9.2: The Inverse Laplace Transform

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Definition of the Inverse Laplace Transform

In Section 9.1 we defined the Laplace transform of \(f\) by

\[F(s)={\cal L}(f)=\int_0^\infty e^{-st}f(t)\,dt. \nonumber\]

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

\[f={\cal L}^{-1}(F). \nonumber\]

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.

Example 9.2.1

Use the table of Laplace transforms to find

1. \[{\cal L}^{-1}\left({1\over s^2-1}\right) \nonumber\]
2. \[{\cal L}^{-1}\left({s\over s^2+9}\right).\nonumber\]

Solution a

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

\[\sinh bt\leftrightarrow {b\over s^2-b^2} \nonumber\]

shows that

\[{\cal L}^{-1}\left({1\over s^2-1}\right)=\sinh t.\nonumber\]

Solution b

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

\[\cos\omega t\leftrightarrow{s\over s^2+\omega^2}\nonumber\]

shows that

\[{\cal L}^{-1}\left({s\over s^2+9}\right)=\cos3t. \nonumber\]

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

Theorem 9.2.1 : Linearity Property

If \(F_1,\) \(F_2,\) …\(,\) \(F_n\) are Laplace transforms and \(c_1,\) \(c_2,\) …, \(c_n\) are constants\(,\) then

\[{\cal L}^{-1}(c_1F_1+c_2F_2+\cdots+c_nF_n)=c_1{\cal L}^{-1}(F_1)+c_2{\cal L}^{-1}(F_2)+\cdots+c_n{\cal L}^{-1}F_n.\nonumber\]

Example 9.2.2

Find

\[{\cal L}^{-1}\left({8\over s+5}+{7\over s^2+3}\right).\nonumber\]

Solution

From the table of Laplace transforms in Section 9.1,

\[e^{at}\leftrightarrow {1\over s-a}\quad\mbox{ and }\quad \sin\omega t\leftrightarrow {\omega\over s^2+\omega^2}. \nonumber\]

Theorem 9.2.1 with \(a=-5\) and \(\omega=\sqrt3\) yields

\[\begin{aligned} {\cal L}^{-1}\left({8\over s+5}+{7\over s^2+3}\right)&= 8{\cal L}^{-1}\left({1\over s+5}\right)+7{\cal L}^{-1}\left({1\over s^2+3}\right)\\ &= 8{\cal L}^{-1}\left({1\over s+5}\right)+{7\over\sqrt3}{\cal L}^{-1}\left({\sqrt3\over s^2+3}\right)\\&= 8e^{-5t}+{7\over\sqrt3}\sin\sqrt3t.\end{aligned}\nonumber\]

Note

We’ll often write inverse Laplace transforms of specific functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms in Section 9.1.

Inverse Laplace Transforms of Rational Functions

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

\[F(s)={P(s)\over Q(s)}, \nonumber\]

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.

Example 9.2.3

Find the inverse Laplace transform of

\[\label{eq:8.2.1} F(s)={3s+2\over s^2-3s+2}.\]

Solution

Factoring the denominator in Equation \ref{eq:8.2.1} yields

\[\label{eq:8.2.2} F(s)={3s+2\over(s-1)(s-2)}.\]

The form for the partial fraction expansion is

\[\label{eq:8.2.3} {3s+2\over(s-1)(s-2)}={A\over s-1}+{B\over s-2}.\]

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

\[3s+2=A(s-2)+B(s-1). \nonumber\]

Taking a limit as \({s\to1}\) yields \(A=-5\) and taking a limit as \({s\to2}\) yields \(B=8\).

\[F(s)=-{5\over s-1}+{8\over s-2} \nonumber\]

and

\[{\cal L}^{-1}(F)=-5{\cal L}^{-1}\left({1\over s-1}\right) +8{\cal L}^{-1}\left({1\over s-2}\right)=-5e^t+8e^{2t}. \nonumber\]

Example 9.2.4

Find the inverse Laplace transform of

\[\label{eq:8.2.7} F(s)={6+(s+1)(s^2-5s+11)\over s(s-1)(s-2)(s+1)}.\]

Solution

The partial fraction expansion of Equation \ref{eq:8.2.7} is of the form

\[\label{eq:8.2.8} F(s)={A\over s}+{B\over s-1}+{C\over s-2}+{D\over s+1}.\]

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

\[6+(s+1)(s^2-5s+11)=A(s-1)(s-2)(s+1)+Bs(s-2)(s+1)+Cs(s-1)(s+1)+Ds(s-1)(s-2). \nonumber\]

Taking a limit as \({s\to0}\) yields \(A={17\over 2}\), the limit as \({s\to1}\) yields \(B=-10\), the limit as \({s\to2}\) yields \(C={7\over 2}\), and the limit as \({s\to-1}\) yields \(D=-1\).

Therefore

\[F(s)={17\over2}\,{1\over s}-{10\over s-1}+{7\over2}\,{1\over s-2}-{1\over s+1} \nonumber\]

and

\[\begin{aligned} {\cal L}^{-1}(F)&= {17\over2}{\cal L}^{-1}\left(1\over s\right)-10{\cal L}^{-1}\left(1 \over s-1\right)+{7\over 2}{\cal L}^{-1}\left(1\over s-2\right)-{\cal L}^{-1}\left(1\over s+1\right)\\&= {17\over2}-10e^t+{7\over2}e^{2t}-e^{-t}.\end{aligned}\nonumber\]

Example 9.2.5

Find the inverse Laplace transform of

\[\label{eq:8.2.17} F(s)={8+3s\over(s^2+1)(s^2+4)}.\]

Solution

The form for the partial fraction expansion is

\[F(s)={A+Bs\over s^2+1}+{C+Ds\over s^2+4}. \nonumber\]

which leads to 

\[{1\over(s^2+1)(s^2+4)}={1\over3}{1\over s^2+1}-{1\over 3}{1\over s^2+4}. \nonumber\]

Multiplying by \(8+3s\) gives us

\[{8+3s\over(s^2+1)(s^2+4)}={8+3s\over3}{1\over s^2+1}-{8+3s\over 3}{1\over s^2+4}. \nonumber\]

Therefore

\[{\cal L}^{-1}(F)={8\over3}\sin t+\cos t-{4\over3}\sin 2t-\cos 2t. \nonumber\]