COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
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This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 11  FOURIER SERIES and the LAPLACE TRANSFORM

a = A = b = B = c = d = s = S = t = T = u = U = v = V = w = W = x = X = y = Y = z = Z = 
                                        0; 
Clear[a,a1,a2,a3,A,A1,A2,b,b1,b2,B1,B2,B,c,d,EqA,EqB,f,f1,f2,f12,F,F1,F2,F12,F3,g,G,h,H,i,int,j,k,L,Ly,L1,L2,Lf,LDf,ℒf,ℒDf,n,PQ,PR,r,res1,res2,s,solset,s1,s2,s3,S,S9,t,table,τ,T,θ,u,U,U1,U2,U7,v,v1,v2,v3,v4,val,V,w,w1,w2,W,x,X,y,y0,y1,Y,Yi,z,z1,z2,Z]; 
Needs["Graphics`Colors`"];

Section 11.1	Fourier Series

Section 11.2	The Dirichlet Problem for the Unit Disk

Section 11.3	Vibrations in Mechanical Systems

Section 11.4	The Fourier Transform

Section 11.5	The Laplace Transform

Section 11.6	Laplace Transforms of Derivatives and Integrals

Section 11.7	Shifting Theorems and the Step Function

Section 11.8	Multiplication and Division by  t

Section 11.9  Inverting the Laplace Transform

So far, most of the applications involving the Laplace transform involves a transform (or part of a transform) that is expressed by 

	Y (s) = ((P (s))/(Q (s)))  

where  P  and  Q  are polynomials that have no common factors.  The inverse of  Y (s)  is found by using its partial fraction representation and referring to Table 11.2 of Laplace Transforms.  We now show how the theory of complex variables can be used to systematically find the partial fraction representation.  The first result is an extension of Lemma 8.1 to  n  linear factors.  The proof is left for the reader.

Theorem 11.19 (Nonrepeated Linear Factors), Page 534. Let P (s) be a polynomial of degree at most n - 1. If Q (s) has degree n, and has distinct complex roots  (Subscript[a, 1]), (Subscript[a, 2]), ... (Subscript[a, n]),  then  

	Y (s) = ((P (s))/(Q (s)))= \(\(P \((s)\)\)\/\(\(\((s - a\_1)\) \((s - a\_2)\) ... \) \((s - a\_n)\)\) = ∑\+\(k = 1\)\%n Res[Y, a\_k]\/\((s - a\_k)\)\).

Theorem 11.20 (A Repeated Linear Factors), Page 534. If P(s) and Q(s) are polynomials of degree μ and ν, respectively and μ < ν + n  and  Q (a) ≠ 0,  then  

	Y (s) = ((P (s))/((((s  -  a))^n) Q (s)))\(\(\(\ \)\(\(=\)\(\ \)\(∑\+\(k = 1\)\%n A\_k\/\((s\  - \ a)\)\^k\  + \ R \((s)\)\)\)\)\),  

where  R (s)  is the sum of all partial fractions that do not involve factors of the form  (((s - a))^j).  Furthermore, the coefficients  (Subscript[A, k])  can be computed with the formula  \(A\_k = \(\(1\/\(\((n - 
                      k)\)!\)\) \(lim\+\(s\  -> \ a\)\ \ d\^\(n - k\)\/\(d\ s\^\(n - k\)\)\ \(P \((s)\)\)\/\(Q \((s)\)\)\ \ for\ \ k\) = 1\), 
      2,  ... , n\).

Proof of Theorem 11.20, see text Page 534.

Example 11.24,  Page 535.  Find the inverse Laplace transform of  Y[s] = (((s^3)  -  4  s  +   1)/(s (((s  -  1))^3))).

Solution 11.24.

Enter Y[s] and find its partial fraction expansion.

Remove[a, b, s, t, 
        Y]; 
Clear[PQ, PR]; 
Y[
          s_]  =  (((s^3)  -  4  s  +  1)/(s (((s - 1))^3))); 
Print["Y[s] = ", Y[s]]; 
Print["Y[s] = ", Apart[Y[s]]];

Traditionally. The partial fraction expansion can be obtained with the residue calculus. 
\(Y[
          s]\  = \(\(\ \)\(a\_3\)\)\/\((s\  - \ 1)\)\^3\  + \(\(\ \)\(a\_2\)\)\/\((s\  - \ 1)\)\^2 + \ 
          a\_1\/\(s\  - \ 1\)\  + \(\(\ \)\(b\_1\)\)\/s\)  
At  s = 1, we compute  (Subscript[a, 1]), (Subscript[a, 2]),  and  (Subscript[a, 3])  as follows.

