4.3 Translation Theorems

INTRODUCTION

It is not convenient to use Definition 4.1.1 each time we wish to find the Laplace transform of a given function f(t). For example, the integration by parts required to determine the transform of, say, f(t) = et t2 sin 3t is formidable to say the least when done by hand. In this section and the next we present several labor-saving theorems that enable us to build up a more extensive list of transforms (see the table in Appendix C) without the necessity of using the definition of the Laplace transform.

4.3.1 Translation on the s-axis

Evaluating transforms such as {e5tt3} and {e−2t cos 4t} is straightforward provided we know {t3} and {cos 4t}, which we do. In general, if we know {f(t)} = F(s) it is possible to compute the Laplace transform of an exponential multiple of the function f, that is, {eat f(t)}, with no additional effort other than translating, or shifting, F(s) to F(s − a). This result is known as the first translation theorem or first shifting theorem.

THEOREM 4.3.1 First Translation Theorem

If exists for and a is any constant, then

for

PROOF:

The proof is immediate, since by Definition 4.1.1

Because exists for exists for ≡

If we consider s a real variable, then the graph of F(s − a) is the graph of F(s) shifted on the s-axis by the amount |a|. If a > 0, the graph of F(s) is shifted a units to the right, whereas if a < 0, the graph is shifted |a| units to the left. See FIGURE 4.3.1.

Two curves are graphed on the s F coordinate plane. The curve labeled F(s) starts at a point on the positive F axis, goes down and to the right, reaches a point in the first quadrant that corresponds to the s value of a, and ends at a point just above the positive s axis. The curve labeled F(s minus a) starts at a point in the first quadrant that corresponds to the s value of a, goes down and to the right, and ends at a point just above the positive s axis. The value of s equals a, a greater than 0 is represented by a dashed vertical line.

FIGURE 4.3.1 Shift on s-axis

For emphasis it is sometimes useful to use the symbolism

where s → s − a means that in the Laplace transform F(s) of f(t) we replace the symbol s where it appears by s − a.

EXAMPLE 1 Using the First Translation Theorem

Find

(a) {e5tt3}

(b) {e−2t cos 4t}.

SOLUTION

The results follow from Theorems 4.1.1 and 4.3.1.

(a)

(b) ≡

Inverse Form of Theorem 4.3.1

To compute the inverse of F(s − a) we must recognize F(s), find f(t) by taking the inverse Laplace transform of F(s), and then multiply f(t) by the exponential function eat. This procedure can be summarized symbolically in the following manner:

(1)

where f(t) = −1{F(s)}.

EXAMPLE 2 Partial Fractions and Completing the Square

Partial fractions: repeated linear factors in (a).

Find

(a) −1

(b) −1

SOLUTION

(a) A repeated linear factor is a term (s − a)n, where a is a real number and n is a positive integer ≥ 2. Recall that if (s − a)n appears in the denominator of a rational expression, then the assumed decomposition contains n partial fractions with constant numerators and denominators s − a, (s − a)2, … , (s − a)n. Hence with a = 3 and n = 2 we write

The numerator obtained by putting the two terms on the right over a common denominator is 2s + 5 = A(s − 3)+ B, and this identity yields A = 2 and B = 11. Therefore,

(2)

and

−1 = 2 −1 + 11 −1 (3)

Now 1/(s − 3)2 is F(s) = 1/s2 shifted 3 units to the right. Since −1{1/s2} = t, it follows from (1) that

−1 = −1 = e3tt.

Finally, (3) is

−1 = 2e3t + 11e3tt. (4)

(b) To start, observe that the quadratic polynomial s2 + 4s + 6 does not have real zeros and so has no real linear factors. In this situation we complete the square:

(5)

Our goal here is to recognize the expression on the right as some Laplace transform F(s) in which s has been replaced throughout by s + 2. What we are trying to do here is analogous to working part (b) of Example 1 backward. The denominator in (5) is already in the correct form; that is, s2 + 2 with s replaced by s + 2. However, we must fix up the numerator by manipulating the constants: s + = (s + 2)+ − = (s + 2)+ .

Now by termwise division, the linearity of −1, parts (d) and (e) of Theorem 4.2.1, and finally from (1),

(6)

(7) ≡

EXAMPLE 3 Initial-Value Problem

Solve y″ − 6y′ + 9y = t2e3t, y(0) = 2, y′(0) = 17.

