# 6.E: The Laplace Transform (Exercises) - Mathematics

These are homework exercises to accompany Libl's "Differential Equations for Engineering" Textmap. This is a textbook targeted for a one semester first course on differential equations, aimed at engineering students. Prerequisite for the course is the basic calculus sequence.

## 6.1: The Laplace transform

Exercise 6.1.5: Find the Laplace transform of (3+ t^5 + sin (pi t)).

Exercise 6.1.6: Find the Laplace transform of (a + bt +ct^2) for some constants (a), (b), and (c).

Exercise 6.1.7: Find the Laplace transform of (A cos (omega t) + B sin (omega t ) ).

Exercise 6.1.8: Find the Laplace transform of ( cos^2 (omega t ) ).

Exercise 6.1.9: Find the inverse Laplace transform of (dfrac{4}{s^2-9}).

Exercise 6.1.10: Find the inverse Laplace transform of ( dfrac{2s}{s^2-1}).

Exercise 6.1.11: Find the inverse Laplace transform of ( dfrac{1}{(s-1)^2(s+1)}).

Exercise 6.1.12: Find the Laplace transform of (f(t)= left{ egin{array}{cc} t & { m{if~}}t geq 1, 0 & { m{if~}}t < 1.end{array} ight.).

Exercise 6.1.13:Find the inverse Laplace transform of ( dfrac{s}{(s^2+s+2)(s+4)}).

Exercise 6.1.14: Find the Laplace transform of (sin left( omega (t-a) ight) ).

Exercise 6.1.15: Find the Laplace transform of ( t sin (omega t) ). Hint: Several integrations by parts.

Exercise 6.1.101: Find the Laplace transform of (4(t+1)^2).

Exercise 6.1.102: Find the inverse Laplace transform of (dfrac{8}{s^3 (s+2)}).

Exercise 6.1.103: Find the Laplace transform of (te^{-t}) (Hint: integrate by parts).

Exercise 6.1.104: Find the Laplace transform of (sin (t) e^{-t}) (Hint: integrate by parts).

## 6.2: Transforms of Derivatives and ODEs

Exercise 6.2.1: Verify Table 6.2.

Exercise 6.2.2: Using the Heaviside function write down the piecewise function that is (0) for (t<0, t^2) for (t) in ([0,1]) and (t) for (t>1).

Exercise 6.2.3: Using the Laplace transform solve

[ mx'' + cx'+kx =0,~~~~~~~ x(0)=a, ~~~~~~~ x'(0)=b.]

where (m>0,c>0,k>0), and (c^2-4km>0) (system is overdamped).

Exercise 6.2.4: Using the Laplace transform solve

[ mx'' + cx'+kx =0,~~~~~~~ x(0)=a, ~~~~~~~ x'(0)=b.]

where (m>0,c>0,k>0), and (c^2-4km<0) (system is underdamped).

Exercise 6.2.5: Using the Laplace transform solve

[ mx'' + cx'+kx =0,~~~~~~~ x(0)=a, ~~~~~~~ x'(0)=b.]

where (m>0,c>0,k>0), and (c^2=4km) (system is critically damped).

Exercise 6.2.6: Solve (x''+x=u(t-1)) for initial conditions (x(0)=0) and (x'(0)=0).

Exercise 6.2.7: Show the diﬀerentiation of the transform property. Suppose (mathcal{L}{f(t)}=F(s)), then show

[ mathcal{L}{-tf(t)}=F'(s).]

Hint: Diﬀerentiate under the integral sign.

Exercise 6.2.8: Solve (x'''+x=t^3u(t-1)) for initial conditions (x(0)=1) and (x'(0)=0), (x''(0)=0).

Exercise 6.2.9: Show the second shifting property: ( mathcal{L}{f(t-a)u(t-a)}=e^{-as}mathcal{L}{f(t)} ).

Exercise 6.2.10: Let us think of the mass-spring system with a rocket from Example 6.2.2. We noticed that the solution kept oscillating after the rocket stopped running. The amplitude of the oscillation depends on the time that the rocket was ﬁred (for 4 seconds in the example). a) Find a formula for the amplitude of the resulting oscillation in terms of the amount of time the rocket is ﬁred. b) Is there a nonzero time (if so what is it?) for which the rocket ﬁres and the resulting oscillation has amplitude 0 (the mass is not moving)?

Exercise 6.2.11: Deﬁne

[ f(t)= left{ egin{array}{ccc} (t-1)^2 & if~1 leq t<2, 3-t & if~2 leq t<3, 0 & otherwise. end{array} ight. ]

a) Sketch the graph of (f(t)). b) Write down (f(t)) using the Heaviside function. c) Solve (x''+x=f(t), x(0)=0,x'(0)=0) using Laplace transform.

Exercise 6.2.12: Find the transfer function for (mx'' + cx'+kx =f(t)) (assuming the initial conditions are zero).

Exercise 6.2.101: Using the Heaviside function (u(t)), write down the function

[ f(t)= left{ egin{array}{ccc} 0 & if~~~~~t<1, t-1 & if~1 leq t<2, if~~~~~2 leq t. ]

Exercise 6.2.102: Solve (x''-x=(t^2-1)u(t-1)) for initial conditions (x(0)=1,x'(0)=2) using the Laplace transform.

Exercise 6.2.103: Find the transfer function for (x'+x=f(t)) (assuming the initial conditions are zero).

## 6.3: Convolution

Exercise 6.3.1: Let (f(t)=t^2) for (t geq 0), and (g(t)=u(t-1)). Compute (f * g).

Exercise 6.3.2: Let (f(t)=t) for (t geq 0), and (g(t)=sin t) for (t geq 0). Compute (f * g).

Exercise 6.3.3: Find the solution to

( mx''+cx'+kx=f(t),~~~~~~x(0)=0,~~~~~~x'(0)=0,)

for an arbitrary function (f(t)), where (m>0,c>0,k>0), and (c^2-4km>0) (system is overdamped). Write the solution as a deﬁnite integral.

Exercise 6.3.4: Find the solution to

( mx''+cx'+kx=f(t),~~~~~~x(0)=0,~~~~~~x'(0)=0,)

for an arbitrary function (f(t)), where (m>0,c>0,k>0), and (c^2-4km<0) (system is underdamped). Write the solution as a deﬁnite integral.

Exercise 6.3.5: Find the solution to

( mx''+cx'+kx=f(t),~~~~~~x(0)=0,~~~~~~x'(0)=0,)

for an arbitrary function (f(t)), where (m>0,c>0,k>0), and (c^2=4km) (system is critically damped). Write the solution as a deﬁnite integral.

Exercise 6.3.6: Solve

( x(t)=e^{-t} +int_0^tcos(t- au)x( au)~d au . )

Exercise 6.3.7: Solve

( x(t)=cos t +int_0^tcos(t- au)x( au)~d au . )

Exercise 6.3.8: Compute (mathcal{L}^{-1} left{ frac{s}{(s^2+4)^2} ight}) using convolution.

Exercise 6.3.9: Write down the solution to (x''-2x=e^{-t^2},x(0)=0,x'(0)=0) as a deﬁnite integral. Hint: Do not try to compute the Laplace transform of (e^{-t^2}).

Exercise 6.3.101: Let (f(t)=cos t) for (t geq 0), and (g(t)=e^{-t}). Compute (f * g).

Exercise 6.3.102: Compute (mathcal{L}^{-1} left{ frac{5}{s^4+s^2} ight}) using convolution.

Exercise 6.3.103: Solve (x''+x=sin t, x(0)=0, x'(0)=0) using convolution.

Exercise 6.3.104: Solve (x'''+x'=f(t), x(0)=0, x'(0)=0,x''(0)=0) using convolution. Write the result as a deﬁnite integral.

## 6.4: Dirac delta and impulse response

Exercise 6.4.1: Solve (ﬁnd the impulse response) ( x'' + x' + x = delta(t),x(0) = 0, x'(0)=0.)

Exercise 6.4.2: Solve (ﬁnd the impulse response) (x'' + 2 x' + x = delta(t), x(0) = 0, x'(0)=0.)

Exercise 6.4.3: A pulse can come later and can be bigger. Solve (x'' + 4 x = 4delta(t-1), x(0) = 0, x'(0)=0.)

Exercise 6.4.4: Suppose that (f(t)) and (g(t)) are diﬀerentiable functions and suppose that (f(t) = g(t) = 0) for all (t leq 0). Show that

[ (f * g)'(t) = (f' * g)(t) = (f * g')(t) .]

Exercise 6.4.5: Suppose that (L x = delta(t), x(0) = 0, x'(0) = 0), has the solution (x = e^{-t}) for (t>0). Find the solution to (Lx = t^2, x(0) = 0, x'(0) = 0) for (t > 0).

Exercise 6.4.6: Compute (mathcal{L}^{-1} left{ frac{s^2+s+1}{s^2} ight}).