PQ  =  (((s^3)  -  4  s  +  1)/s); 
(Subscript[a, 3])  =  Limit[PQ, s -> 1]; 
(Subscript[a, 2])  =  Limit[(∂_s) PQ, s -> 1]; 
(Subscript[a, 1])  = 
        (1/(2!)) Limit[(∂_(s, s))PQ, s -> 1]; 
\(\(Print[\*""\!\(a\_3\) = \!\(lim\+\(s -> 1\)\) "", 
        PQ, " = ", a\_3];\)\ \)
\(\(Print[\*""\!\(a\_2\) = \!\(lim\+\(s -> 1\)\) "", 
        Together[∂\_s\ PQ], " = ", a\_2];\)\ \)
\(\(Print[\*""\!\(a\_1\) = \!\(lim\+\(s -> 1\)\) "", 
        Together[∂\_\(s, s\)PQ], " = ", 
        a\_1];\)\ \)

At  s = 0, we compute  (Subscript[b, 1])  as follows.

PR  =  (((s^3)  -  4  s  +  1)/(((s - 1))^3)); 
(Subscript[b, 1])  =  Limit[PR, s -> 0]; 
\(\(Print[\*""\!\(b\_1\) = \!\(lim\+\(s -> 0\)\)"", 
        PR, " = ", b\_1];\)\ \)

Using the residue calculus, we construct the partial fraction expansion.

Y[
          s_]  = (( (Subscript[a, 3]))/(((s  -  1))^3))  + (( (Subscript[a, 2]))/(((s  -  1))^2)) +  
          ((Subscript[a, 1])/(s  -  1))  + (( (Subscript[b, 1]))/s); 
Print["Y[s] = ", Y[s]];

Use a table of Laplace transforms to solve for  y[t]=(ℒ^(-1))(Y[s]).

Ys = 
        Table[0, {i, Length[Y[s]]}]; 
For[i = 1, i ≤ Length[Y[s]], i++, 
        (Subscript[Ys, [[i]]])  =  (Subscript[(Y[s]), [[i]]])]; 
yt = 
        InverseLaplaceTransform[Ys,  s,  t]; 
table = Transpose[{yt, Ys}]; 
y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  
          t]; 
\(\(Print[
        TableForm[table, 
          TableHeadings -> {None, {"f[t]", "ℒ(f[t])"}}]];\)\ \)
Print[""]; 
Print["Y[s] = ", Y[s]]; 
Print[""]; 
Print["y[t] = ", y[t]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  t]; 
Print["Y[s] = ", Apart[Y[s]]]; 
Print[""]; 
Print["y[t] = ", Expand[y[t]]];

Theorem 11.21 (Irreducible Quadratic Factors), Page 536. Let P(s) and Q(s) be polynomials with real coefficients such that the degree of P (s) is at most 1 larger than the degree of Q (s).   If  T (s)  does not have a factor of the form  (((s - a))^2) + (b^2),  then  

	\(Y \((s)\) = \(\(P \((s)\)\)\/\(Q \((s)\)\) = \(P \((s)\)\)\/\(\((\((s\  - \ a)\)\^2\  + \ b\^2)\) T \((s)\)\)\)\)\(\(\(=\)\(\(2  A \((s\  - \ a)\)\  - \ 2  B\ b\)\/\(\((s\  - \ a)\)\^2\  + \ b\^2\) + R \((s)\)\)\)\),  where  \(A\  + \ i\ B = \(P \((a\  + \ i\ b)\)\)\/\(Q' \((a\  + \ i\ b)\)\)\).

Proof of Theorem 11.21, see text Page 536.

Example 11.25,  Page 537.  Find the inverse Laplace transform of  Y[s] = ((5  s)/(((s^2)  +  4) ((s^2)  +  9))).

Solution 11.25.

Enter Y[s] and find its partial fraction expansion.

Remove[a, A, b, B, s, t, Y, 
        z]; 
Clear[i, res1, res2, Yi]; 
Y[
          s_]  =  ((5  s)/(((s^2) + 4) ((s^2) + 9))); 
Print["Y[s] = ", Y[s]]; 
Print["Y[s] = ", Apart[Y[s]]];

Traditionally. The partial fraction expansion can be obtained with the residue calculus. 
The simple poles in the upper half plane occur at  (Subscript[z, 1]) = 2  i and  (Subscript[z, 2]) = 3  i.  First determine their residues.

