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Mathematics LibreTexts

8.2E: The Inverse Laplace Transform (Exercises)

  • Page ID
    18506
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    [exer:8.2.1] Use the table of Laplace transforms to find the inverse Laplace transform.

    1. \( {3\over(s-7)^4}\) &
    2. \( {2s-4\over s^2-4s+13}\) &
    3. \( {1\over s^2+4s+20}\)
    4. \( {2\over s^2+9}\) &
    5. \( {s^2-1\over(s^2+1)^2}\) &
    6. \( {1\over(s-2)^2-4}\)
    7. \( {12s-24\over(s^2-4s+85)^2}\) &
    8. \( {2\over(s-3)^2-9}\) &
    9. \( {s^2-4s+3\over(s^2-4s+5)^2}\)

    [exer:8.2.2] Use Theorem [thmtype:8.2.1} and the table of Laplace transforms to find the inverse Laplace transform.

    1. \( {2s+3\over(s-7)^4}\) &
    2. \( {s^2-1\over(s-2)^6}\) &
    3. \( {s+5\over s^2+6s+18}\)
    4. \( {2s+1\over s^2+9}\) &
    5. \( {s\over s^2+2s+1}\) &
    6. \( {s+1\over s^2-9}\)
    7. \( {s^3+2s^2-s-3\over(s+1)^4}\) &
    8. \( {2s+3\over(s-1)^2+4}\) &
    9. \( {1\over s}-{s\over s^2+1}\)
    10. \( {3s+4\over s^2-1}\) &
    11. \( {3\over s-1}+{4s+1\over s^2+9}\) &
    12. \( {3\over(s+2)^2}-{2s+6\over s^2+4}\)

    [exer:8.2.3] Use Heaviside’s method to find the inverse Laplace transform.

    1. \( {3-(s+1)(s-2)\over(s+1)(s+2)(s-2)}\)&
    2. \( {7+(s+4)(18-3s)\over(s-3)(s-1)(s+4)}\)
    3. \( {2+(s-2)(3-2s)\over(s-2)(s+2)(s-3)}\) &
    4. \( {3-(s-1)(s+1)\over(s+4)(s-2)(s-1)}\)
    5. \( {3+(s-2)(10-2s-s^2)\over(s-2)(s+2)(s-1)(s+3)}\) &
    6. \( {3+(s-3)(2s^2+s-21)\over(s-3)(s-1)(s+4)(s-2)}\)

    [exer:8.2.4] Find the inverse Laplace transform.

    1. \( {2+3s\over(s^2+1)(s+2)(s+1)}\)&
    2. \( {3s^2+2s+1\over(s^2+1)(s^2+2s+2)}\)
    3. \( {3s+2\over(s-2)(s^2+2s+5)}\) &
    4. \( {3s^2+2s+1\over(s-1)^2(s+2)(s+3)}\)
    5. \( {2s^2+s+3\over(s-1)^2(s+2)^2}\) &
    6. \( {3s+2\over(s^2+1)(s-1)^2}\)

    [exer:8.2.5] Use the method of Example [example:8.2.9} to find the inverse Laplace transform.

    1. \( {3s+2\over(s^2+4)(s^2+9)}\) &
    2. \( {-4s+1\over(s^2+1)(s^2+16)}\) &
    3. \( {5s+3\over(s^2+1)(s^2+4)}\)
    4. \( {-s+1\over(4s^2+1)(s^2+1)}\) &
    5. \( {17s-34\over(s^2+16)(16s^2+1)}\) &
    6. \( {2s-1\over(4s^2+1)(9s^2+1)}\)

    [exer:8.2.6] Find the inverse Laplace transform.

    1. \( {17 s-15\over(s^2-2s+5)(s^2+2s+10)}\) &
    2. \( {8s+56\over(s^2-6s+13)(s^2+2s+5)}\)
    3. \( {s+9\over(s^2+4s+5)(s^2-4s+13)}\) &
    4. \( {3s-2\over(s^2-4s+5)(s^2-6s+13)}\)
    5. \( {3s-1\over(s^2-2s+2)(s^2+2s+5)}\) &
    6. \( {20s+40\over(4s^2-4s+5)(4s^2+4s+5)}\)

    [exer:8.2.7] Find the inverse Laplace transform.

    1. \( {1\over s(s^2+1)}\) &
    2. \( {1\over(s-1)(s^2-2s+17)}\)
    3. \( {3s+2\over(s-2)(s^2+2s+10)}\) &
    4. \( {34-17s\over(2s-1)(s^2-2s+5)}\)
    5. \( {s+2\over(s-3)(s^2+2s+5)}\) &
    6. \( {2s-2\over(s-2)(s^2+2s+10)}\)

    [exer:8.2.8] Find the inverse Laplace transform.

    1. \( {2s+1\over(s^2+1)(s-1)(s-3)}\) &
    2. \( {s+2\over(s^2+2s+2)(s^2-1)}\)
    3. \( {2s-1\over(s^2-2s+2)(s+1)(s-2)}\) &
    4. \( {s-6\over(s^2-1)(s^2+4)}\)
    5. \( {2s-3\over s(s-2)(s^2-2s+5)}\) &
    6. \( {5s-15\over(s^2-4s+13)(s-2)(s-1)}\)\

    [exer:8.2.9] Given that \(f(t)\leftrightarrow F(s)\), find the inverse Laplace transform of \(F(as-b)\), where \(a>0\).

    [exer:8.2.10]

    If \(s_1\), \(s_2\), …, \(s_n\) are distinct and \(P\) is a polynomial of degree less than \(n\), then

    \[{P(s)\over(s-s_1)(s-s_2)\cdots(s-s_n)}= {A_1\over s-s_1}+{A_2\over s-s_2}+\cdots+{A_n\over s-s_n}.\]

    Multiply through by \(s-s_i\) to show that \(A_i\) can be obtained by ignoring the factor \(s-s_i\) on the left and setting \(s=s_i\) elsewhere.

    Suppose \(P\) and \(Q_1\) are polynomials such that \(\mbox{degree}(P)\le\mbox{degree}(Q_1)\) and \(Q_1(s_1)\ne0\). Show that the coefficient of \(1/(s-s_1)\) in the partial fraction expansion of

    \[F(s)={P(s)\over(s-s_1)Q_1(s)}\]

    is \(P(s_1)/Q_1(s_1)\).

    Explain how the results of (a) and (b) are related.