8.2 The Inverse Laplace Transform

Definition of the Inverse Laplace Transform

In Section 8.1 we defined the Laplace transform of \(f\) by

We’ll also say that \(f\) is an inverse Laplace Transform of \(F\), and write

To solve differential equations with the Laplace transform, we must be able to obtain \(f\) from its transform \(F\). There’s a formula for doing this, but we can’t use it because it requires the theory of functions of a complex variable. Fortunately, we can use the table of Laplace transforms to find inverse transforms that we’ll need.

Solution (a) Setting \(b=1\) in the transform pair

shows that

Solution (b) Setting \(\omega=3\) in the transform pair

shows that

The next theorem enables us to find inverse transforms of linear combinations of transforms in the table. We omit the proof.

Solution From the table of Laplace transforms in Section 8.8,,

Theorem 8.2.1 with \(a=-5\) and \(\omega=\sqrt3\) yields

Solution Completing the square in the denominator yields

Because of the form of the denominator, we consider the transform pairs

and write

Inverse Laplace Transforms of Rational Functions

Using the Laplace transform to solve differential equations often requires finding the inverse transform of a rational function

where \(P\) and \(Q\) are polynomials in \(s\) with no common factors. Since it can be shown that \(\lim_{s\to\infty}F(s)=0\) if \(F\) is a Laplace transform, we need only consider the case where \(\mbox{degree}(P)<\mbox{degree}(Q)\). To obtain \({\cal L}^{-1}(F)\), we find the partial fraction expansion of \(F\), obtain inverse transforms of the individual terms in the expansion from the table of Laplace transforms, and use the linearity property of the inverse transform. The next two examples illustrate this.

Solution (Method 1) Factoring the denominator in (8.2.1) yields

The form for the partial fraction expansion is

Multiplying this by \((s-1)(s-2)\) yields

Setting \(s=2\) yields \(B=8\) and setting \(s=1\) yields \(A=-5\). Therefore

and

Solution (Method 2) We don’t really have to multiply (8.2.3) by \((s-1)(s-2)\) to compute \(A\) and \(B\). We can obtain \(A\) by simply ignoring the factor \(s-1\) in the denominator of (8.2.2) and setting \(s=1\) elsewhere; thus,

Similarly, we can obtain \(B\) by ignoring the factor \(s-2\) in the denominator of (8.2.2) and setting \(s=2\) elsewhere; thus,

To justify this, we observe that multiplying (8.2.3) by \(s-1\) yields

and setting \(s=1\) leads to (8.2.4). Similarly, multiplying (8.2.3) by \(s-2\) yields

and setting \(s=2\) leads to (8.2.5). (It isn’t necesary to write the last two equations. We wrote them only to justify the shortcut procedure indicated in (8.2.4) and (8.2.5).) ∎

The shortcut employed in the second solution of Example 8.2.4 is Heaviside’s method. The next theorem states this method formally. For a proof and an extension of this theorem, see Exercise 10.

Solution The partial fraction expansion of (8.2.7) is of the form

To find \(A\), we ignore the factor \(s\) in the denominator of (8.2.7) and set \(s=0\) elsewhere. This yields

Similarly, the other coefficients are given by

and

Therefore

and

Solution The form for the partial fraction expansion is

Because of the repeated factor \((s+2)^2\) in (8.2.9), Heaviside’s method doesn’t work. Instead, we find a common denominator in (8.2.10). This yields

If (8.2.9) and (8.2.11) are to be equivalent, then

The two sides of this equation are polynomials of degree two. From a theorem of algebra, they will be equal for all \(s\) if they are equal for any three distinct values of \(s\). We may determine \(A\), \(B\) and \(C\) by choosing convenient values of \(s\).

The left side of (8.2.12) suggests that we take \(s=-2\) to obtain \(C=-8\), and \(s=-1\) to obtain \(A=2\). We can now choose any third value of \(s\) to determine \(B\). Taking \(s=0\) yields \(4A+2B+C=-12\). Since \(A=2\) and \(C=-8\) this implies that \(B=-6\). Therefore

and

Solution The form for the partial fraction expansion is

The easiest way to obtain \(A\), \(B\), and \(C\) is to expand the numerator in powers of \(s+2\). This yields

Therefore

and

Solution One form for the partial fraction expansion of \(F\) is

However, we see from the table of Laplace transforms that the inverse transform of the second fraction on the right of (8.2.14) will be a linear combination of the inverse transforms

of

respectively. Therefore, instead of (8.2.14) we write

Finding a common denominator yields

If (8.2.13) and (8.2.16) are to be equivalent, then

This is true for all \(s\) if it’s true for three distinct values of \(s\). Choosing \(s=0\), \(-1\), and \(1\) yields the system