(Subscript[z, 1])  =  2  I; 
(Subscript[z, 2])  =  3  I; 
res1  =  Residue[Y[s], {s, (Subscript[z, 1])}]; 
res2  =  Residue[Y[s], {s, (Subscript[z, 2])}]; 
Print["Y[s] = ", Y[s]]; 
Print["Res[Y[s],", (Subscript[z, 1]), "] = ", res1]; 
Print["Res[Y[s],", (Subscript[z, 2]), "] = ", 
        res2];

Using the residue calculus, we construct the partial fraction expansion.

(Subscript[a, 1])  =  Re[(Subscript[z, 1])]; 
(Subscript[b, 1])  =  Im[(Subscript[z, 1])]; 
(Subscript[A, 1])  =  Re[res1]; 
(Subscript[B, 1])  =  Im[res1]; 
(Subscript[a, 2])  =  Re[(Subscript[z, 2])]; 
(Subscript[b, 2])  =  Im[(Subscript[z, 2])]; 
(Subscript[A, 2]) =  Re[res2]; 
(Subscript[B, 2])  =  Im[res2]; 
Y1[
          s_]  = (((2 (Subscript[A, 1]) (s  -  (Subscript[a, 1]))  -  2 (Subscript[B, 1]) (Subscript[b, 1])))/(((((s  -  (Subscript[a, 1])))^2)  +  ((Subscript[b, 1])^2)))); 
Y2[
          s_]  = (((2 (Subscript[A, 2]) (s  -  (Subscript[a, 2]))  -  2 (Subscript[B, 2]) (Subscript[b, 2])))/(((((s  -  (Subscript[a, 2])))^2)  +  ((Subscript[b, 2])^2)))); 
Y[s_] = Y1[s]  +  Y2[s]; 
\(\(Print[\*""\!\(Y\_1\)[s] = "", Y1[s]];\)\ \)
\(\(Print[\*""\!\(Y\_2\)[s] = "", 
        Y2[s]];\)\ \)
Print[""]; 
Print[" Y[s] = ", Y[s] ];

Use a table of Laplace transforms to solve for  y[t]=(ℒ^(-1))(Y[s]).

Ys = 
        Table[0, {i, Length[Y[s]]}]; 
For[i = 1, i ≤ Length[Y[s]], i++, 
        (Subscript[Ys, [[i]]])  =  (Subscript[(Y[s]), [[i]]])]; 
yt = 
        InverseLaplaceTransform[Ys,  s,  t]; 
table = Transpose[{yt, Ys}]; 
y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  
          t]; 
\(\(Print[
        TableForm[table, 
          TableHeadings -> {None, {"f[t]", "ℒ(f[t])"}}]];\)\ \)
Print[""]; 
Print["Y[s] = ", Y[s]]; 
Print[""]; 
Print["y[t] = ", y[t]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  t]; 
Print["Y[s] = ", Apart[Y[s]]]; 
Print[""]; 
Print["y[t] = ", y[t]];

Example 11.26,  Page 537.  Find the inverse Laplace transform of  \(Y[
          s] = \(s\^3\  + \ 3  s\^2\  - \ s\  + \ 1\)\/\(s \(\((s\  + \ 1)\)\^2\) \((s\^2\  + \ 1)\)\)\).

Solution 11.26.

Enter Y[s] and find its partial fraction expansion.

Remove[s, t, y, Y]; 
Clear[i, Yi]; 
Y[
          s_]  =  (((s^3) + 3  (s^2) - s + 1)/(s (((s + 1))^2) ((s^2) + 1))); 
Print["Y[s] = ", Y[s]]; 
Y[s_]  = Apart[Y[s]]; 
Print["Y[s] = ", Y[s]]; 
Y[s_]  = 
        Map[Simplify, ExpandAll[Y[s]]]; 
Print["Y[s] = ", Y[s]];

Use a table of Laplace transforms to solve for  y[t]=(ℒ^(-1))(Y[s]).