SOLUTION

Before transforming the DE note that its right-hand side is similar to the function in part (a) of Example 1. We use Theorem 4.3.1, the initial conditions, simplify, and then solve for Y(s) = {f(t)}:

The first term on the right has already been decomposed into individual partial fractions in (2) in part (a) of Example 2:

Y(s) =

Thus

(8)

From the inverse form (1) of Theorem 4.3.1, the last two terms are

and so (8) is . ≡

EXAMPLE 4 An Initial-Value Problem

Solve y″ + 4y′ + 6y = 1 + e−t, y(0) = 0, y′(0) = 0.

SOLUTION

Since the quadratic term in the denominator does not factor into real linear factors, the partial fraction decomposition for Y(s) is found to be

Moreover, in preparation for taking the inverse transform, we have already manipulated the last term into the necessary form in part (b) of Example 2. So in view of the results in (6) and (7) we have the solution

≡

4.3.2 Translation on the t-axis

Unit Step Function

In engineering, one frequently encounters functions that are either “off ” or “on.” For example, an external force acting on a mechanical system or a voltage impressed on a circuit can be turned off after a period of time. It is convenient, then, to define a special function that is the number 0 (off) up to a certain time t = a and then the number 1 (on) after that time. This function is called the unit step function or the Heaviside function. See the Remarks at the end of this section.

DEFINITION 4.3.1 Unit Step Function

The unit step function (t − a) is defined to be

Notice that we define (t − a) only on the nonnegative t-axis since this is all that we are concerned with in the study of the Laplace transform. In a broader sense (t − a) = 0 for t < a. The graphs of (t − a) and (t) are given in (a) and (b), respectively, of FIGURE 4.3.2.

Two graphs a and b represent U(t minus a) and U(t), respectively. Graph (a). Two pieces of the function U(t minus a) are graphed on the t U plane. The first piece is a straight line that starts at the origin (0, 0), goes to the right, and ends at the point (a, 0). The second piece is also a straight line that starts at the point (a, 1), goes to the right, and exits the right of the first quadrant. The point (a, 1) is marked with a dot. The positive t axis below the point (a, 0) is labeled a greater than 0. Graph (b). The function U(t) is a straight line graphed on the t U plane. The line starts at the point (0, 1), goes to the right, and ends almost at the middle of the first quadrant. The point (0, 1) is marked with a dot. The positive t axis below the line is labeled a equals 0.

FIGURE 4.3.2 Graphs of unit step functions

When a function f defined for t ≥ 0 is multiplied by (t − a), the unit step function “turns off ” a portion of the graph of that function. For example, consider the function f(t) = 2t − 3. To “turn off ” the portion of the graph of f on, say, the interval 0 ≤ t < 1, we simply form the product (2t − 3)(t − 1). See FIGURE 4.3.3. In general, the graph of f(t) (t − a) is 0 (off) for 0 ≤ t < a and is the portion of the graph of f(on) for t ≥ a.

The graph of function f(t) has two pieces graphed on the t y plane. The first piece is a straight line that starts at the origin, goes to the right, and ends at 1 unit on the positive t axis. The second piece is also a straight line that starts at a point marked with a dot in the fourth quadrant, goes up and to the right through the positive t axis, and exits the top right of the first quadrant. The dot corresponds to a point 1 unit to the right of negative y axis and 1 unit below the positive t axis.

FIGURE 4.3.3 Function can be written f(t) = (2t − 3)(t − 1)

The unit step function can also be used to write piecewise-defined functions in a compact form. For example, by considering 0 ≤ t < 2, 2 ≤ t < 3, t ≥ 3, and the corresponding values of (t − 2) and (t − 3), it should be apparent that the piecewise-defined function shown in FIGURE 4.3.4 is the same as f(t) = 2 − 3 (t − 2) + (t − 3). Also, a general piecewise-defined function of the type

The graph of function f(t) has three pieces graphed on the t f(t) plane. The first piece is a straight line that starts at the point (0, 2), goes to the right, and ends at the point (2, 2). The second piece is also a straight line that starts at the point (2, negative 1), goes to the right, and ends at the point (3, negative 1). The third piece is a straight line that starts at the point (3, 0), goes to the right, and ends at the right on the positive t axis.