Exercise 6.4.7 (challenging): Solve Example 6.4.3 via integrating 4 times in the (x) variable.

Exercise 6.4.8: Suppose we have a beam of length (1) simply supported at the ends and suppose that force (F=1) is applied at (x=frac{3}{4}) in the downward direction. Suppose that (EI=1) for simplicity. Find the beam deﬂection (y(x)).

Exercise 6.4.101: Solve (ﬁnd the impulse response) (x'' = delta(t), x(0) = 0, x'(0)=0).

Exercise 6.4.102: Solve (ﬁnd the impulse response) (x' + a x = delta(t), x(0) = 0, x'(0)=0).

Exercise 6.4.103: Suppose that (L x = delta(t), x(0) = 0, x'(0) = 0), has the solution (x(t) = cos(t)) for (t>0). Find (in closed form) the solution to (Lx = sin(t), x(0) = 0, x'(0) = 0 for t > 0).

Exercise 6.4.104: Compute ({mathcal{L}}^{-1} left{ frac{s^2}{s^2+1} ight}).

Exercise 6.4.105: Compute ({mathcal{L}}^{-1} left{ frac{3 s^2 e^{-s} + 2}{s^2} ight}).

## Dimensional Laplace Equation

The two- dimensional Laplace equation ∂ 2 u/∂x 2 + ∂ 2 u/∂y 2 = 0 is satisfied by the cubic u(x, y) = − x 3 – y 3 + 3xy 2 + 3x 2 y. It can be used to define the exact essential (Dirichlet) boundary conditions on the edges of any two-dimensional shape. For a unit square with a corner at the origin the boundary conditions are

Obtain a finite element solution and sketch it along with the exact values along lines of constant x or y, or compare to exact_case 20 in the MODEL code. Solve using a domain of: a) the above unit square, or b) a rectangle over 0 ≤ x ≤ 3 and 0 ≤ y ≤ 2, or c) a quarter circle of radius 2 centered at the origin.

Modify the above problem to have normal flux (Neumann) boundary conditions on the two edges where y is constant (corresponding to the same exact solution):

Consider a two-dimensional Poisson equation on a unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 that contains a local high gradient peak on its interior centered at x = β, y = β. The amplitude of the peak is set by a parameter a. Let A = (x – β)/α, B = (y – β)/α, C = (1 – β)/a, and D = β/α. Then for the source, Q, defined by

Consider the partial differential equation ∇ · ( E ∇ ϕ ) + ν · ∇ ϕ + F ϕ + Q = 0 . The second term is new. In 2-D it is v . ∇ ϕ = < v x v y >[ ∂ ϕ ∂ x ∂ ϕ ∂ y ] . Apply the Galerkin method to get the additional matrix, say U v e , that appears because of this term. State the result in matrix integral form. Is the result symmetric? The data ν are often given at the nodes and we employ standard interpolation for those data:

Outline how you would numerically integrate in that case. Give the linear line element matrix for a 1-D problem with vx given constant data.

Sketch the boundary conditions for the ideal fluid flow around a cylinder given in Fig. 11.95 for an approach based on the use of a) the velocity potential, b) the stream function.

Solve the fin example of Fig. 11.26 by hand a) with insulated edges, b) with edge convection having the same convection coefficient.

For the solid conducting bar of Fig. 11.2 assume a length L = 8 cm, a width 2W = 4 cm, and a thickness of 2H = 1 cm. The material has a thermal conductivity of k = 3W/cm C, is surrounded by a fluid at a temperature of = 20 C, and has a surface convection coefficient of h = 0.1 W/cm 2 C. Employ two equal length conduction elements and associated face, line, or point convection elements to obtain a solution for the temperature distribution and the convection heat loss. Use conduction elements that are: a) linear line elements, b) bilinear rectangular elements, c) trilinear brick elements.

How would you generalize the post-processing related to Eq. 11.46 to also locate the x position of the resultant force?

For the conduction mesh in Figs. 11.9 and 11 prepare a combined surface plot of the element and SCP flux for: a) x, b) y, c) RMS values.

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## Integral Transforms

### Bromwich Integral

We now develop an expression for the inverse Laplace transform L − 1 appearing in the equation

One approach lies in the Fourier transform, for which we know the inverse relation. There is a difficulty, however. Our Fourier transformable function had to satisfy the Dirichlet conditions. In particular, we required that in order for g ( ω ) to be a valid Fourier transform,

so that the infinite integral would be well defined. 13 Now we wish to treat functions F (t) that may diverge exponentially. To surmount this difficulty, we extract an exponential factor, e β t , from our (possibly) divergent F (t) and write

If F (t) diverges as e α t , we require β to be greater than α so that G (t) will be convergent. Now, with G ( t ) = 0 for t < 0 and otherwise suitably restricted so that it may be represented by a Fourier integral, as in Eq. (20.22) , we have

Inserting Eq. (20.201) into Eq. (20.200) , we have

We now make a change of variable to s = β + i u , causing the integral over v in Eq. (20.202) to assume the form of a Laplace transform:

The variable s is now complex, but must be restricted to ℜ e ( s ) ≥ β in order to guarantee convergence. Note that the Laplace transform has extended a function specified on the positive real axis onto the complex plane, ℜ e s ≥ β . 14

We now need to rewrite Eq. (20.202) using the variable s in place of u. The range − ∞ < u < ∞ corresponds to a contour in the complex plane of s, which is a vertical line from β − i ∞ to β + i ∞ we also need to substitute d u = d s / i . Making these changes, Eq. (20.202) becomes

Here is our inverse transform. The path has become an infinite vertical line in the complex plane. Note that the constant β was chosen so that f (s) would be nonsingular for s ≥ β . It can be shown that the nonsingularity of f (s) extends to complex s provided that ℜ e s ≥ β , so the integrand of Eq. (20.203) can have singularities only to the left of the integration path. See Fig. 20.20 . Figure 20.20 . Possible singularities of e s t f ( s ) .

The inverse transformation given by Eq. (20.203) is known as the Bromwich integral, although sometimes it is referred to as the Fourier-Mellin theorem or Fourier-Mellin integral. This integral may now be evaluated by the regular methods of contour integration ( Chapter 11 ). If t > 0 and f (s) is analytic except for isolated singularities (and no branch points), and is also small at large | s | , the contour may be closed by an infinite semicircle in the left half-plane that does not contribute to the integral. Then by the residue theorem ( Section 11.8 ),

It is worth mentioning that in many cases of interest f (s) may become large in the left half-plane or have branch points, and evaluation of the Bromwich integral may then present significant challenges.

Possibly this means of evaluation with ℜ e s ranging through negative values seems paradoxical in view of our previous requirement that ℜ e s ≥ β . The paradox disappears when we recall that the requirement ℜ e s ≥ β was imposed to guarantee convergence of the Laplace transform integral that defined f (s). Once f (s) is obtained, we may then proceed to exploit its properties as an analytic function in the complex plane wherever we choose.

Perhaps a pair of examples may clarify the evaluation of Eq. (20.203) .

### Inversion via Calculus of Residues

If f ( s ) = a / ( s 2 − a 2 ) , then the integrand for the Bromwich integral will be

### Multiregion Inversion

If f ( s ) = ( 1 − e − a s ) / s , the Bromwich integral then has integrand

Considering first t > a , we may close the contour for the Bromwich integral in the left half-plane without changing its value. Our integrand is an entire function (analytic everywhere in the finite s -plane note that the s in the denominator cancels when the numerator is expanded in a Maclaurin series). Since no singularities are enclosed, we conclude that for t > a , F ( t ) = 0 .

For t in the range 0 < t < a , a different situation is encountered. Expanding the integrand into the two terms Figure 20.21 . Contours for Example 20.10.2 .

Finally, for t < 0 , the entire integrand becomes small in the right half-plane, the contour (for the entire integrand) surrounds no singularities, and the integral is zero. Summarizing these three cases, Figure 20.22 . Finite-length step function u ( t ) − u ( t − a ) .

Two general comments may be in order. First, these two examples hardly begin to show the usefulness and power of the Bromwich integral. It is always available for inverting a complicated transform when the tables prove inadequate.

Second, this derivation is not presented as a rigorous one. Rather, it is given more as a plausibility argument, although it can be made rigorous. The determination of the inverse transform is somewhat similar to the solution of a differential equation. It makes little difference how you get the inverse transform. Guess at it if you want. It can always be checked by verifying that

Two alternate derivations of the Bromwich integral are the subjects of Exercises 20.10.1 and (20.10.2) .

Derive the Bromwich integral from Cauchy's integral formula.

Hint. Apply the inverse transform L − 1 to

Derive the Laplace transformation convolution theorem by use of the Bromwich integral.

by a partial fraction expansion.

Repeat, using the Bromwich integral.

by using a partial fraction expansion.

Repeat using the convolution theorem.

Repeat using the Bromwich integral.