Ys = 
        Table[0, {i, Length[Y[s]]}]; 
For[i = 1, i ≤ Length[Y[s]], i++, 
        (Subscript[Ys, [[i]]])  =  (Subscript[(Y[s]), [[i]]])]; 
yt = 
        InverseLaplaceTransform[Ys,  s,  t]; 
table = Transpose[{yt, Ys}]; 
y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  
          t]; 
\(\(Print[
        TableForm[table, 
          TableHeadings -> {None, {"f[t]", "ℒ(f[t])"}}]];\)\ \)
Print[""]; 
Print["Y[s] = ", Y[s]]; 
Print[""]; 
Print["y[t] = ", y[t]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  t]; 
Print["Y[s] = ", Y[s]]; 
Print[""]; 
Print["y[t] = ", y[t]];

Example 11.27,  Page 538.	 Solve the initial value problem  

	\(y' \((t)\) = \ \ \ \(y \((t)\)\  - \ \ x \((t)\)\ \ \ \ \ \ with\ \ \ y \((0)\) = 
          1\), 
x' \((t)\) = \(5  y \((t)\) - 
            3  x \((t)\)\ \ \ \ \ \ \ \ and\ \ \ x \((0)\) = 2. \)\)

Solution 11.27.

Enter the initial conditions and the system of  D.E.s.  Then Laplace transform the D.E.s and solve for the transforms Y[s] any X[s] and the functions y[t] and x[t].

Remove[s, t, x, X, y, Y]; 
Clear[eqn1, eqn2, solset, x0, y0]; 
de1  =  y'[
            t]  ==    y[t]   -   x[
              t] ; 
de2  =  x'[t]  ==  
          5  y[t]  -  3  x[t] ; 
y0  =  1; 
x0  =  2; 
eqn1 = 
        ReplaceAll[
          de1, {x[t] -> X[s], 
            y[t] -> Y[s], y'[t] -> 
              s Y[s] - y0}]; 
eqn2 = 
        ReplaceAll[
          de2, {x[t] -> X[s], 
            y[t] -> Y[s], x'[t] -> 
              s X[s] - x0}]; 
Print[de1]; 
Print[de2]; 
Print["y[0] = ", y0]; 
Print["x[0] = ", x0]; 
Print[""]; 
Print[eqn1]; 
Print[eqn2]; 
Print[""]; 
solset  =  Solve[{eqn1, eqn2},  {X[s], Y[s]}]; 
Print[solset]; 
X[s_]  =  
        (Subscript[solset, [[1, 1, 2]]]); 
Y[s_]  =  
        (Subscript[solset, [[1, 2, 2]]]); 
y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  
          t]; 
y[t_]  = FullSimplify[y[t]]; 
x[t_]  =  
        InverseLaplaceTransform[X[s],  s,  
          t]; 
x[t_]  = FullSimplify[x[t]]; 
Print["Y[s] = ", Y[s]]; 
Print["X[s] = ", X[s]]; 
Print[" "]; 
Print["y[t] = ", y[t]]; 
Print["x[t] = ", x[t]];

Aside. We can substitute y[t] and x[t] into the D.E.'s and verify that they are the solution.

Print[" "]; 
Print["        y'[t] = ", y'[t]]; 
Print["  y[t] - x[t] = ", 
        Expand[y[t] - x[t]] ]; 
Print["  Does  y'[t] = y[t] - x[t] ? ", 
        ExpandAll[y'[t] == y[t] - x[t]]]; 
Print[" "]; 
Print["        x'[t] = ", x'[
          t]]; 
Print["        x'[t] = ", Expand[x'[t]]]; 
Print["5y[t] - 3x[t] = ", 
        Expand[5  y[t]  -  3  x[t]]]; 
Print["Does    x'[t] = 5y[t] - 3x[t] ? ", 
        ExpandAll[x'[t] == 
            5  y[t]  -  3  x[t]]];

Aside. We use Mathematica/s DSolve package to solve the system of D.E.'s and plot the solution.

Clear[eqn1, eqn2, solset, t, x, y]; 
eqn1  =  y'[t]  ==   y[t]  -  x[t]; 
eqn2  =  x'[t]  ==  
          5  y[t]  -  3  x[t]; 
Print[eqn1]; 
Print[eqn2]; 
\(\(solset\  = \ 
        DSolve[{eqn1, eqn2, \ x[0] == 2, \ 
            y[0] == 1}, \ 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {y[t], \ 
            x[t]}, \ t];\)\ \)
Print[solset]; 
y[t_]  =  
        FullSimplify[
          (Subscript[solset, [[1, 1, 2]]])]; 
x[t_]  =  
        FullSimplify[
          (Subscript[solset, [[1, 2, 2]]])]; 
ParametricPlot[{x[t], y[t]}, {t, 0, 5}, 
        PlotRange -> {{-0.1, 2}, {-0.1, 1}}, 
        PlotStyle -> Magenta, AxesLabel -> {"x", "y"}, 
        Ticks -> {Range[0, 2, 0.5], Range[0, 1, 0.25]}]; 
Print["y[t] = ", y[t]]; 
Print["x[t] = ", x[t]]; 
Print["y[0] = ", y0]; 
Print["x[0] = ", x0];