FIGURE 4.3.4 Function can be written f(t) = 2 − 3(t − 2) + (t − 3)

(9)

is the same as

(10)

Similarly, a function of the type

(11)

can be written

(12)

EXAMPLE 5 A Piecewise-Defined Function

Express f(t) = in terms of unit step functions. Graph.

SOLUTION

The graph of f is given in FIGURE 4.3.5. Now from (9) and (10) with a = 5, g(t) = 20t, and h(t) = 0, we get . ≡

The graph of function f(t) has two pieces graphed on the t f(t) plane. The first piece is a straight line that starts at the origin, goes up and to the right, and ends at the point (5, 100). The second piece is also a straight line that starts at the point (5, 0), goes to the right, and ends at the right on the positive t axis. The point (5, 0) is marked with a dot.

FIGURE 4.3.5 Function in Example 5

Consider a general function y = f(t) defined for t ≥ 0. The piecewise-defined function

(13)

plays a significant role in the discussion that follows. As shown in FIGURE 4.3.6, for a > 0 the graph of the function y = f(t − a) (t − a) coincides with the graph of y = f(t − a) for t ≥ a (which is the entire graph of y = f(t), t ≥ 0, shifted a units to the right on the t-axis) but is identically zero for 0 ≤ t < a.

Two graphs a and b. Graph (a) is titled f(t), t greater than or equal to 0. A curve is graphed on the t f(t) plane. It starts at a point on the positive f(t) axis, goes up and to the right, and reaches a high point. Then, it goes down and to the right, reaches a low point, and ends at the right of the first quadrant. The starting point is marked with a dot. Graph (b) is titled f(t minus a) U(t minus a). Two pieces of the function are graphed on the t f(t) plane. The first piece is a straight line that starts at the origin, goes to the right, and ends at the point (a, 0). The second piece is a curve that starts at a point in the first quadrant that corresponds to the t value of a, goes up and to the right, and reaches a high point. Then, it goes down and to the right, reaches a low point, and ends at the right of the first quadrant. The starting point of the curve is marked with a dot.

FIGURE 4.3.6 Shift on t-axis

We saw in Theorem 4.3.1 that an exponential multiple of f(t) results in a translation of the transform F(s) on the s-axis. As a consequence of the next theorem we see that whenever F(s) is multiplied by an exponential function e−as, a > 0, the inverse transform of the product e−asF(s) is the function f shifted along the t-axis in the manner illustrated in Figure 4.3.6(b). This result, presented next in its direct transform version, is called the second translation theorem or second shifting theorem.

THEOREM 4.3.2 Second Translation Theorem

If exists for and a is a positive constant, then

for s > c.

PROOF:

By the additive interval property of integrals, e−stf(t − a) (t − a) dt can be written as two integrals:

Now if we let v = t − a, dv = dt in the last integral, then

≡

EXAMPLE 6 Using the Second Translation Theorem

Find

SOLUTION

Because the function has period we can write

Then with the identification , it follows from Theorem 4.3.2 that

≡

We often wish to find the Laplace transform of just a unit step function. This can be found from either Definition 4.1.1 or Theorem 4.3.2. If we identify f(t) = 1 in Theorem 4.3.2, then f(t − a) = 1, F(s) = {1} = 1/s, and so

(14)

EXAMPLE 7 Figure 4.3.4 Revisited

Find the Laplace transform of the function f whose graph is given in Figure 4.3.4.

SOLUTION

We use f expressed in terms of the unit step function

and the result given in (14):

≡

Inverse Form of Theorem 4.3.2

If f(t) = −1{F(s)}, the inverse form of Theorem 4.3.2, with a > 0, is

(15)

EXAMPLE 8 Using Formula (15)

Find (a) −1 (b) −1

SOLUTION

(a) With the identifications a = 2, F(s) = 1/(s − 4), −1{F(s)} = e4t, we have from (15)

(b) With a = π/2, F(s) = s/(s2 + 9), −1{F(s)} = cos 3t, (15) yields

.