Use the Bromwich integral to find the function whose transform is f ( s ) = s − 1 / 2 . Note that f ( s ) has a branch point at s = 0 . The negative x-axis may be taken as a cut line. See Fig. 20.23 . Figure 20.23 . Contour for Exercise 20.10.6 .

Hint. A portion of the path needed to close the contour will yield nonzero contributions to the contour integral. These will need to be taken into account to get the proper value for the Bromwich integral.

Hint. Convert your Bromwich integral into an integral representation of J 0 ( t ) . Figure 20.24 shows a possible contour. Figure 20.24 . A possible contour for the inversion of J0(t).

Evaluate the inverse Laplace transform

Expansion in a series and term-by-term inversion.

Direct evaluation of the Bromwich integral.

Change of variable in the Bromwich integral: s = ( a / 2 ) ( z + z − 1 ) .

Evaluate the Bromwich integral for

Heaviside expansion theorem. If the transform f ( s ) may be written as a ratio

Using the Bromwich integral, invert f ( s ) = s − 2 e − k s . Express F ( t ) = L − 1 f ( s ) in terms of the (shifted) unit step function u ( t − k ) .

ANS. F ( t ) = ( t − k ) u ( t − k ) .

You have a Laplace transform:

Partial fractions and use of tables,

ANS. F ( t ) = e − b t − e − a t a − b , a ≠ b .

## 6.E: The Laplace Transform (Exercises) - Mathematics

elementary differential equations 6e, rainville, bedient [pdf]

### Elementary Differential Equations (6E) by Earl D. Rainville, Phillip E. Bedient, Richard E. Bedient

MathSchoolinternational contain 5000+ of Mathematics Free PDF Books and Physics Free PDF Books . Which cover almost all topics for students of Mathematics, Physics and Engineering. Here is extisive list of Differential Equations ebooks . We hope students and teachers like these textbooks , notes and solution manuals. Elementary Differential Equations (6E) written by Earl D. Rainville, Phillip E. Bedient, Richard E. Bedient .
The new geometric material appears mainly in Sections 1.4 and 11.8. In the former we introduce students to the idea of a family of curves as solutions to a differential equation, and in the latter, the concept of the phase plane of a system of equations is presented. We have moved the treatment of systems of equations to an earlier place in the text.
Of all areas of mathematics covered in a typical undergraduate curriculum, the field of differential equations is arguably the most affected by the computer. Numerous packages have been produced which are either designed specifically for differential equations, or have subpackages designed for that material. We have made the rather arbitrary decision to present our computer examples using the Maple package. We could have equally well chosen any of the other Computer Algebra Systems such as Mathematica, Matlab, or Derive. Further, there are a number of numerical graphing packages that are more efficient for producing geometric results. Among the more commonly available are MacMath and Phaser.
Each computer supplement contains an example from the corresponding chapter, worked out with the aid of Maple. Subsequently a set of computer exercises is presented which can be solved using whichever package is available to the student. It is our hope that these brief introductions will encourage users to look beyond the text for further computer explorations.
(Earl D. Rainville)

Book Detail :-
Title: Elementary Differential Equations
Edition: 6th
Author(s): Earl D. Rainville, Phillip E. Bedient, Richard E. Bedient
Publisher: Macmillan
Series:
Year: 1981
Pages: 544
Type: PDF
Language: English
ISBN: 9780023977701,0-02-397770-1
Country: US

About Author :- The author Earl David Rainville (1907 – 1966) was the professor taught in the Department of Engineering Mathematics at the University of Michigan, where he began as an assistant professor in 1941. He studied at the University of Colorado, receiving his B.A. there in 1930 before going on to graduate studies at Michigan, where he received his Ph.D. in 1939 under the supervision of Ruel Churchill. He died on April 29, 1966.

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Book Contents :-
Elementary Differential Equations (6E) written by Earl D. Rainville, Phillip E. Bedient, Richard E. Bedient cover the following topics. '
1. Definitions, Elimination of Arbitrary Constants
2. Equations of Order One
3. Elementary Applications
4. Additional Topics on Equations of Order One
5. Linear Differential Equations
6. Linear Equations with Constant Coefficients
7. Nonhomogeneous Equations: Undetermined Coefficients
8. Variation of Parameters
9. Inverse Differential Operators
10. Applications
11. The Laplace Transform
12. Inverse Transforms
13. Linear Systems of Equations
14. Electric Circuits and Networks
15. The Existence and Uniqueness of Solutions
16. Nonlinear Equations
17. Power Series Solutions
18. Solutions Near Regular Singular Points
19. Equations of Hypergeometric Type
20. Numerical Methods
21. Partial Differential Equations
22. Orthogonal Sets
23. Fourier Series
24. Boundary Value Problems
25. Additional Properties of the Laplace Transform
26. Partial Differential Equations Transform Methods
Index

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The preparatory reading for this session is Chapter 2 of Karris which

• defines the Laplace transformation
• gives the most useful properties of the Laplace transform with proofs
• presents the Laplace transforms of the elementary signals discussed in the last session
• presents the transforms of the more common system response types that are found in basic signals and systems.

There is some intellectual benefit to being aware of the properties of the Laplace transformation and their proofs but being a pragmatic breed, we engineers typically prefer to make use of quick references of these properties and transforms, relying on Mathematics only when facing a problem not before encountered.

In our practice, we want to encourage you to use of the properties and transform tables to solve problems so I will present only the properties and not the proofs.

By Solution Manual (SM) we mean Comprehensive solutions to end of each chapter’s problems which also called as Instructor Solution Manual (ISM). It’s totally different with student solution manuals (SSM). Solutions Manuals are official instructor chapter wise files and are available in digital word/pdf formats.

By Test Bank (TB) we mean Multiple choice questions, Essay questions, True/False questions, … with corresponding correct answers. Test Banks files are official instructor chapter wise items and are available in digital word/pdf formats.

Test Banks play an important role in your midterm and final exams. In other word only those professors who are assigned a course in universities have access to these test bank files for making exams tests ready for students without spending much time.

Tutorial 4.3 First Shifting Theorem 143Solution. We have = L−1 s+5 L−1 5 + s s2 + 4s + 20 s2 + 4s + 4 + 16 = L−1 (s + 2) + 3 (s + 2)2 + 16 = e−2tL−1 s + 3 (by the ﬁrst shifting theorem) s2 + 16 = e−2tL−1 s3 + s2 + 16 s2 + 16 = e−2t L−1 s + L−1 3 s2 + 16 s2 + 16 = e−2t 3 . cos 4t + sin 4t 4ExercisesExercise 4.3.1. Find the Laplace transforms of (i) t2e−3t, (ii) 2e−t cos t . 2Solution.Exercise 4.3.2. Find the Laplace transform of e−3t[2 cos 5t − 3 sin 5t]. [GTU, June 2014]Solution.

144 Chapter 4 Laplace Transforms and ApplicationsExercise 4.3.3. Evaluate L−1 10 . [GTU, Jan. 2013] (s−2)4Solution.Exercise 4.3.4. Find L−1 3 . [GTU, Dec. 2009] s2+6s+18Solution.Exercise 4.3.5. Find L−1 s+2 . s2−4s+13Solution.

Tutorial 4.3 First Shifting Theorem 145 [GTU, Dec. 2013]Exercise 4.3.6. Find L−1 s . s4+a4Solution.Viva QuestionsQuestion 4.3.7. What is ﬁrst shifting theorem ? When to use it ?Question 4.3.8. What are the Laplace transforms of the following functions ?(i) te−4t (ii) e−2t sin 5t (iii) et cosh 2t (iv) e2t sin 2t cos 2tQuestion 4.3.9. What are the inverse Laplace transforms of the following functions ?(i) 1 (ii) s−1 (iii) 3 (iv) 4 (s−2)2 (s−1)2+4 (s+4)2+9 s2−2s−3Answers4.3.1 (i) 2 , (ii) 8(s+1) 4.3.2 2s−9 4.3.3 5e2tt3 (s+3)3 4s2+8s+5 s2+6s+34 34.3.4 e−3t sin 3t 4.3.5 e2t cos 3t + 4 sin 3t 4.3.6 1 sin at sinh at 3 2a24.3.8 (i) 1 (ii) 5 (iii) s (iv) 2 (s+4)2 s2+4s+29 s2−2s+3 s2−4s+204.3.9 (i) te2t (ii) et cos 2t (iii) e−4t sin 3t (iv) 2et sinh 2t