Solving this system yields

Hence, from (8.2.15),

Therefore

Solution The form for the partial fraction expansion is

The coefficients \(A\), \(B\), \(C\) and \(D\) can be obtained by finding a common denominator and equating the resulting numerator to the numerator in (8.2.17). However, since there’s no first power of \(s\) in the denominator of (8.2.17), there’s an easier way: the expansion of

can be obtained quickly by using Heaviside’s method to expand

and then setting \(x=s^2\) to obtain

Multiplying this by \(8+3s\) yields

Therefore

Using Technology

Some software packages that do symbolic algebra can find partial fraction expansions very easily. We recommend that you use such a package if one is available to you, but only after you’ve done enough partial fraction expansions on your own to master the technique.

8.2 Exercises

1. Use the table of Laplace transforms to find the inverse Laplace transform.

   (a) \(\dst{3\over(s-7)^4}\)	(b) \(\dst{2s-4\over s^2-4s+13}\)	(c) \(\dst{1\over s^2+4s+20}\)
   (d) \(\dst{2\over s^2+9}\)	(e) \(\dst{s^2-1\over(s^2+1)^2}\)	(f) \(\dst{1\over(s-2)^2-4}\)
   (g) \(\dst{12s-24\over(s^2-4s+85)^2}\)	(h) \(\dst{2\over(s-3)^2-9}\)	(i) \(\dst{s^2-4s+3\over(s^2-4s+5)^2}\)

   Show answer

   (a) \(\dst{\frac{t^{3}e^{7t}}{2}}\)  (b) \(2e^{2t}\cos 3t\)  (c) \(\dst{e^{-2t}\over4}\sin 4t\)  (d) \(\dst{2\over3}\sin 3t\)  (e) \(t\cos t\) 

      (f) \(\dst{e^{2t}\over2}\sinh 2t\)  (g) \(\dst{2te^{2t}\over3} \sin 9t\)  (h) \(\dst{2e^{3t}\over3}\sinh 3t\)  (i) \(e^{2t} t\cos t\)

2. Use Theorem 8.2.1 and the table of Laplace transforms to find the inverse Laplace transform.

   (a) \(\dst{2s+3\over(s-7)^4}\)	(b) \(\dst{s^2-1\over(s-2)^6}\)	(c) \(\dst{s+5\over s^2+6s+18}\)
   (d) \(\dst{2s+1\over s^2+9}\)	(e) \(\dst{s\over s^2+2s+1}\)	(f) \(\dst{s+1\over s^2-9}\)
   (g) \(\dst{s^3+2s^2-s-3\over(s+1)^4}\)	(h) \(\dst{2s+3\over(s-1)^2+4}\)	(i) \(\dst{1\over s}-{s\over s^2+1}\)
   (j) \(\dst{3s+4\over s^2-1}\)	(k) \(\dst{3\over s-1}+{4s+1\over s^2+9}\)	(l) \(\dst{3\over(s+2)^2}-{2s+6\over s^2+4}\)

   Show answer

   (a) \(\dst{t^2e^{7t}+{17\over6}t^3e^{7t}}\)  (b) \(\dst{e^{2t}\left(\frac{1}{6}t^3+\frac{1}{6}t^4+\frac{1}{40}t^5 \right)}\)  (c) \(\dst{e^{-3t}\left(\cos 3t+{2\over3} \sin 3t\right)}\) 

      (d) \(\dst{2\cos 3t+{1\over3}\sin 3t}\)  (e) \((1-t)e{^{-t}}\)  (f) \(\dst{\cosh 3t+{1\over3}\sinh 3t}\)  (g) \(\dst{\left(1-t-t^2-{1\over6} t^3\right)e^{-t}}\) 

      (h) \(\dst{e^t\left(2\cos 2t+{5\over2}\sin 2t\right)}\)  (i) \(1-\cos t\)  (j) \(3\cosh t+4\sinh t\)  (k) \(\dst{3e^t+4\cos 3t+{1\over3}\sin 3t}\) 

      (l) \(3te^{-2t}-2\cos 2t-3\sin 2t\)