The last expression can be simplified somewhat using the addition formula for the cosine. Verify that the result is the same as . ≡

Alternative Form of Theorem 4.3.2

We are frequently confronted with the problem of finding the Laplace transform of a product of a function g and a unit step function (t − a) where the function g lacks the precise shifted form f(t − a) in Theorem 4.3.2. To find the Laplace transform of g(t)(t − a), it is possible to fix up g(t) into the required form f(t − a) by algebraic manipulations. For example, if we want to use Theorem 4.3.2 to find the Laplace transform of t2 (t − 2), we would have to force g(t) = t2 into the form f(t − 2). You should work through the details and verify that

is an identity. Therefore

by Theorem 4.3.2. But since these manipulations are time-consuming and often not obvious, it is simpler to devise an alternative version of Theorem 4.3.2. Using Definition 4.1.1, the definition of (t − a), and the substitution u = t − a, we obtain

(16)

That is,

(16)

With a = 2 and g(t) = t2 we have by (16),

which agrees with our previous calculation.

EXAMPLE 9 Second Translation Theorem—Alternative Form

Find {cos t (t − π)}.

SOLUTION

With g(t) = cos t, a = π, then g(t + π) = cos(t + π) = −cos t by the addition formula for the cosine function. Hence by (16),

In the next two examples, we solve in turn an initial-value problem and a boundary-value problem. Both problems involve a piecewise-linear differential equation. ≡

EXAMPLE 10 An Initial-Value Problem

Solve y′ + y = f(t), y(0) = 5, where f(t) =

SOLUTION

The function f can be written as f(t) = 3 cos t (t − π) and so by linearity, the results of Example 9, and the usual partial fractions, we have

(17)

Now proceeding as we did in Example 8, it follows from (15) with a = π that the inverses of the terms in the bracket are

and

−1 = cos(t − π) (t − π).

Thus the inverse of (17) is

(18)

With the aid of a graphing utility we get the graph of (18), shown in FIGURE 4.3.7. ≡

Two pieces of a function are graphed on the t y plane. The first piece is a curve that starts at the point (0, 5), goes down and to the right, and ends approximately at the point (pi, 0.2). The second piece is also a curve that follows an oscillatory pattern. It starts at the approximate point (pi, 0.2), goes down and to the right through the positive t axis, and reaches a point in the fourth quadrant. Then, it goes up and to the right through the positive t axis, reaches a point in the first quadrant, again goes down and to the right through the positive t axis, and ends at the bottom right of the fourth quadrant.

FIGURE 4.3.7 Graph of function (18) in Example 10

Beams

In Section 3.9 we saw that the static deflection y(x) of a uniform beam of length L carrying load w(x) per unit length is found from the linear fourth-order differential equation

,(19)

where E is Young’s modulus of elasticity and I is a moment of inertia of a cross section of the beam. The Laplace transform is particularly useful when w(x) is piecewise defined, but in order to use the transform, we must tacitly assume that y(x) and w(x) are defined on (0, ∞) rather than on (0, L). Note, too, that the next example is a boundary-value problem rather than an initial-value problem.

EXAMPLE 11 A Boundary-Value Problem

A beam of length L is embedded at both ends as shown in FIGURE 4.3.8. Find the deflection of the beam when the load is given by

where w0 is a constant.

A beam of length L is embedded between walls. A vertical line along the wall at the left end is labeled y. A horizontal line along the axis of the beam is labeled x. A variable load w(x) acts over the beam from the left end to almost half of the length of the beam.

FIGURE 4.3.8 Embedded beam with a variable load in Example 11

SOLUTION

Recall that, since the beam is embedded at both ends, the boundary conditions are y(0) = 0, y′(0) = 0, y(L) = 0, y′(L) = 0. Now by (10) we can express w(x) in terms of the unit step function:

Transforming (19) with respect to the variable x gives

EI

or

.

If we let c1 = y″(0) and c2 = y‴(0), then

Y(s) = ,

and consequently

Applying the conditions y(L) = 0 and y′(L) = 0 to the last result yields a system of equations for c1 and c2:

Solving, we find c1 = 23w0 L2/960EI and c2= −9w0 L/40EI. Thus the deflection is

≡

REMARKS

A photo of Oliver Heaviside.