146 Chapter 4 Laplace Transforms and Applications4.4 Tutorial : Method of Partial FractionsThe method of partial fractions can be described as follows:Case-1. Distinct Linear Factors φ(s) φ(a) φ(b) φ(c) (s − a)(s − b)(s − c) = (s − a)(a − b)(a − c) + (b − a)(s − b)(b − c) + (c − a)(c − b)(s − c) (This is called short cut method.)Case-2. Repeated Linear Factors φ(s) = A1 + A2 + A3 + A4 + A5 + A6 (s − a)(s − b)2(s − c)3 s − a s − b (s − b)2 s − c (s − c)2 (s − c)3Case-3. Distinct Quadratic Factors φ(s) = A1s + B1 + A2s + B2 (s2 + as + b)(s2 + cs + d) s2 + as + b s2 + cs + dCase-4. Repeated Quadratic Factors (s2 + as φ(s) + cs + d)2 = A1s + B1 + A2s + B2 + A3s + B3 + b)(s2 s2 + as + b s2 + cs + d (s2 + cs + d)2Solved ExamplesExample 4.4.1. Find L−1 5s2+3s−16 . [GTU, Dec. 2011] (s−1)(s−2)(s+3)Solution. Since the denominator has distinct linear factors, by short cut method of partialfractions, we obtain 5s2 + 3s − 16 (s − 1)(s − 2)(s + 3) 5(1)2 + 3(1) − 16 5(2)2 + 3(2) − 16 5(−3)2 + 3(−3) − 16 = (s − 1)(1 − 2)(1 + 3) + (2 − 1)(s − 2)(2 + 3) + (−3 − 1)(−3 − 2)(s + 3) −8 10 20 = −4(s − 1) + 5(s − 2) + 20(s + 3) 221 = + +. s−1 s−2 s+3ThusL−1 5s2 + 3s − 16 = 2L−1 1 +2L−1 1 +L−1 1 = 2et+2e2t+e−3t. (s − 1)(s − 2)(s + 3) s−1 s−2 s+3

Tutorial 4.4 Method of Partial Fractions 147Example 4.4.2. Find L−1 5s+3 . [GTU- May 2012, June 2014] (s2+2s+5)(s−1)Solution. By the partial fraction method, 5s + 3 As + B C (s2 + 2s + 5)(s − 1) = s2 + 2s + 5 + s − 1 ⇒ 5s + 3 = (As + B)(s − 1) + C(s2 + 2s + 5) ⇒ 5s + 3 = (A + C)s2 + (−A + B + 2C)s + (−B + 5C).Equating the corresponding coeﬃcients of s2, s and 1 on both sides, we obtain A + C = 0, −A + B + 2C = 5, −B + 5C = 3.Solving, we obtain A = −1, B = 2, C = 1.Thus 5s + 3 −s + 2 1 (s2 + 2s + 5)(s − 1) = s2 + 2s + 5 + s − 1 = − + s−2 + 4 + s 1 1 s2 2s + 1 − = − s−2 1 + (s + 1)2 + 4 s − 1 = − (s + 1) − 3 + s 1 1 (s + 1)2 + 4 − = − s+1 − 3 1 +. (s + 1)2 + 4 (s + 1)2 + 4 s − 1Hence,L−1 5s + 3 = −L−1 s+1 − 3L−1 1 + L−1 1 (s2 + 2s + 5)(s − 1) (s + 1)2 + 4 (s + 1)2 + 4 s−1Exercises = −e−tL−1 s − 3e−t 1 + et s2 + 4 s2 + 4 = −e−t cos 2t + 3 e−t sin 2t + et. 2Exercise 4.4.1. Find L−1 √1 √ . [GTU, Dec. 2010] (s+ 2)(s+ 3)Solution.

148 Chapter 4 Laplace Transforms and ApplicationsExercise 4.4.2. Find L−1 s+10 . [GTU, March 2010] s2−s−2Solution.Exercise 4.4.3. Find L−1 s . [GTU, Jan. 2015] (s+1)(s−1)2Solution.

Tutorial 4.4 Method of Partial Fractions 149Exercise 4.4.4. Find L−1 s+2 . (s+1)(s2+4)Solution.Answers e− √√ 4.4.2 4e2t − 3e−t 1 √ 1√ 2t − e− 3t 4 (et e−t 2tet)4.4.1 3− 2 4.4.3 − +4.4.4 1 [e−t − cos 2t + 3 sin 2t] 5

150 Chapter 4 Laplace Transforms and Applications4.5 Tutorial : Diﬀerentiation of Laplace Transform Formula for Multiplication by Power of tIf f (t) is any function whose Laplace transform exists, then L[tnf (t)] = (−1)n dn L[f (t)] (n = 0, 1, 2, . . .). dsn Inverse Laplace Transform by DiﬀerentiationIf F (s) is any function whose inverse Laplace transform exists, then L−1[F (s)] = − 1 L−1[F (s)]. tThis formula is used when F (s) involves any of the functions log, tan−1, cot−1.Solved ExamplesExample 4.5.1. Find the Laplace transform of t2 cosh πt. [GTU, Jan. 2015]Solution. It is known that L[tnf (t)] = (−1)n dn L[f (t)] (n = 0, 1, 2, . . .). dsnFor n = 2 and f (t) = cosh πt, we obtainL[t2 cosh πt] = (−1)2 d2 L[cosh πt] ds2 d2 s = ds2 s2 − π2 d (s2 − π2)(1) − s(2s) = ds (s2 − π2)2 d −s2 − π2 = ds (s2 − π2)2 = −d s2 + π2 ds (s2 − π2)2 =− (s2 − π2)2(2s) − (s2 + π2) · 2(s2 − π2)2s (s2 − π2)4 = − (s2 − π2) <2s(s2 − π2) − 4s(s2 + π2)>(s2 − π2)4 =− (s2 − π2)(2s3 − 2π2s − 4s3 − 4π2s) (s2 + π2)4 = − −2s3 − 6π2s (s2 − π2)3 2s(s2 + 3π2) = (s2 − π2)3 .

Tutorial 4.5 Diﬀerentiation of Laplace Transform 151Example 4.5.2. Find L[te−t cos t].Solution. First we ﬁnd L[t cos t]. It is known that L[tnf (t)] = (−1)n dn L[f (t)] (n = 0, 1, 2, . . .). dsnFor n = 1 and f (t) = cos t, we obtain L[t cos t] = (−1)1 d L[cos t] ds ds = − ds s2 + 1 = − (s2 + 1) · 1 − s · 2s (s2 + 1)2 −s2 + 1 = − (s2 + 1)2 s2 − 1 =. (s2 + 1)2Now by First shifting theorem L[e−tt cos t] = (s + 1)2 − 1 = s2 + 2s = s(s + 2) . [(s + 1)2 + 1]2 (s2 + 2s + 2)2 (s2 + 2s + 2)2Example 5.5.3. Evaluate L−1 ln ω2 . [GTU-March 2010, June 2014] 1 + s2 = log(s2 + 1) − 2 log s. s2 + 1Solution. Observe that s2 1 F (s) = log 1 + = log s2Therefore, 2s 2 L−1[F (s)] = 2 cos t − 2t. F (s) = s2 + 1 − s ⇒Now, L−1[F (s)] = − 1 L−1[F (s)] ⇒ 1 = 2 − cos t). t log 1 + (t s2 tExample 5.5.4. Evaluate L−1 cot−1 s . ωSolution. Here F (s) = cot−1 s ⇒ 11 ω ω F (s) = − · = − s2 + ω2 . 1+ s2 ω ω2Therefore, L−1[F (s)] = −L−1 ω = − sin ωt. s2 + ω2Now, L−1[F (s)] = − 1 L−1[F (s)] ⇒ L−1 cot−1 s sin ωt =. t ωt

152 Chapter 4 Laplace Transforms and Applications [GTU, May 2012] ExercisesExercise 4.5.1. Find L[t2 sinh at].Solution.Exercise 4.5.2. Find L[eatt sin at]. [GTU, Dec. 2013]Solution.

Tutorial 4.5 Diﬀerentiation of Laplace Transform 153 [GTU, May 2011]Exercise 5.5.3. Obtain L−1 log 1 . sSolution.Exercise 5.5.4. Evaluate L−1 log s + a . [GTU-Dec. 2010, June 2013] s+bSolution.Exercise 5.5.5. Evaluate L−1 tan−1 2 .Solution. s

154 Chapter 4 Laplace Transforms and ApplicationsExercise 5.5.6. Find L−1 log 1 − π2 . s2Solution.Viva QuestionsQuestion 5.5.7. What is the formula for L[tnf (t)] ?Question 5.5.8. What is the Laplace transform of t sin t ?Question 5.5.9. What is the formula for L−1[F (s)] ?Answers4.5.1 2a(3s2+a2) 4.5.2 2ωs 5.5.3 1 5.5.4 e−bt−e−at (s2−a2)3 (s2+ω2)2 t t5.5.5 sin 2t 5.5.6 2 (1 − cosh πt) 5.5.8 2s t t (s2+1)2

Tutorial 4.6 Integration of Laplace Transform 1554.6 Tutorial : Integration of Laplace Transform Formula for Division by tIf f (t) is any function whose Laplace transform exists, then f (t) ∞ L = L[f (t)]ds. ts Solved ExamplesExample 4.6.1. Find L sin ωt . [GTU, May 2011] tSolution. It is known that f (t) ∞ L = L[f (t)]ds. tsFor f (t) = sin ωt, we obtain sin ωt ∞ L = L[sin ωt]ds t s ∞ω = ds s s2 + ω2 = tan−1 s ∞ ωs = tan−1 ∞ − tan−1 s ω = π − tan−1 s 2ω = cot−1 s . ωExercisesExercise 4.6.1. Find L 1−cos t . tSolution.

