(I) Partial Fraction Decomposition

Let \( R(s) = \frac{P(s)}{Q(s)} \) be a rational fraction where the degree of \( P(s) \) is strictly less than the degree of \( Q(s) \). To find the inverse Laplace transform \(\mathcal{L}^{-1}\{R(s)\}\):

  1. Perform partial fraction decomposition on \( R(s) \).
  2. Find the inverse Laplace transform of each individual term.

Rule 1: Linear Factor Partial Fractions

The portion of the partial fraction decomposition of \( R(s) \) corresponding to a linear factor \( (s-a) \) of multiplicity \( n \) is:

\[ \frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + \dots + \frac{A_n}{(s-a)^n} \]

where \( A_i's \) are constants.

Rule 2: Quadratic Factor Partial Fractions

The portion of the partial fraction decomposition of \( R(s) \) corresponding to the irreducible quadratic factor \( (s-a)^2 + b^2 \) of multiplicity \( n \) is:

\[ \frac{A_1 s + B_1}{(s-a)^2 + b^2} + \frac{A_2 s + B_2}{[(s-a)^2 + b^2]^2} + \dots + \frac{A_n s + B_n}{[(s-a)^2 + b^2]^n} \]

where \( A_i's\) and \( B_i's \) are constants.

(II) Translation on the s-axis

Step 2 is often based on the following property of the Laplace transform.

Theorem: Translation on the s-axis

If \( F(s) = \mathcal{L}\{f(t)\} \) exists for \( s > c \), then \( \mathcal{L}\{e^{at}f(t)\} \) exists for \( s > a + c \) and:

\[ \mathcal{L}\{e^{at}f(t)\} = F(s-a) \]

Equivalently, the inverse is:

\[ \mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t) \]

Sketch of proof:

\( \begin{aligned} F(s-a) &= \int_{0}^{\infty} e^{-(s-a)t} f(t) \, dt \\ &= \int_{0}^{\infty} e^{-st} (e^{at} f(t)) \, dt \\ &= \mathcal{L}\{e^{at}f(t)\} \end{aligned} \)

Common Translated Transforms

Function \( f(t) \) Transform \( F(s) \)
\( e^{at}t^n \) \( \frac{n!}{(s-a)^{n+1}} \quad (s > a) \)
\( e^{at}\cos(kt) \) \( \frac{s-a}{(s-a)^2 + k^2} \quad (s > a) \)
\( e^{at}\sin(kt) \) \( \frac{k}{(s-a)^2 + k^2} \quad (s > a) \)

(III) Applied Examples

Example 1: Initial Value Problem

Solve the Initial Value Problem: \( x'' + 6x' + 34x = 0; \quad x(0) = 3, \quad x'(0) = 1 \).

Solution: Apply the Laplace transform to the equation:

\( \begin{aligned} [s^2 X(s) - 3s - 1] + 6[sX(s) - 3] + 34X(s) &= 0 \\ (s^2 + 6s + 34)X(s) - 3s - 19 &= 0 \\ X(s) &= \frac{3s + 19}{s^2 + 6s + 34} \end{aligned} \)

Complete the square in the denominator:

\( \begin{aligned} X(s) &= \frac{3(s+3) + 10}{(s+3)^2 + 5^2} \\ &= \frac{3(s+3)}{(s+3)^2 + 5^2} + \frac{10}{(s+3)^2 + 5^2} \end{aligned} \)

Take the inverse Laplace transform using the translated trigonometric formulas:

\[ x(t) = 3e^{-3t}\cos(5t) + 2e^{-3t}\sin(5t) = e^{-3t}(3\cos(5t) + 2\sin(5t)) \]
Example 2: Linear Factors

Find the inverse Laplace transform of \( F(s) = \frac{5s - 4}{s^3 - s^2 - 2s} \).

Solution: Factor the denominator and perform partial fraction decomposition:

\( \begin{aligned} \frac{5s - 4}{s(s+1)(s-2)} &= \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s-2} \\ 5s - 4 &= A(s+1)(s-2) + Bs(s-2) + Cs(s+1) \end{aligned} \)

Substitute strategic values for \( s \):

  • If \( s = 0 \implies -4 = -2A \implies A = 2 \)
  • If \( s = -1 \implies -9 = 3B \implies B = -3 \)
  • If \( s = 2 \implies 6 = 6C \implies C = 1 \)
\( \begin{aligned} F(s) &= \frac{2}{s} - \frac{3}{s+1} + \frac{1}{s-2} \end{aligned} \)
\[ f(t) = 2 - 3e^{-t} + e^{2t} \]
Example 3: Repeated Linear Factors (Differentiation Method)

Find the inverse Laplace transform of \( F(s) = \frac{s^3}{(s-4)^4} \).