© The Picture Art Collection/Alamy Stock Photo

Oliver Heaviside

Many of the results presented in this chapter were devised by the English electrical engineer Oliver Heaviside (1850–1925) and are set forth in his 1899 treatise Electromagnetic Theory. The Heaviside operational calculus, as his procedures came to be called, were a means of solving linear differential equations with constant coefficients that arose in his investigation of transmission lines. Since many of his results lacked formal proof, Heaviside’s procedures initially met with scorn from mathematicians. A curmudgeon, Heaviside, in turn, called these “establishment” mathematicians “woodenheaded.” When Heaviside, using his symbolic methods, was able to obtain answers to problems that mathematicians could not solve, their scorn turned to censure and his papers were no longer published in mathematics journals. In 1902 Heaviside predicted the existence of a layer of ionized gas in the upper atmosphere that would reflect radio waves back to Earth. After the discovery of this layer in 1924 it came to be called the Heaviside layer. Heaviside lived the last years of his life as a recluse in poverty, forgotten by the scientific community. He died in an unheated house in 1925.

In an attempt to put Heaviside’s operational calculus on a sound foundation, mathematicians discovered that the rules of his calculus matched many properties of the Laplace transform. Over time, Heaviside’s operational calculus disappeared, to be replaced by the theory and applications of the Laplace transform.

4.3 Exercises Answers to selected odd-numbered problems begin on page ANS-10.

4.3.1 Translation on the s-axis

In Problems 1–20, find either F(s) or f(t), as indicated.

  1. {te10t}
  2. {te−6t}
  3. {t3e−2t}
  4. {t10e−7t}
  5. {t(et + e2t)2}
  6. {e2t(t − 1)2}
  7. {et sin 3t}
  8. {e−2t cos 4t}
  9. {(1 − et + 3e−4t) cos 5t}
  11. −1
  12. −1
  13. −1
  14. −1
  15. −1
  16. −1
  17. −1
  18. −1
  19. −1
  20. −1

In Problems 21–24, write the hyperbolic functions in terms of exponential functions, and then find using Theorem 4.3.1.

In Problems 25–34, use the Laplace transform to solve the given initial-value problem.

  1. y′ + 4y = e−4t, y(0) = 2
  2. y′ − y = 1+ tet, y(0) = 0
  3. y″ + 2y′ + y = 0, y(0) = 1, y′(0) = 1
  4. y″ − 4y′ + 4y = t3e2t, y(0) = 0, y′(0) = 0
  5. y″ − 6y′ + 9y = t, y(0) = 0, y′(0) = 1
  6. y″ − 4y′ + 4y = t3, y(0) = 1, y′(0) = 0
  7. y″ − 6y′ + 13y = 0, y(0) = 0, y′(0)= −3
  8. 2y″ + 20y′ + 51y = 0, y(0) = 2, y′(0) = 0
  9. y″ − y′ = et cos t, y(0) = 0, y′(0) = 0
  10. y″ − 2y′ + 5y = 1+ t, y(0) = 0, y′(0) = 4

In Problems 35 and 36, use the Laplace transform and the procedure outlined in Example 11 to solve the given boundary-value problem.

  1. y″ + 2y′ + y = 0, y′(0) = 2, y(1) = 2
  2. y″ + 8y′ + 20y = 0, y(0) = 0, y′(π) = 0
  3. A 4-lb weight stretches a spring 2 ft. The weight is released from rest 18 in. above the equilibrium position, and the resulting motion takes place in a medium offering a damping force numerically equal to times the instantaneous velocity. Use the Laplace transform to find the equation of motion x(t).
  4. Recall that the differential equation for the instantaneous charge q(t) on the capacitor in an LRC-series circuit is

    (20)

    See Section 3.8. Use the Laplace transform to find q(t) when L = 1 h, R = 20 Ω, C = 0.005 F, E(t) = 150 V, t > 0, q(0) = 0, and i(0) = 0. What is the current i(t)?

  5. Consider the battery of constant voltage E0 that charges the capacitor shown in FIGURE 4.3.9. Divide equation (20) by L and define 2λ = R/L and ω2 = 1/LC. Use the Laplace transform to show that the solution q(t) of q″ + 2λq′ + ω2q = E0/L, subject to q(0) = 0, i(0) = 0, is

    The circuit consisting of the following components: a battery of constant voltage E subscript 0, a resistor R, a capacitor C, and an inductor L. The inductor L and capacitor C are placed parallel to each other. The voltage E subscript 0 is connected between the inductor and the capacitor. The resistor R is parallel to the voltage E subscript 0 and is connected between the inductor and the capacitor.