156 Chapter 4 Laplace Transforms and ApplicationsExercise 4.6.2. Find L sin 2t . [GTU- Jan. 2013, June 2014] tSolution.Exercise 4.6.3. Find L sin3 t . tSolution.

Tutorial 4.6 Integration of Laplace Transform 157Additional ExercisesExercise 4.6.4. Evaluate (i) L sin 3t sin t , (ii) L e−3t sin 2t . t tViva Questions f (t)Question 4.6.5. What is the formula for L t ?Question 4.6.6. Find the Laplace transform of et−e−t . tAnswers √ s2+1 s s+3 s 2 24.6.1 log 4.6.2 cot−1 4.6.3 (i) 3 cot−1 s − 1 cot−1 s , (ii) cot−1 4 3 3 14.6.4 log s2+16 4 4.6.6 log s+1 s2+4 s−1

158 Chapter 4 Laplace Transforms and Applications4.7 Tutorial : Unit Step Function and Second Shifting TheoremUnit Step FunctionThe unit step function or Heaviside function u(t − a) is deﬁned as u(t − a) = 0, t < a (a ≥ 0) 1, t ≥ a.The Laplace transform of the unit step function is given by e−as L[u(t − a)] = . s Second Shifting Theorem for Laplace TransformIf f (t) is any function whose Laplace transform exists, then L[f (t)u(t − a)] = e−asL[f (t + a)]. Second Shifting Theorem for Inverse Laplace TransformIf L−1[F (s)] = f (t), then L−1[e−asF (s)] = f (t − a)u(t − a).Solved ExamplesExample 4.7.1. Find L[(t − 1)2u(t − 1)].Solution. By the Second Shifting Theorem, L[f (t)u(t − a)] = e−asL[f (t + a)].For f (t) = (t − 1)2 and a = 1, we obtainL[(t − 1)2u(t − 1)] = e−sL[t2] = e−s 2! 2e−s s3 =. s3Example 4.7.2. Find L[e−3tu(t − 2)]. [GTU, May 2012]Solution. By the Second Shifting Theorem, L[f (t)u(t − a)] = e−asL[f (t + a)].For f (t) = e−3t and a = 2, we obtainL[e−3tu(t − 2)] = e−2sL[e−3(t+2)] = e−2sL[e−3te−6] = e−2se−6L[e−3t] = e−2s−6 . s+3Example 4.7.3. Find L−1 3e−πs . s2+9

Tutorial 4.7 Unit Step Function and Second Shifting Theorem 159Solution. The given function is of the form e−asF (s) with a = π and F (s) = 3 . Therefore, s2+9 f (t) = L−1[F (s)] = sin 3t.By the second shifting theorem, we have L−1[e−asF (s)] = f (t − a)u(t − a) ⇒ L−1 3e−πs = sin 3(t − π)u(t − π) s2 + 9 = sin(3t − 3π)u(t − π) = − sin 3t u(t − π). ExercisesExercise 4.7.1. Find L[(t − 1)u(t − 1)].Solution.Exercise 4.7.2. Find the Laplace transform of et−2u(t − 2).Solution.

160 Chapter 4 Laplace Transforms and ApplicationsExercise 4.7.3. Find L[e−2tu(t − 3)].Solution.Exercise 4.7.4. Find the Laplace transform of t2u(t − 2). [GTU, Jan. 2015]Solution.Exercise 4.7.5. Find L−1 se−2s . [GTU- Dec. 2009, Dec. 2010] s2+π2Solution.

Tutorial 4.7 Unit Step Function and Second Shifting Theorem 161Exercise 4.7.6. Find L−1 e−3s . s3Solution.Exercise 4.7.7. Find L−1 1+e− sπ . 2 s2+4Solution.Additional ExercisesExercise 4.7.8. Find the Laplace transforms of (i) cos tu(t − π), (ii) L[sin t u t − π ]. 2Exercise 4.7.9. Evaluate (i) L−1 e−3s , (ii) L−1 se−2s . (s−1)3 s2+16

162 Chapter 4 Laplace Transforms and Applications Viva QuestionsQuestion 4.7.10. Deﬁne unit step function.Question 4.7.11. What is the Laplace transform of unit step function ?Question 4.7.12. What is the second shifting theorem ?Question 4.7.13. Evaluate (i) L[tu(t − 4)], (ii) L[sin tu(t − π)], (iii) L[(t − 2)2u(t − 2)].Answers4.7.1 e−s 4.7.2 e−2s 4.7.3 e−3(s+2) 4.7.4 e−2s(2+4s+4s2) s2 s−1 s+2 s34.7.5 u(t − 2) cos πt 4.7.6 1 (t − 3)2u(t − 3) 4.7.7 1 sin 2t 1−u t − π 2 2 24.7.8 (i) − se−πs , (ii) se−πs/2 4.7.9 (i) e−(t−3) (t − 3)2u(t − 3) , (ii) u(t − 2) cos 4(t − 2) s2+1 s2+1 24.7.13 (i) e−4s(1+4s) , (ii) e−πs , (iii) 2e−2s s2 s2+1 s3

Tutorial 4.8 Laplace Transforms of Periodic Functions 1634.8 Tutorial : Laplace Transforms of Periodic FunctionsFormula for Periodic FunctionsLet f (t) be a periodic function with period p, then 1 p L[f (t)] = 1 − e−ps e−stf (t)dt. 0Solved ExamplesExample 4.8.1. Find the Laplace transform of Half wave rectiﬁer f (t) = sin ωt, 0 < t < π , f 2π = f (t). [GTU, March 2010] ω t+ 0, π < t < 2π ω ω ωSolution. The given function is a periodic function with period p = 2π . ω 1 2π ∴ L[f (t)] = 1 − e− 2πs ω e−stf (t)dt ω 0 1 π = 1 − e− 2πs ω e−st sin ωtdt ω 0 = 1 e−st π ω 1 − e− 2πs s2 + ω2 ω (−s sin ωt − ω cos ωt) 0 = 1 e− πs (−s sin π − ω cos π) + ω ω 1 − e− 2πs s2 + ω2 s2 + ω2 ω = 1 ωe− sπ +ω ω 1 − e− 2πs s2 + ω2 ω 1 ω 1 + e− sπ ω = 1 − e− πs 1 + e− πs s2 + ω2 ω ω = ω . (s2 + ω2) 1 − e− sπ ωExercisesExercise 4.8.1. Find the Laplace transform of Full wave rectiﬁer f (t) = | sin ωt| (t ≥ 0). [GTU, May 2011]Solution.

164 Chapter 4 Laplace Transforms and ApplicationsAdditional ExercisesExercise 4.8.2. Find the Laplace transform of f (t) = k, 0 < t < a and f (t + 2a) = f (t). −k, a < t < 2aViva QuestionsQuestion 4.8.3. What is the Laplace transform of a periodic function with period p ? Answers4.8.1 ω coth πs 4.8.2 k tanh as s2+ω2 2ω s 2

Tutorial 4.9 Convolution 1654.9 Tutorial : Convolution Convolution ProductLet f (t) and g(t) be two functions whose Laplace transforms exist. Then the convolutionproduct of f and g is denoted by f ∗ g and is deﬁned as t (f ∗ g)(t) = f (u)g(t − u)du. 0 Convolution TheoremSuppose that L−1[F (s)] = f (t) and L−1[G(s)] = g(t). Then L−1 F (s)G(s) = f ∗ g.Solved ExamplesExample 4.9.1. Find the convolution of t and et. [GTU, Dec. 2010]Solution. The convolution of f and g is given by t (f ∗ g)(t) = f (u)g(t − u)du. 0For f (t) = t and g(t) = et, we obtain t ∗ et = t uet−udu 0 t = et ue−udu 0 t = et u(−e−u) − (1)(e−u) 0 = −et ue−u + e−u t 0 = −et (te−t + e−t) − (0 + e0) = −et(te−t + e−t − 1) = et(1 − te−t − e−t) = et − t − 1.Example 4.9.2. Using convolution theorem, obtain the value of L−1 1 . s(s2+4) [GTU- Dec. 2009, Jan. 2013]