3. Use Heaviside’s method to find the inverse Laplace transform.

   (a) \(\dst{3-(s+1)(s-2)\over(s+1)(s+2)(s-2)}\)	(b) \(\dst{7+(s+4)(18-3s)\over(s-3)(s-1)(s+4)}\)
   (c) \(\dst{2+(s-2)(3-2s)\over(s-2)(s+2)(s-3)}\)	(d) \(\dst{3-(s-1)(s+1)\over(s+4)(s-2)(s-1)}\)
   (e) \(\dst{3+(s-2)(10-2s-s^2)\over(s-2)(s+2)(s-1)(s+3)}\)	(f) \(\dst{3+(s-3)(2s^2+s-21)\over(s-3)(s-1)(s+4)(s-2)}\)

   Show answer

   (a) \(\dst{{1\over4}e^{2t}- {1\over4}e^{-2t}-e^{-t}}\)  (b) \(\dst{{1\over5}e^{-4t}-{41\over5}e^t+5e^{3t}}\)  (c) \(\dst{-{1\over2}e^{2t}-{13\over10}e^{-2t}\; -{1\over5}e^{3t}}\) 

      (d) \(\dst{-{2\over5}e^{-4t}-{3\over5}e^t}\)  (e) \(\dst{{3\over20}e^{2t}-{37\over12}e^{-2t}\; +{1\over3}e^t+{8\over5}e^{-3t}}\) (f) \(\dst{{39\over10}e^t+{3\over14}e^{3t}+{23\over105}e^{-4t} -{7\over3} e^{2t}}\) 

4. Find the inverse Laplace transform.

   (a) \(\dst{2+3s\over(s^2+1)(s+2)(s+1)}\)	(b) \(\dst{3s^2+2s+1\over(s^2+1)(s^2+2s+2)}\)
   (c) \(\dst{3s+2\over(s-2)(s^2+2s+5)}\)	(d) \(\dst{3s^2+2s+1\over(s-1)^2(s+2)(s+3)}\)
   (e) \(\dst{2s^2+s+3\over(s-1)^2(s+2)^2}\)	(f) \(\dst{3s+2\over(s^2+1)(s-1)^2}\)

   Show answer

     (a) \(\dst{{4\over5}e^{-2t}-{1\over2}e^{-t}-{3\over10}\cos t} +{11\over10}\sin t\)  (b) \(\dst{{2\over5}\sin t+{6\over5}\cos t +{7\over5}e^{-t}\sin t-{6\over5}e^{-t}\cos t}\)

      (c) \(\dst{{8\over13}e^{2t}-{8\over13}e^{-t}\cos 2t+{15\over26} e^{-t}\sin 2t}\)  (d) \(\dst{{1\over2}te^t+{3\over8}e^t+e^{-2t} -{11\over8}e^{-3t}}\)

      (e) \(\dst{{2\over3}te^t+{1\over9}e^t +te^{-2t}-{1\over9}e^{-2t}}\) (f) \(\dst{-e^t+{5\over2}te^t+\cos t-{3\over2}\sin t}\)

5. Use the method of Example 8.2.9 to find the inverse Laplace transform.

   (a) \(\dst{3s+2\over(s^2+4)(s^2+9)}\)	(b) \(\dst{-4s+1\over(s^2+1)(s^2+16)}\)	(c) \(\dst{5s+3\over(s^2+1)(s^2+4)}\)
   (d) \(\dst{-s+1\over(4s^2+1)(s^2+1)}\)	(e) \(\dst{17s-34\over(s^2+16)(16s^2+1)}\)	(f) \(\dst{2s-1\over(4s^2+1)(9s^2+1)}\)

   Show answer

     (a) \(\dst{{3\over5}\cos 2t+{1\over5}\sin 2t -{3\over5}\cos 3t-{2\over 15}\sin 3t}\)  (b) \(\dst{-{4\over 15}\cos t+{1\over 15}\sin t +{4\over 15}\cos 4t-{1\over60}\sin 4t}\) 

      (c) \(\dst{{5\over3}\cos t+\sin t-{5\over3} \cos 2t-{1\over2}\sin 2t}\)  (d) \(\dst{-{1\over3}\cos{t\over2}+{2\over3} \sin{t\over2}+{1\over3}\cos t-{1\over3}\sin t}\) 

      (e) \(\dst{{1\over 15}\cos{t\over4}-{8\over 15}\sin{t\over4} -{1\over 15}\cos 4t+{1\over30}\sin4t}\)  (f) \(\dst{{2\over5}\cos{t\over3}-{3\over5}\sin{t\over3} -{2\over5}\cos{t\over2}+{2\over5}\sin{t\over2}}\)