    FIGURE 4.3.9 Circuit in Problem 39

  6. Use the Laplace transform to find the charge q(t) in an RC-series when q(0) = 0 and E(t) = E0e−kt, k > 0. Consider two cases: k ≠ 1/RC and k = 1/RC.

4.3.2 Translation on the t-axis

In Problems 41–52, find either F(s) or f(t), as indicated.

  1. {(t − 1) (t − 1)}
  2. {e2 − t (t − 2)}
  3. {t (t − 2)}
  4. {(3t + 1) (t − 1)}
  5. {cos 2t (t − π)}
  7. −1
  8. −1
  9. −1
  10. −1
  11. −1
  12. −1

In Problems 53–58, match the given graph with one of the given functions in (a)–(f). The graph of f(t) is given in FIGURE 4.3.10.

A curve is graphed on the t f(t) plane. It starts at a point on the positive f(t) axis, goes down and to the right, reaches a point in the first quadrant, goes up and to the right, and reaches a high point. Then, it goes down and to the right, becomes smooth at the end, and ends at the right of the first quadrant just above the t axis. The starting point of the curve is marked with a dot. The starting point of the rise of the curve corresponds to the t value of a and the end of fall of the peak corresponds to the t value of b.

FIGURE 4.3.10 Graph for Problems 53–58

(a) f(t)− f(t) (t − a)

(b) f(t − b) (t − b)

(c) f(t) (t − a)

(d) f(t)− f(t) (t − b)

(e) f(t) (t − a)− f(t) (t − b)

(f) f(t − a) (t − a)− f(t − a) (t − b)

  1. Two pieces of the function are graphed on the t f(t) plane. The first piece is a straight line that starts at the origin, goes to the right, and ends at (a, 0). The second piece is a curve that starts at a point in the first quadrant that corresponds to the t value of a, goes up and to the right, reaches a high point, then goes down and to the right, and reaches a point in the first quadrant that corresponds to the t value of b. Then, it goes down and to the right, becomes smooth at the end, and ends at the right of the first quadrant just above the t axis. The starting point of the curve is marked with a dot.

    FIGURE 4.3.11 Graph for Problem 53

  2. Three pieces of the function are graphed on the t f(t) plane. The first piece is a straight line that starts at the origin, goes to the right, and ends at (a, 0). The second piece is a curve that starts at a point in the first quadrant that corresponds to the t value of a, goes up and to the right, and reaches a high point. Then it goes down and to the right, and ends at a point in the first quadrant that corresponds to the t value of b. The starting point of the curve is marked with a dot. The third piece is a straight line that starts at the point (b, 0), goes to the right, and ends at the right on the positive t axis.

    FIGURE 4.3.12 Graph for Problem 54

  3. Three pieces of the function are graphed on the t f(t) plane. The first piece is a straight line that starts at the origin, goes to the right, and ends at (a, 0). The second piece is a curve that starts at a point in the first quadrant that corresponds to the t value of a, goes down and to the right, and reaches a low point in the first quadrant. Then it goes up and to the right, and ends at a point that corresponds to the t value of b. The starting point of the curve is marked with a dot. The third piece is a straight line that starts at the point (b, 0), goes to the right, and ends at the right on the positive t axis.

    FIGURE 4.3.13 Graph for Problem 55

  4. Three pieces of the function are graphed on the t f(t) plane. The first piece is a curve that starts at a point on the positive f(t) axis, goes down and to the right, and reaches a low point in the first quadrant. Then it goes up and to the right, and ends at a point that corresponds to the t value of a. The starting point of the curve is marked with a dot. The second piece is a straight line that starts at the point (a, 0), goes to the right, and ends at (b, 0). The third piece is also a straight line that starts at the point (b, 0), goes to the right, and ends at the right on the positive t axis.

    FIGURE 4.3.14 Graph for Problem 56

  5. Three pieces of the function are graphed on the t f(t) plane. The first piece is a curve that starts at a point on the positive f(t) axis, goes down and to the right, and reaches a low point in the first quadrant. Then it goes up and to the right, and ends at a point that corresponds to the t value of a. The starting point of the curve is marked with a dot. The second piece is a straight line that starts at the point (a, 0), goes to the right, and ends at (b, 0). The third piece is also a straight line that starts at the point (b, 0), goes to the right, and ends at the right on the positive t axis.