Theorem 11.22 (Inverse Laplace Transform), Page 539. Let Y (s) = ((P (s))/(Q (s))), where P(s) and Q(s) are polynomials of degree m and n, respectively, and  n > m.  The inverse Laplace transform of  Y (s)  is given by  

	y (t) = (ℒ^(-1))(Y (s)) = ∑ 
          Res[F (s) (e^(s t)), (Subscript[s, k])],  

where the sum is taken over all the residues of the complex function  F (s) (e^(s t)).

Proof of Theorem 11.22, see text Page 539.

Using the residue calculus, the English mathematician Oliver Heaviside discovered a method for inverting the Laplace transform.

Theorem 11.23 (Heaviside Expansion Theorem), Page 540. Let P (s) and Q (s) be polynomials of degree m and n, respectively, where n > m. If Q (s) has n distinct simple zeros at the points  (Subscript[s, 1]),(Subscript[s, 2]),⋯,(Subscript[s, n]),  then  ((P (s))/(Q (s)))  is the Laplace transform of the function  y (t)  given by

	y (t)=(ℒ^(-1))(((P (s))/(Q (s))))=\(∑\+\(k = 1\)\%n\(\( 
                    P \((s\_k)\)\)\/\(Q' \((s\_k)\)\)\) e\^\(s\_k\ t\)\).

Proof of Theorem 11.23, see text Page 540.

Example 11.28,  Page 541.  	Find the inverse Laplace transform of  \(Y[
          s] = \(4  s\  + \ 3\)\/\(\(\ \)\(s\^3\  + \ 2  s\^2\  + \ s\  + \ 2\)\)\).

Solution 11.28.

Enter Y[s] and use the Heaviside Expansion Theorem to find y[t].

Remove[s, t, y, Y]; 
Clear[P, Q, solset]; 
P[s_]  =  4  s  +  3; 
Q[s_]  =  (s^3)  +  2  (s^2)  +  s  +  2; 
Y[s_]  =  P[s]/Q[s]; 
solset  =  Solve[Q[s] == 0, s]; 
(Subscript[s, 1])  =  
        (Subscript[solset, [[1, 1, 2]]]); 
(Subscript[s, 2])  =  
        (Subscript[solset, [[2, 1, 2]]]); 
(Subscript[s, 3])  =  
        (Subscript[solset, [[3, 1, 2]]]); 
Print["P[s] = ", P[s]]; 
Print["Q[s] = ", Q[s]]; 
Print[""]; 
Print["Y[s] = ", Y[s]]; 
Print[""]; 
Print["Solve the equation Q[s] = 0"]; 
Print[Q[s]  == 0 ]; 
Print[Factor[Q[s]] == 0]; 
Print[solset]; 
\(\(Print[\*""\!\(s\_1\) = "", s\_1];\)\ \)
\(\(Print[\*""\!\(s\_2\) = "", s\_2];\)\ \)
\(\(Print[\*""\!\(s\_3\) = "", 
        s\_3];\)\ \)

Q[s] has simple zeros at the points  (Subscript[s, 1])  = -2,  (Subscript[s, 2])  = -i,  
      and  (Subscript[s, 3])  = i.  Construct the three terms  P[(Subscript[s, k])]/(Q')[
                (Subscript[s, k])] (e^((Subscript[s, k]) t))  for  k  =  1, 
      2, 3.