Solution: Set up the partial fraction decomposition:

\( \begin{aligned} F(s) &= \frac{A}{s-4} + \frac{B}{(s-4)^2} + \frac{C}{(s-4)^3} + \frac{D}{(s-4)^4} \end{aligned} \)

Multiply both sides by \( (s-4)^4 \):

\( \begin{aligned} s^3 &= A(s-4)^3 + B(s-4)^2 + C(s-4) + D \end{aligned} \)

Use differentiation to find the coefficients systematically:

  1. Let \( s = 4 \implies D = 4^3 = 64 \).
  2. Differentiate once: \( 3s^2 = 3A(s-4)^2 + 2B(s-4) + C \). Let \( s = 4 \implies C = 3(16) = 48 \).
  3. Differentiate again: \( 6s = 6A(s-4) + 2B \). Let \( s = 4 \implies 24 = 2B \implies B = 12 \).
  4. Differentiate again: \( 6 = 6A \implies A = 1 \).
\( \begin{aligned} F(s) &= \frac{1}{s-4} + \frac{12}{(s-4)^2} + \frac{48}{(s-4)^3} + \frac{64}{(s-4)^4} \end{aligned} \)

Recall that \( \mathcal{L}^{-1}\left\{\frac{n!}{(s-a)^{n+1}}\right\} = e^{at}t^n \). Applying this to each term:

\( \begin{aligned} f(t) &= e^{4t} + 12e^{4t}t + 48\frac{e^{4t}t^2}{2!} + 64\frac{e^{4t}t^3}{3!} \end{aligned} \)
\[f(t) = e^{4t} \left( 1 + 12t + 24t^2 + \frac{32}{3}t^3 \right) \]
Example 4: Irreducible Quadratic with Translation

Find the inverse Laplace transform \( f(t) \) of:

\[ F(s) = \frac{s + 7}{(s - 2)[(s - 2)^2 + 9]} \]

Solution:

Step 1: Partial Fraction Decomposition

Using Rule 2 for irreducible quadratic factors, we set up the decomposition using the shift \( (s-2) \) as our base to simplify the final inverse transform:

\[ \frac{s + 7}{(s - 2)[(s - 2)^2 + 9]} = \frac{A}{s - 2} + \frac{B(s - 2) + C}{(s - 2)^2 + 9} \]

Multiply by the common denominator and solve for coefficients:

\[ s + 7 = A[(s - 2)^2 + 9] + [B(s - 2) + C](s - 2) \]
  • Find A: Let \( s = 2 \implies 9 = 9A \implies A = 1 \)
  • Find B and C: Expand or use substitution. Let \( u = s - 2 \):
    \( u + 9 = 1(u^2 + 9) + (Bu + C)u \)
    \( u + 9 = (1 + B)u^2 + Cu + 9 \)
  • Equating \( u^2 \): \( 0 = 1 + B \implies B = -1 \)
  • Equating \( u^1 \): \( 1 = C \implies C = 1 \)
\[ F(s) = \frac{1}{s - 2} - \frac{s - 2}{(s - 2)^2 + 3^2} + \frac{1}{(s - 2)^2 + 3^2} \] Step 2: Inverse Laplace Transform (Translation)

Applying the Translation Theorem \( \mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t) \) where \( a = 2 \):

\[ f(t) = \mathcal{L}^{-1}\left\{\frac{1}{s - 2}\right\} - \mathcal{L}^{-1}\left\{\frac{s - 2}{(s - 2)^2 + 3^2}\right\} + \frac{1}{3}\mathcal{L}^{-1}\left\{\frac{3}{(s - 2)^2 + 3^2}\right\} \] \[ f(t) = e^{2t} - e^{2t} \cos(3t) + \frac{1}{3} e^{2t} \sin(3t) \]

Factoring out the exponential term gives the final result:

\[ f(t) = e^{2t} \left( 1 - \cos(3t) + \frac{1}{3} \sin(3t) \right) \]
Example 5: Irreducible Quadratic Factors

Find the inverse Laplace transform of \( F(s) = \frac{s^2 - 2s}{s^4 + 5s^2 + 4} \).

Solution: Factor the denominator to \( (s^2+1)(s^2+4) \) and set up the decomposition:

\( \begin{aligned} \frac{s^2 - 2s}{(s^2+1)(s^2+4)} &= \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+4} \\ s^2 - 2s &= (As+B)(s^2+4) + (Cs+D)(s^2+1) \\ s^2 - 2s &= (A+C)s^3 + (B+D)s^2 + (4A+C)s + (4B+D) \end{aligned} \)

Equating coefficients yields a system of equations:

  • \( s^3: A + C = 0 \implies C = -A \)
  • \( s^1: 4A + C = -2 \implies 4A - A = -2 \implies 3A = -2 \implies A = -2/3 \) (so \( C = 2/3 \))
  • \( s^2: B + D = 1 \)
  • \( s^0: 4B + D = 0 \implies D = -4B \implies B - 4B = 1 \implies -3B = 1 \implies B = -1/3 \) (so \( D = 4/3 \))
\( \begin{aligned} F(s) &= \frac{-\frac{2}{3}s - \frac{1}{3}}{s^2+1} + \frac{\frac{2}{3}s + \frac{4}{3}}{s^2+4} \end{aligned} \)

Take the inverse transform (note that the sine term corresponding to \( s^2 + 4 \) requires a factor of 2 in the numerator):

\[f(t) = -\frac{2}{3}\cos(t) - \frac{1}{3}\sin(t) + \frac{2}{3}\cos(2t) + \frac{2}{3}\sin(2t) \]