    FIGURE 4.3.15 Graph for Problem 57

  6. Two pieces of the function are graphed on the t f(t) plane. The first piece is a curve that starts at a point on the positive f(t) axis, goes down and to the right, and reaches a low point in the first quadrant. Then it goes up and to the right at a point that corresponds to the t value of a, reaches a high point, again goes down and to the right, and ends at a point that corresponds to the t value of b. The starting point of the curve is marked with a dot. The second piece is a straight line that starts at the point (b, 0), goes to the right, and ends at the right on the positive t axis.

    FIGURE 4.3.16 Graph for Problem 58

In Problems 59–70, write each function in terms of unit step functions. Find the Laplace transform of the given function.

  9. Three pieces of a function are graphed on the t f(t) plane. The first piece is a straight line that starts at the origin, goes to the right, and ends at the point (a, 0). The second piece is a straight line that starts at the point (a, 1), goes to the right, and ends at the point (b, 1). The third piece is also a straight line that starts at the point (b, 0), goes to the right, and ends at the right on the positive t axis. The graph is titled rectangular pulse.

    FIGURE 4.3.17 Graph for Problem 67

  10. Four pieces of a function are graphed as straight lines on the t f(t) plane. The first piece starts at the point (0, 1), goes to the right, and ends at the point (1, 1). The second piece starts at the point (1, 2), goes to the right, and ends at the point (2, 2). The third piece starts at the point (2, 3), goes to the right, and ends at the point (3, 3). The fourth piece starts at the point (3, 4), goes to the right, and ends at the right of the first quadrant. The graph is titled staircase function.

    FIGURE 4.3.18 Graph for Problem 68

  11. Three pieces of a function are graphed on the t f(t) plane. The first piece is a straight line that starts at the point (0, 2), goes down and to the right, and ends at the point (1, 1). The second piece is a straight line that starts at the point (1, 1), goes to the right, and ends at the point (2, 1). The third piece is also a straight line that starts at the point (2, 0), goes to the right, and ends at the right on the positive t axis.

    FIGURE 4.3.19 Graph for Problem 69

  12. Three pieces of a function are graphed on the t f(t) plane. The first piece is a straight line that starts at the point (0, 0), goes down and to the right, and ends at the point (1, negative 1). The second piece is a straight line that starts at the point (1, negative 1), goes up and to the right through the point (2, 0), and ends at the point (3, 1). The third piece is also a straight line that starts at the point (3, 0), goes to the right, and ends at the right on the positive t axis.

    FIGURE 4.3.20 Graph for Problem 70

In Problems 71–78, use the Laplace transform to solve the given initial-value problem.

  1. y′ + y = f(t), y(0) = 0, where f(t) =
  2. y′ + y = f(t), y(0) = 0, where f(t) =
  3. y′ + 2y = f(t), y(0) = 0, where f(t) =
  4. y″ + 4y = f(t), y(0) = 0, y′(0)= −1, where

  5. y″ + 4y = sin t (t − 2π), y(0) = 1, y′(0) = 0
  6. y″ − 5y′ + 6y = (t − 1), y(0) = 0, y′(0) = 1
  7. y″ + y = f(t), y(0) = 0, y′(0) = 1, where

  8. y″ + 4y′ + 3y = 1− (t − 2)− (t − 4)+ (t − 6), y(0) = 0, y′(0) = 0
  9. Suppose a mass weighing 32 lb stretches a spring 2 ft. If the weight is released from rest at the equilibrium position, find the equation of motion x(t) if an impressed force f(t) = 20t acts on the system for 0 ≤ t < 5 and is then removed (see Example 5). Ignore any damping forces. Use a graphing utility to obtain the graph x(t) on the interval [0, 10].
  10. Solve Problem 79 if the impressed force f(t) = sin t acts on the system for 0 ≤ t < 2π and is then removed.

In Problems 81 and 82, use the Laplace transform to find the charge q(t) on the capacitor in an RC-series circuit subject to the given conditions.