166 Chapter 4 Laplace Transforms and ApplicationsSolution. The function 1 can be written as s(s2+4) 1 11 = · . s(s2 + 4) s2 + 4 sLet 11Then F (s) = s2 + 4 and G(s) = .By the convolution theorem, s 1 f (t) = sin 2t and g(t) = 1. 2 L−1 1 = L−1 11 = L−1 F (s)G(s) = (f ∗ g)(t). (4.9.1) s(s2 + 4) · s2 + 4 sNow, t (f ∗ g)(t) = f (u)g(t − u)du 0 1 t = 20 sin 2u · 1du = 1 − cos 2u t 2 20 = −1 cos 2t − 1 4 = 1 1 − cos 2t . 4Putting this value in (4.9.1), we obtain L−1 1 = 1 1 − cos 2t . s(s2 + 4) 4Example 4.9.3. Find the inverse Laplace transform of s . [GTU, Jan. 2015] (s2+1)2Solution. Observe that s = 1 · sLet .Then (s2 + 1)2 s2 + 1 s2 + 1 F (s) = 1s . and G(s) = s2 + 1 s2 + 1 f (t) = sin t and g(t) = cos t.By the convolution theorem, L−1 s = L−1 1 · s = L−1[F (s)G(s)] = (f ∗ g)(t). (s2 + 1)2 s2 + 1 s2 + 1

Tutorial 4.9 Convolution 167Now, t (f ∗ g)(t) = f (u)g(t − u)du 0 t = sin u cos(t − u)du 0 1t = sin t + sin(2u − t) du 20 1 cos(2u − t) t = u sin t − 2 20 = 1 t sin t − cos t − 0 − cos(−t) 22 2 = 1 t sin t − cos t + cos t 2 22 1 = t sin t. 2 Exercises [GTU, Dec. 2009]Exercise 4.9.1. Find 1 ∗ 1.Solution.Exercise 4.9.2. Find the convolution of et and e−t.Solution.

168 Chapter 4 Laplace Transforms and ApplicationsExercise 4.9.3. State convolution theorem and use it to obtain the Laplace inverse of 1 . s2(s2+1)Solution.Exercise 4.9.4. Find L−1 1 . [GTU, Dec. 2013] s(s+a)3Solution.

Tutorial 4.9 Convolution 169Exercise 4.9.5. Using convolution theorem, obtain L−1 1 . (s2+a2)2 [GTU-Dec. 2010, May 2011]Solution.Exercise 4.9.6. Using convolution theorem, obtain L−1 s . (s2+π2)2Solution.

170 Chapter 4 Laplace Transforms and ApplicationsAdditional ExercisesExercise 4.9.7. Find 1 ∗ cos ωt.Exercise 4.9.8. State convolution theorem and use it to evaluate the inverse Laplace of as2(s2+a2) . [GTU, March 2010]Viva QuestionsQuestion 4.9.9. Deﬁne convolution f ∗ g and ﬁnd t ∗ t.Question 4.9.10. What is convolution theorem ?Answers4.9.1 t 4.9.2 sinh t 4.9.3 t − sin t 4.9.4 1 2 − t2 + 2t + 2 e−at 2 a3 a a2 a34.9.5 1 sin at − t cos at 4.9.6 t sin πt 4.9.7 sin ωt 4.9.8 at−sin at 4.9.9 t3 2a2 a 2π ω a2 6

Tutorial 4.10 Laplace Transform of Integral 1714.10 Tutorial : Laplace Transform of IntegralIf L[f (t)] = F (s), then t t F (s) t F (s) (ii) L f (u)dudu = and so on. s2(i) L f (u)du = , 0 0 0sSolved Examples tExample 4.10.1. Find the Laplace transform of e−u cos udu. 0 [GTU- May 2012, June 2013]Solution. It is known that t F (s) L f (u)du = . 0sLet f (u) = e−u cos u. Then F (s) = s+1 . Hence, (s+1)2+1 t = 1 s+1 1 s+1 = . e−u cos udu 0 s (s + 1)2 + 1 s s2 + 2s + 2Exercises tt [GTU, June 2013]Exercise 4.10.1. Find L sin aududu . 00Solution.Answers4.10.1 a s2(s2+a2)

172 Chapter 4 Laplace Transforms and Applications4.11 Tutorial : Solution of ODEs by Laplace Transforms Laplace Transforms of DerivativesIf L[y(t)] = Y (s), then(1) L[y ] = sY − y(0)(2) L[y ] = s2Y − sy(0) − y (0)(3) L[y ] = s3Y − s2y(0) − sy (0) − y (0).and so on. Solved ExamplesExample 4.11.1. Solve by Laplace transform y + 6y = 1 y(0) = 2, y (0) = 0. [GTU, Jan. 2013]Solution. Applying Laplace transform to the given diﬀerential equation, we obtain L[y ] + 6L[y] = L ⇒ [s2Y − sy(0) − y (0)] + 6Y = 1 s ⇒ s2Y − 2s + 6Y = 1 s ⇒ (s2 + 6)Y = 1 + 2s s 1 2s ⇒ Y = s(s2 + 6) + s2 + 6 .Applying inverse Laplace transform, we obtain L−1[Y ] = L−1 1 +2 s . (4.11.1) s(s2 + 6) s2 + 6 (4.11.2)Observe that 1 11 s(s2 + 6) = s2 + 6 · . sLet F (s) = 11Then and G(s) = . s2 + 6 s f (t) = √16 √ and g(t) = 1. sin 6tBy the convolution theorem, L−1 1 = L−1 1 · 1 = L−1[F (s)G(s)] = (f ∗ g)(t). s(s2 + 6) s2 + 6 s

Tutorial 4.11 Solution of ODEs by Laplace Transforms 173Now,(f ∗ g)(t) = t f (u)g(t − u)du = √1 t √ √1 √ t √ sin 6u −cos√ 6u 1 − cos 6t · 1du = 1 . = 60 666 0 0Thus by (4.11.2), we get L−1 1 = − 1 cos √ s(s2 + 6) 6u. 6Putting this value in (4.11.1), we get y = 1 − 1 √ + √ = 1 + 11 √ cos 6t 2 cos 6t cos 6t. 66 66Example 4.11.2. Using the method of Laplace transform, solve the IVP y + 2y + y = e−t y(0) = −1, y (0) = 1.Solution. Applying Laplace transform to the given diﬀerential equation, we obtain L[y ] + 2L[y ] + L[y] = L[e−t] ⇒ [s2Y − sy(0) − y (0)] + 2[sY − y(0)] + Y = 1 s+1 ⇒ s2Y + s − 1 + 2sY + 2 + Y = 1 s+1 ⇒ (s2 + 2s + 1)Y + s + 1 = 1 s+1 ⇒ (s + 1)2Y = 1 − (s + 1) s+1 ⇒ Y= 1 − 1 . (s + 1)3 s + 1Applying inverse Laplace transform, we obtain L−1[Y ] = L−1 1 − L−1 1 = e−t t2 − e−t = e−t t2 (s + 1)3 s+1 2! −1 . 2Example 4.11.3. Using Laplace transform solve the initial value problem y + y = sin 2t y(0) = 2, y (0) = 1. [GTU- Dec. 2010, Jan. 2015]Solution. Applying Laplace transform to the given diﬀerential equation, we obtain L[y ] + L[y] = L[sin 2t] ⇒ s2Y − sy(0) − y (0) + Y = 2 s2 + 4 ⇒ s2Y − 2s − 1 + Y = 2 s2 + 4 ⇒ (s2 + 1)Y = 2 + 2s + 1 s2 + 4

174 Chapter 4 Laplace Transforms and Applications ⇒ Y= 2 2s 1 ++ (s2 + 4)(s2 + 1) s2 + 1 s2 + 1 ⇒ 2 (s2 + 4) − (s2 + 1) + 2s + 1 Y= 3 (s2 + 4)(s2 + 1) s2 + 1 s2 + 1 ⇒ 2 1 −2 1 2s 1 Y= ++ 3 s2 + 1 3 s2 + 4 s2 + 1 s2 + 1 ⇒ Y = 51 − 12 s 3 s2 + 1 3 s2 + 4 + 2 s2 + 1 .Applying inverse Laplace transform, we obtain L−1[Y ] = 5 L−1 1 − 1 L−1 2 +2 s 3 s2 + 1 3 s2 + 4 s2 + 1 ⇒ y = 5 sin t − 1 sin 2t + 2 cos t. 33ExercisesExercise 4.11.1. Using Laplace transform solve the diﬀerential equation y + y − 6y = 1 y(0) = 0, y (0) = 1.Solution.

Tutorial 4.11 Solution of ODEs by Laplace Transforms 175Exercise 4.11.2. Solve the diﬀerential equation by Laplace transform method y + 4y + 3y = e−t, y(0) = y (0) = 1. [GTU, Dec. 2013]Solution.