Print["P[", (Subscript[s, 1]), "]/Q'[", 
        (Subscript[s, 1]), "] = (", P[(Subscript[s, 1])], ")/(", Q'[(Subscript[s, 1])], ") = ", 
        P[(Subscript[s, 1])]/(Q')[(Subscript[s, 1])]]; 
Print["P[", (Subscript[s, 2]), "]/Q'[", (Subscript[s, 2]), "] = (", 
        P[(Subscript[s, 2])], ")/(", Q'[(Subscript[s, 2])], ") = ", 
        P[(Subscript[s, 2])]/(Q')[(Subscript[s, 2])]]; 
Print["P[", (Subscript[s, 3]), "]/Q'[", (Subscript[s, 3]), "] = (", 
        P[(Subscript[s, 3])], ")/(", Q'[(Subscript[s, 3])], ") = ", 
        P[(Subscript[s, 3])]/(Q')[(Subscript[s, 3])]]; 
Print[""]; 
Print["P[", (Subscript[s, 1]), "]/Q'[", 
        (Subscript[s, 1]), "] ", (E^((Subscript[s, 1]) t)), " = ", P[
                (Subscript[s, 1])]/(Q')[
                (Subscript[s, 1])] (E^((Subscript[s, 1]) t))]; 
Print[""]; 
Print["P[", (Subscript[s, 2]), "]/Q'[", 
        (Subscript[s, 2]), "] ", (E^((Subscript[s, 2]) t)), " = ", P[
                (Subscript[s, 2])]/(Q')[
                (Subscript[s, 2])] (E^((Subscript[s, 2]) t))]; 
Print["P[", (Subscript[s, 2]), "]/Q'[", 
        (Subscript[s, 2]), "] ", (E^((Subscript[s, 2]) t)), " = ", 
        ComplexExpand[P[(Subscript[s, 2])]/(Q')[
                  (Subscript[s, 2])] (E^((Subscript[s, 2]) t))]]; 
Print[""]; 
Print["P[", (Subscript[s, 3]), "]/Q'[", 
        (Subscript[s, 3]), "] ", (E^((Subscript[s, 3]) t)), " = ", P[
                (Subscript[s, 3])]/(Q')[
                (Subscript[s, 3])] (E^((Subscript[s, 3]) t))]; 
Print["P[", (Subscript[s, 3]), "]/Q'[", 
        (Subscript[s, 3]), "] ", (E^((Subscript[s, 3]) t)), " = ", 
        ComplexExpand[P[(Subscript[s, 3])]/(Q')[
                  (Subscript[s, 3])] (E^((Subscript[s, 3]) t))]];

Use the Heaviside Expansion Theorem to construct y[t].

y[
          t_]  = P[(Subscript[s, 1])]/(Q')[
                  (Subscript[s, 1])] (E^((Subscript[s, 1]) t)) + P[(Subscript[s, 2])]/(Q')[
                  (Subscript[s, 2])] (E^((Subscript[s, 2]) t)) + P[(Subscript[s, 3])]/(Q')[
                  (Subscript[s, 3])] (E^((Subscript[s, 3]) t)); 
Print["y[t] = P[", (Subscript[s, 1]), "]/Q'[", 
        (Subscript[s, 1]), "] ", (E^((Subscript[s, 1]) t)), " + ", "P[", 
        (Subscript[s, 2]), "]/Q'[", 
        (Subscript[s, 2]), "] ", (E^((Subscript[s, 2]) t)), " + ", "P[", 
        (Subscript[s, 3]), "]/Q'[", 
        (Subscript[s, 3]), "] ", (E^((Subscript[s, 3]) t))]; 
Print["y[t] = (", P[(Subscript[s, 1])]/(Q')[
                (Subscript[s, 1])] (E^((Subscript[s, 1]) t)), ") + (", P[
                (Subscript[s, 2])]/(Q')[
                (Subscript[s, 2])] (E^((Subscript[s, 2]) t)), ") + (", P[
                (Subscript[s, 3])]/(Q')[
                (Subscript[s, 3])] (E^((Subscript[s, 3]) t)), ")"]; 
Print["y[t] = (", 
        ComplexExpand[P[(Subscript[s, 1])]/(Q')[
                  (Subscript[s, 1])] (E^((Subscript[s, 1]) t))], ")"]; 
Print["     + (", 
        ComplexExpand[P[(Subscript[s, 2])]/(Q')[
                  (Subscript[s, 2])] (E^((Subscript[s, 2]) t))], ")"]; 
Print["     + (", 
        ComplexExpand[P[(Subscript[s, 3])]/(Q')[
                  (Subscript[s, 3])] (E^((Subscript[s, 3]) t))], ")"]; 
y[t_]  = ComplexExpand[y[t]]; 
Print[""]; 
Print["y[t] = ", y[t]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  t]; 
Print["F[s] = ", 
        Map[Expand, Apart[Y[s]]]]; 
Print[""]; 
Print["y[t] = ", y[t]];

Section 11.9  Exercises for Inverting the Laplace Transform
See textbook page 541.

Section 11.10	Convolution

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