  1. q(0) = 0, R = 2.5 Ω, C = 0.08 F, E(t) given in FIGURE 4.3.21
    Two pieces of the function are graphed on the t E(t) plane. The first piece is a straight line that starts at the point (0, 0), goes to the right, and ends at the point (3, 0). The second piece is also a straight line that starts at the point (3, 5), goes to the right, and ends at the right of the first quadrant. The point (3, 5) is marked with a dot.

    FIGURE 4.3.21 E(t) in Problem 81

  2. q(0) = q0, R = 10 Ω, C = 0.1 F, E(t) given in FIGURE 4.3.22
    Two pieces of the function are graphed on the t E(t) plane. The first piece is a curve that starts at the point (0, 30), goes up and to the right, and ends at a point in the first quadrant that corresponds to the t value of 1.5. The curve is labeled 30 times e^t. The second piece is a straight line that starts at the point (1.5, 0), goes to the right, and ends at the right on the positive t axis.

    FIGURE 4.3.22 E(t) in Problem 82

    1. Use the Laplace transform to find the current i(t) in a single-loop LR-series circuit when i(0) = 0, L = 1 h, R = 10 Ω, and E(t) is as given in FIGURE 4.3.23.
    2. Use a computer graphing program to graph i(t) for 0 ≤ t ≤ 6. Use the graph to estimate imax and imin, the maximum and minimum values of the current, respectively.
      Two pieces of the function are graphed on the t E(t) plane. The first piece is a curve that starts at the point (0, 0), goes up and to the right, and reaches a high (pi over 2, 1). Then it goes down and to the right through the point (pi, 0), and ends at the point (3 pi over 2, negative 1). The curve is labeled sin t, 0 less than or equal to t less than 3 pi over 2. The second piece is a straight line that starts at the point (3 pi over 2, 0), goes to the right, and ends at the right on the positive t axis.

      FIGURE 4.3.23 E(t) in Problem 83

    1. Use the Laplace transform to find the charge q(t) on the capacitor in an RC-series circuit when q(0) = 0, R = 50 Ω, C = 0.01 F, and E(t) is as given in FIGURE 4.3.24.
    2. Assume E0 = 100 V. Use a computer graphing program to graph q(t) for 0 ≤ t ≤ 6. Use the graph to estimate qmax, the maximum value of the charge.
      Two pieces of the function are graphed on the t E(t) plane. The first piece is a straight line that starts at the point (1, E subscript 0), goes to the right, and ends at the point (3, E subscript 0). The second piece is also a straight line that starts at the point (3, 0), goes to the right, and ends at the right on the positive t axis.

      FIGURE 4.3.24 E(t) in Problem 84

  5. A cantilever beam is embedded at its left end and free at its right end. Use the Laplace transform to find the deflection y(x) when the load is given by
  6. Solve Problem 85 when the load is given by

  7. Find the deflection y(x) of a cantilever beam embedded at its left end and free at its right end when the load is as given in Example 11.
  8. A beam is embedded at its left end and simply supported at its right end. Find the deflection y(x) when the load is as given in Problem 85.
  9. Cake Inside an Oven Reread Example 4 in Section 2.7 on the cooling of a cake that is taken out of an oven.
    1. Devise a mathematical model for the temperature of a cake while it is inside the oven based on the following assumptions: At t = 0 the cake mixture is at the room temperature of 70°; the oven is not preheated so that at t = 0, when the cake mixture is placed into the oven, the temperature inside the oven is also 70°; the temperature of the oven increases linearly until t = 4 minutes, when the desired temperature of 300° is attained; the oven temperature is a constant 300° for t ≥ 4.
    2. Use the Laplace transform to solve the initial-value problem in part (a).

Discussion Problems

  1. Discuss how you would fix up each of the following functions so that Theorem 4.3.2 could be used directly to find the given Laplace transform. Check your answers using (16) of this section.

    (a) {(2t + 1) (t − 1)}

    (b) {et (t − 5)}

    (c) {cos t (t − π)}

    (d) {(t2 − 3t) (t − 2)}

  2. Assume that Theorem 4.3.1 holds when the symbol a is replaced by ki, where k is a real number and i2 = −1. Show that {tekti} can be used to deduce

    and

  3. Use the Laplace transform to solve the initial-value problem

    x″ + ω2x = cos ωt, x(0) = 0, x′(0) = 0.