176 Chapter 4 Laplace Transforms and ApplicationsExercise 4.11.3. Using Laplace transform solve d2x + dx + 5x = e−t sin t, where x(0) = 0, x (0) = 1. [GTU, May 2012] 2 dt2 dtSolution.Viva QuestionsQuestion 4.11.4. Give the formulas for L[y ] and L[y ].Question 4.11.5. How to solve the diﬀerential equation using Laplace transform ?Question 4.11.6. Using Laplace transform, solve the IVP y + 4y = 0 y(0) = 1, y (0) = 6.Answers4.11.1 − 1 + 3 e2t − 2 e−3t 4.11.2 1 [7e−t + 2te−t − 3e−3t] 4.11.3 1 e−t(sin t + sin 2t) 6 10 15 4 34.11.6 cos 2t + 3 sin 2t

5ChapterPartial Diﬀerential Equations andApplicationsPartial diﬀerential equations arise in the problems involving the phenomena of two or morevariables such as sound, heat transfer, electrostatics, electrodynamics, ﬂuid ﬂow, elasticityetc. The functions in such type of phenomena usually depend on time t and one or severalspace variables. The range of applications of partial diﬀerential equations is wide as compareto ordinary diﬀerential equations because most of engineering problems involve multivariatefunctions.DeﬁnitionA diﬀerential equation which involves partial derivatives with respect to two or more indepen-dent variables is called a partial diﬀerential equation (PDE). For examples, ∂2u ∂2u (Laplace Equation)(1) ∂x2 + ∂y2 = 0(2) ∂2u = c2 ∂2u (Wave Equation) ∂t2 ∂x2 (Heat Equation)(3) ∂u = c2 ∂2u ∂t ∂x2(4) ∂u + ∂u − ∂2u = u ∂t ∂x ∂x∂tNotationsThe following are the standard notations for the ﬁrst and second ordered partial derivatives ofa variable z considered as a function of variables x and y: ∂z ∂z ∂2z ∂2z ∂2z p= , q= , r = ∂x2 , s= , t = ∂y2 . ∂x ∂y ∂x∂yIn the case of other variables, a partial derivative is denoted by a suﬃx. For instant, ut =∂u∂t . The dependence of a variable is understood from the nature of the problem as well as thecontext. 177

178 Chapter 5 Partial Diﬀerential Equations and Applications5.1 Tutorial : Formation of PDEsA partial diﬀerential equation can be formed in two ways:(1) By eliminating arbitrary constants from the given relation(2) By eliminating arbitrary functions from the given relation.Solved ExamplesExample 5.1.1. Form the partial diﬀerential from z = (x − 2)2 + (y − 3)2. [GTU- June 2013, June 2014]Solution. The given relation is z = (x − 2)2 + (y − 3)2.Diﬀerentiating z partially with respect to x and y, we get ∂z = 2(x − 2) and ∂z = 2(y − 3). ∂x ∂ySquaring and adding these equations, we obtain ∂z 2 ∂z 2 = 4z. + = 4 (x − 2)2 + (y − 3)2 ∂x ∂yHence, the required partial diﬀerential equation is 4z = p2 + q2.Example 5.1.2. Form the partial diﬀerential equation from the relation 2z = (ax + y)2 + b.Solution. The given relation is 2z = (ax + y)2 + b. (5.1.1)Diﬀerentiating equation (5.1.1) partially w. r. t. x, we obtain ∂z = 2(ax + y) · a ⇒ ∂z = (ax + y) · a. (5.1.2) 2 ∂x ∂xDiﬀerentiating equation (5.1.1) partially w. r. t. y, we obtain ∂z ⇒ ∂z (5.1.3) 2 = 2(ax + y) = (ax + y). ∂y ∂yBy equations (5.1.2) and (5.1.3), we obtain ∂z ∂z (5.1.4) =a . ∂x ∂yBy equation (5.1.3), 1 ∂z − y . a= x ∂y

Tutorial 5.1 Formation of PDEs 179Hence, by equation (5.1.4), ∂z 1 ∂z − y ∂z = . ∂x x ∂y ∂y ⇒ p = 1 − y)q (q x ⇒ px = q2 − qy ⇒ px + qy = q2.Example 5.1.3. Form a partial diﬀerential equation by eliminating the arbitrary function fromz = f (x2 − y2). [GTU, Dec. 2013]Solution. The given relation is z = f (x2 − y2). (5.1.5)Diﬀerentiating equation (5.1.5) partially w. r. t. x, we obtain ∂z = f (x2 − y2)(2x) ⇒ f (x2 − y2) = 1 ∂z (5.1.6) , ∂x 2x ∂xwhere f denotes diﬀerential of f w. r. t. (x2 − y2). Diﬀerentiating equation (5.1.5) partially w. r. t. y, we obtain ∂z = f (x2 − y2)(−2y) ⇒ f (x2 − y2) = − 1 ∂z (5.1.7) . ∂y 2y ∂yBy equations (5.1.6) and (5.1.7), we obtain1 ∂z = − 1 ∂z ⇒ ∂z = −x ∂z ⇒ yp = −xq ⇒ yp + xq = 0. y2x ∂x 2y ∂y ∂x ∂yExample 5.1.4. Form a partial diﬀerential equation by eliminating the arbitrary function fromf (x2 − y2, xyz) = 0. [GTU, Jan. 2015]Solution. Let u = x2 − y2 and v = xyz. Then the given relation becomes f (u, v) = 0 (5.1.8)Diﬀerentiating equation (5.1.8) partially w. r. t. x, we obtain ∂f =0 ∂x ⇒ ∂f ∂u ∂f ∂v + =0 ∂u ∂x ∂v ∂x ⇒ fu · (2x) + fv · y ∂z =0 x +z ∂x ⇒ fu · (2x) = −fv · (xyp + yz) ⇒ fu = − xyp + yz . (5.1.9) fv 2xDiﬀerentiating equation (5.1.8) partially w. r. t. y, we obtain ∂f =0 ∂y

180 Chapter 5 Partial Diﬀerential Equations and Applications ⇒ ∂f ∂u ∂f ∂v + =0 ∂u ∂y ∂v ∂y ⇒ fu · (−2y) + fv · x ∂z =0 y +z ∂y ⇒ fu · (2y) = fv · (xyq + xz) ⇒ fu = xyq + xz (5.1.10) . fv 2y (5.1.11) (5.1.12)By equations (5.1.9) and (5.1.10), we obtain (5.1.13) (5.1.14) −xyp − yz xyq + xz = 2x 2y ⇒ −xy2p − y2z = x2yq + x2z ⇒ xy2p + x2yq = −(x2 + y2)z.Example 5.1.5. Form the partial diﬀerential equation from the relation z = f (x + it) + g(x − it).Solution. The given relation is z = f (x + it) + g(x − it). So we obtain∂z = f (x + it) + g (x − it)∂x∂z = if (x + it) − ig (x − it)∂y∂2z =f (x + it) + g (x − it)∂x2∂2z = i2f (x + it) + i2g (x − it) = i2 f (x + it) + g (x − it)∂y2From (5.1.13) and (5.1.14), we obtain∂2z = i2 ∂2z ⇒ ∂2z = −∂2z ⇒ ∂2z ∂2z∂y2 ∂x2 ∂y2 ∂x2 + = 0. ∂x2 ∂y2 ExercisesExercise 5.1.1. Form the partial diﬀerential equation from z = (x − a)2 + (y − b)2.Solution.

Tutorial 5.1 Formation of PDEs 181Exercise 5.1.2. Eliminate the arbitrary constants from the relation 2z = x2 + y2 a2 b2 .Solution.Exercise 5.1.3. Form a partial diﬀerential equation from Z = f x . [GTU, Jan. 2013] ySolution.Exercise 5.1.4. Form a partial diﬀerential equation from f (x + y + z, x2 + y2 + z2) = 0. [GTU- Jan. 2013, June 2104]Solution.

182 Chapter 5 Partial Diﬀerential Equations and ApplicationsExercise 5.1.5. Form the partial diﬀerential equation from the relation xyz = φ(x + y + z). [GTU, Dec. 2013]Solution.