6. Find the inverse Laplace transform.

   (a) \(\dst{17 s-15\over(s^2-2s+5)(s^2+2s+10)}\)	(b) \(\dst{8s+56\over(s^2-6s+13)(s^2+2s+5)}\)
   (c) \(\dst{s+9\over(s^2+4s+5)(s^2-4s+13)}\)	(d) \(\dst{3s-2\over(s^2-4s+5)(s^2-6s+13)}\)
   (e) \(\dst{3s-1\over(s^2-2s+2)(s^2+2s+5)}\)	(f) \(\dst{20s+40\over(4s^2-4s+5)(4s^2+4s+5)}\)

   Show answer

     (a) \(\dst{e^t(\cos 2t+\sin 2t)-e^{-t}\left(\cos 3t+{4\over3}\sin 3t\right)}\)  (b) \(\dst{e^{3t}\left(-\cos 2t+{3\over2}\sin 2t\right)+e^{-t}\left(\cos 2t+{1\over2}\sin 2t\right)}\)

      (c) \(\dst{e^{-2t}\left({1\over8}\cos t+{1\over 4}\sin t\right)-e^{2t}\left({1\over8}\cos 3t-{1\over 12}\sin 3t\right)}\)  (d) \(\dst{e^{2t}\left(\cos t+{1\over2}\sin t\right)-e^{3t}\left(\cos 2t-{1\over4}\sin 2t\right)}\)

      (e) \(\dst{e^t\left({1\over5}\cos t+{2\over5} \sin t\right) -e^{-t}\left({1\over5}\cos 2t+{2\over5}\sin 2t\right)}\)  (f) \(\dst{e^{t/2}\left(-\cos t+{9\over8}\sin t\right)+e^{-t/2}\left(\cos t-{1\over8}\sin t\right)}\)

7. Find the inverse Laplace transform.

   (a) \(\dst{1\over s(s^2+1)}\)	(b) \(\dst{1\over(s-1)(s^2-2s+17)}\)
   (c) \(\dst{3s+2\over(s-2)(s^2+2s+10)}\)	(d) \(\dst{34-17s\over(2s-1)(s^2-2s+5)}\)
   (e) \(\dst{s+2\over(s-3)(s^2+2s+5)}\)	(f) \(\dst{2s-2\over(s-2)(s^2+2s+10)}\)

   Show answer

   (a) \(1-\cos t\)  (b) \(\dst{{e^t\over 16}(1-\cos 4t)}\)  (c) \(\dst{{4\over9}e^{2t}+{5\over9}e^{-t} \sin 3t-{4\over9}e^{-t}\cos 3t}\) (d) \(\dst{3e^{t/2}-{7\over2}e^t\sin 2t-3e^t\cos 2t}\)

      (e) \(\dst{{1\over4}e^{3t}-{1\over4}e^{-t}\cos 2t}\)  (f) \(\dst{{1\over9}e^{2t}-{1\over9}e^{-t}\cos 3t+{5\over 9}e^{-t}\sin 3t}\)

8. Find the inverse Laplace transform.

   (a) \(\dst{2s+1\over(s^2+1)(s-1)(s-3)}\)	(b) \(\dst{s+2\over(s^2+2s+2)(s^2-1)}\)
   (c) \(\dst{2s-1\over(s^2-2s+2)(s+1)(s-2)}\)	(d) \(\dst{s-6\over(s^2-1)(s^2+4)}\)
   (e) \(\dst{2s-3\over s(s-2)(s^2-2s+5)}\)	(f) \(\dst{5s-15\over(s^2-4s+13)(s-2)(s-1)}\)

   Show answer

     (a) \(\dst{-{3\over 10}\sin t+{2\over5}\cos t- {3\over4}e^t+{7\over20}e^{3t}}\)  (b) \(\dst{-{3\over5}e^{-t}\sin t+{1\over5} e^{-t}\cos t-{1\over2}e^{-t}+{3\over 10}e^{t}}\) 

      (c) \(\dst{-{1\over 10}e^t\sin t-{7\over 10}e^t \cos t+{1\over5} e^{-t}+{1\over2}e^{2t}}\)  (d) \(\dst{-{1\over2}e^t+{7\over 10}e^{-t}-{1\over 5}\cos 2t+{3\over5}\sin 2t}\)

      (e) \(\dst{{3\over 10}+{1\over 10}e^{2t}+{1\over 10}e^t\sin 2t-{2\over5}e^t\cos 2t}\)  (f) \(\dst{-{4\over9}e^{2t}\cos 3t+{1\over3} e^{2t}\sin 3t-{5\over9} e^{2t}+e^t}\)

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

   Show answer

   \(\dst{{1\over a}e^{{b\over a}t}f\left({t\over a}\right)}\)

10. 1. 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.

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

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