial, we get

1 Y (s) H(s)F(s) where H(s) . = = as2 bs c + + So, 1 1 y(t) L− [Y (s)] L− [H(s)F(s)] , = |t = |t and the convolution identity tells us that

y(t) h f (t) (27.11a) = ∗ where 1 1 1 h(x) L− [H(s)] L− . (27.11b) = |x = as2 bs c + + t

The fact that the formula for y in equation set (27.11) is the solution to

ay′′ by′ cy f (t) with y(0) 0 and y′(0) 0 + + = = = is often called Duhamel’s principle. The function H(s) is usually referred to as the transfer function, and its inverse transform, h(t) , is usually called the impulse response function.1 Keep in mind that a , b and c are constants, and that we assumed f is Laplace transformable.

1 The reason why h is called the “impulse response function” will be revealed in chapter 29. A few authors also refer to h as a “weight” function. Convolution and Differential Equations (Duhamel’s Principle) 551

As illustrated in our example, Duhamel’s principle makes it easy to write down solutions to the given initial-value problem once we have found h . This is especially useful if we need to ﬁnd solutions to

ay′′ by′ cy f (t) with y(0) 0 and y′(0) 0 + + = = = for a number of different choices of f . Burwhystopatsecond-orderproblems? Itshouldbeclearthattheabovedifferentialequation did not have to be second order. A completely analogous derivation can be done starting with any nonhomogeneous linear differential equation with constant coefﬁcients, provided all the appropriate initial values are zero. Doing so, leads to the following theorem:

Theorem 27.2 (Duhamel’s principle) Let N be any positive integer, let a0 , a1 , ... and aN be any collection of constants, and let f (t) be any Laplace transformable function. Then, the solution to

(N) (N 1) a0 y a1 y − aN 2 y′′ aN 1 y′ aN y f (t) + + ··· + − + − + = satisfying the N th-order “zero” initial conditions,

N 1 y(0) 0 , y′(0) 0 , y′′(0) 0 , ... and y − (0) 0 , = = = = is given by y(t) h f (t) = ∗ where L 1 1 h(x) − [H(s)] x and H(s) n n 1 . = | = a0s a1s an + − +···+ Three quick notes:

1. As noted a few pages ago, the convolution h f can be computed using either ∗ t t h(x) f (t x) dx or h(t x) f (x) dx . − − Z0 Z0 In practice, use whichever appears easier to compute given the h and f involved. In the examples here, we used the ﬁrst. Later, when we re-examine “resonance” in mass/spring systems (section 28.7), we will use the other integral formula.

2. It turns out that the f (t) in Duhamel’s principle (as described above) does not have to be Laplace transformable. By applying the above theorem with

f (t) if t < T fT (t) , = ( 0 if t T ≥ and then letting T , you can show that y h f is well deﬁned and satisﬁes the →∞ = ∗ initial-value problem even when f is merely piecewise continuous on (0, ) . ∞ 3. It is not hard to show that, if you want the solution to

(N) (N 1) a0 y a1 y − aN 2 y′′ aN 1 y′ aN y f (t) , + + ··· + − + − + = 552 Convolution and Laplace Transforms

but satisfying nonzeroinitialconditions,thenyousimplyneedtoaddthesolution obtained by Duhamel’s principle to the solution to the corresponding homogeneous differential equation

(N) (N 1) a0 y a1 y − aN 2 y′′ aN 1 y′ aN y 0 + + ··· + − + − + = that satisﬁes the desired initial conditions.

Additional Exercises

27.2. Compute the convolution f g(t) of each of the following pairs of functions: ∗ 1 a. f (t) e3t and g(t) e5t b. f (t) and g(t) t2 = = = √t = c. f (t) √t and g(t) 6 d. f (t) t and g(t) e3t = = = = e. f (t) t2 and g(t) t2 f. f (t) sin(t) and g(t) t = = = = g. f (t) sin(t) and g(t) sin(t) h. f (t) step (t) and g(t) e2t = = = 3 = i. f (t) step (t) and g(t) t2 = 3 = 27.3. Verify the associative property of convolution. That is, verify equation (27.5) on page 542.

27.4. Usingconvolution, computetheinverseLaplacetransformofeachofthefollowing: 1 1 1 a. b. c. (s 4)(s 3) s(s 3) s(s2 4) − − − + 1 1 s2 d. e. f. (s 3)(s2 1) (s2 9)2 (s2 4)2 − + + + e 4s 1 g. − h. (leave in integral form) s(s2 1) √s(s 3) + − 27.5. Foreachofthefollowinginitial-valueproblems,ﬁndthecorresponding transfer function H and the impulse response function h , and write down the corresponding convolution integral formula for the solution:

a. y′′ 4y f (t) with y(0) 0 and y′(0) 0 + = = =

b. y′′ 4y f (t) with y(0) 0 and y′(0) 0 − = = =

c. y′′ 6y′ 9y f (t) with y(0) 0 and y′(0) 0 − + = = =

d. y′′ 6y′ 18y f (t) with y(0) 0 and y′(0) 0 − + = = =

e. y′′′ 16y′ f (t) with y(0) 0 and y′(0) 0 + = = = Additional Exercises 553

27.6. Using the results from exercise 27.5 a, ﬁnd the solution to

y′′ 4y f (t) with y(0) 0 and y′(0) 0 + = = = for each of the following choices of f : a. f (t) 1 b. f (t) t c. f (t) e3t = = = d. f (t) sin(2t) e. f (t) sin(αt) where α 3 = = 6= 27.7. Using the results from exercise 27.5 c, ﬁnd the solution to

y′′ 6y′ 9y f (t) with y(0) 0 and y′(0) 0 − + = = = for each of the following choices of f : a. f (t) 1 b. f (t) t c. f (t) e3t = = = 3t αt d. f (t) e− e. f (t) e where α 3 = = 6= 27.8. Using the results from exercise 27.5 e, ﬁnd the solution to

y′′′ 16y′ f (t) with y(0) 0 and y′(0) 0 + = = = for each of the following choices of f : a. f (t) 1 b. f (t) t c. f (t) e3t = = = d. f (t) sin(4t) e. f (t) sin(at) where α 4 = = 6= 27.9. Let f and g be two piecewise continuous functions on the positive real line satisfying, for all t > 0 , f (t) < M es0t and g(t) < M es0t | | f | | g for some constants M f , Mg and s0 .

s0t a. Show that f g(t) < M f Mge t whenever t > 0 . | ∗ | b. Why does this tell us that f g is of exponential order s1 for any s1 > s0 ? ∗