Tutorial 5.1 Formation of PDEs 183Exercise 5.1.6. Form a partial diﬀerential equation from z = f (x + 6y) + g(x − 6y).Solution.Additional ExercisesExercise 5.1.7. Form the partial diﬀerential equation from z = y2 + 2f 1 + log y . xExercise 5.1.8. Eliminate the function f from the relation f (xy + z2, x + y + z) = 0. [GTU, June 2013]Viva QuestionsQuestion 5.1.9. Deﬁne partial diﬀerential equation.Question 5.1.10. How to form a partial diﬀerential equation ?Question 5.1.11. Form the partial diﬀerential equation from the relation z = (x + a)(x + b).Question 5.1.12. Form the partial diﬀerential equation from the relation z = y + f (x + ln y).Answers5.1.1 z = p2+q2 5.1.2 2z = px + qy 5.1.3 xp + yq = 0 5.1.4 (y − z)p − (x − z)q = x − y 4 y(xp+z) x(yq+z) ∂2z ∂2z5.1.5 1+p = 1+q 5.1.6 ∂y2 = 36 ∂x2 5.1.7 px2 + qy = 2y25.1.8 (y + 2pz)(1 + q) − (x + 2pz)(1 + p) = 0 5.1.11 z = pq 5.1.12 p = qy

184 Chapter 5 Partial Diﬀerential Equations and Applications5.2 Tutorial : Solution of PDEs by Direct IntegrationThis method is applicable to the PDE which contains only one partial derivative. Solved ExamplesExample 5.2.1. Solve ∂2z = sin x sin y given that ∂z = −2 sin y when x = 0 and z = 0 when ∂x∂y ∂y πy is an odd multiple of 2 . [GTU, June 2013]Solution. The given PDE is ∂2z = sin x sin y. ∂x∂yIntegrating w. r. t. x, keeping y constant, we get ∂z = − cos x sin y + f (y). (5.2.1) ∂yNow ∂z = −2 sin y when x = 0. Thus ∂y −2 sin y = − cos 0 sin y + f (y) ⇒ f (y) = − sin y.Therefore, by equation (5.2.1), we get ∂z = − cos x sin y − sin y. ∂yIntegrating w. r. t. y, keeping x constant, we get z = cos x cos y + cos y + g(x). (5.2.2)Now z = 0 when y = (2n + 1) π . Thus 2 ππ ⇒ g(x) = 0. 0 = cos x cos(2n + 1) + cos(2n + 1) + g(x) 22Hence, by equation (5.2.2), we get z = cos x cos y + cos y.Example 5.2.2. Solve ∂2z = z. [GTU- Jan. 2013, Jan. 2105] ∂x2Solution. The given PDE is ∂2z ⇒ ∂2z − z = 0. ∂x2 = z ∂x2This equation looks like linear ODE, Hence, it will be solved using the method described earlier.The symbolic form is (D2 − 1)z = 0, where D = ∂ . ∂x

Tutorial 5.2 Solution of PDE by Direct Integration 185Therefore the auxiliary equation is m2 − 1 = 0 ⇒ (m − 1)(m + 1) = 0 ⇒ m = 1, −1.Hence, the complete solution is given by z = f1(y)ex + f2(y)e−x,where f1(y) and f2(y) are arbitrary functions of y, since y was regarded as a constant thisfunction act as arbitrary constants.ExercisesExercise 5.2.1. Solve ∂2u = e−t cos x given that u = 0 when t = 0 and ∂u = 0 when x = 0. ∂x∂t ∂tSolution.Exercise 5.2.2. Solve ∂2z +z =0 given that when x = 0, z = ey and ∂z = 1. ∂x2 ∂xSolution.

186 Chapter 5 Partial Diﬀerential Equations and ApplicationsAdditional ExercisesExercise 5.2.3. Solve ∂2z = 4z given that when x = 0, ∂z = 2 sin y and ∂z = 0. ∂x2 ∂x ∂yViva QuestionsQuestion 5.2.4. Which type of PDEs can be solved by direct integration method ?Question 5.2.5. Solve ∂2z = sin(x + 2y). ∂x∂y Answers5.2.1 u(x, t) = (1 − e−t) sin x 5.2.2 z = ey cos x + sin x 5.2.3 sin y sinh 2x5.2.5 z = − sin(x + 2y) + F (y) + g(x)

Tutorial 5.3 First Order Linear PDEs 1875.3 Tutorial : First Order Linear PDEs DeﬁnitionA partial diﬀerential equation of the form P p + Qq = R,where P, Q, R are functions of x, y, z or constants is called the Lagrange’s linear equation ofﬁrst order.Lagrange’s MethodStep-1. Form the auxiliary equations dx = dy = dz . P Q RStep-2. The auxiliary equations can be solved using the grouping method or the multiplier method (described below). Many times a combination of these two is required. Solutions of auxiliary equations are denoted by u(x, y) = c1, v(x, y) = c2.Step-3. Write the complete solution as F (u, v) = 0.• Grouping Method: In this method, we compare any two fractions which makes the inte- gration possible. In other words, to compare ﬁrst two fractions, the remaining third variable must be absent from them or it is possible to eliminate it by appropriate operations.• Multipliers Method: In this method, we ﬁnd two sets of multipliers l, m, n and l , m , n (either constants or functions of x, y, z) such that lP + mQ + nQ = 0 and l P + m Q + n R = 0or the selection makes the integration possible.Remark. Whenever it is required to apply multipliers method, always give the ﬁrst priorityto x, y, z as multipliers. If they don’t work, then search for the others. Solved ExamplesExample 5.3.1. Solve x(y2 − z2)p + y(z2 − x2)q = z(x2 − y2).Solution. The given equation is of the form P p + Qq = R, where P = x(y2 − z2), Q = y(z2 − x2), R = z(x2 − y2).Therefore, the Lagrange’s auxiliary equations are dx dy dz x(y2 − z2) = y(z2 − x2) = z(x2 − y2) .Using x, y, z as multipliers, we get each ratio as dx dy dz xdx + ydy + zdzx(y2 − z2) = y(z2 − x2) = z(x2 − y2) = x2y2 − x2z2 + y2z2 − x2y2 + x2z2 − y2z2 .

188 Chapter 5 Partial Diﬀerential Equations and ApplicationsFrom the last two fractions, we get dz xdx + ydy + zdz ⇒ xdx + ydy + zdz = 0. z(x2 − y2) = 0Integrating, we get x2 + y2 + z2 = c1 ⇒ x2 + y2 + z2 = c1. 2222Thus u(x, y) = x2 + y2 + z2 = c1.Now, for the second solution we proceed as follows. Using 1 , 1 , 1 as multipliers for each of x y zratios in the Lagrange’s auxiliary equations respectively we get dx dy dz 1 dx + 1 dy + 1 dz . === x y z x(y2 − z2) y(z2 − x2) z(x2 − y2) y2 − z2 + z2 − x2 + x2 − y2Comparing last two fractions, we get 1 dx + 1 dy + 1 dz = dz ⇒ 1 11 x y z z(x2 − dx + dy + dz = 0. 0 y2) xyzIntegrating, we obtain ln x + ln y + ln z = ln c2 ⇒ ln(xyz) = ln c2 ⇒ xyz = c2.Thus v(x, y) = xyz = c2.Hence, the complete solution is given by F (u, v) = 0 ⇒ F x2 + y2 + z2, xyz = 0.Example 5.3.2. Solve x2p + y2q = (x + y)z.Solution. The given equation is of the form P p + Qq = R, where P = x2, Q = y2, R = (x + y)z.Therefore, the Lagrange’s auxiliary equations are dx dy dz == . x2 y2 (x + y)zFrom the ﬁrst two fractions, we get dx dyIntegrating, we get =. x2 y2 −1 = −1 + c1 ⇒ 1 − 1 = c1 ⇒ x−y x y y x xy = c1.

Tutorial 5.3 First Order Linear PDEs 189Thus x−y u(x, y) = xy = c1.For the second solution using 1 , 1 ,− 1 as multipliers, we get x y z dx = dy = dz = 1 dx + 1 dy − 1 dz . x2 y2 (x + y)z x y z x+y − (x + y)Comparing the ﬁrst and last fractions, we get dx = 1 dx + 1 dy − 1 dz ⇒ 1 dx + 1 − 1 = 0. x2 x y z dy dz 0 xyzIntegrating, we get xy xy ln x + ln y − ln z = ln c2 ⇒ ln z = ln c2 ⇒ z = c2.Thus xy v(x, y) = z = c2.Hence, the complete solution is given by F (u, v) = 0 ⇒ F x − y xy = 0. , xy zExample 5.3.3. Solve pz − qz = z2 + (x + y)2. [GTU, Dec. 2013]Solution. The given equation is of the form P p + Qq = R, where P = z, Q = −z, R = z2 + (x + y)2.Therefore, the Lagrange’s auxiliary equations are dx dy dz == . z −z z2 + (x + y)2From the ﬁrst two fractions, we get dx dy ⇒ dx + dy = 0. z = −zIntegrating, we get x + y = c1. (5.3.1)Thus u(x, y) = x + y = c1.Again, using the last two fractions − dy = z2 + dz z (x + y)2 .

190 Chapter 5 Partial Diﬀerential Equations and ApplicationsFrom (5.3.1), we have x + y = c1. So we obtain dy dz zdz 2zdz − = z2 + c21 ⇒ −dy = z2 + c21 ⇒ −2dy = z2 + c21 . zIntegrating, we get−2y = log(z2 + c12) − c2 ⇒ log(z2 + c12) + 2y = c2 ⇒ log z2 + (x + y)2 + 2y = c2.Thus v(x, y) = log z2 + (x + y)2 + 2y = c2.Hence, the complete solution is given by F x + y, log z2 + (x + y)2 + 2y = 0. Exercises [GTU, June 2013]Exercise 5.3.1. Solve x(y − z)p + y(z − x)q = z(x − y).Solution.

Tutorial 5.3 First Order Linear PDEs 191Exercise 5.3.2. Solve x(y2 + z)p − y(x2 + z)q = z(x2 − y2).Solution.

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