Math Problem Statement

Given ( F(s) = \frac{4s + 6}{s^2(s+1)} ), find the partial fraction decomposition laplace transform

Solution

To find the partial fraction decomposition of the given Laplace transform F(s)=4s+6s2(s+1)F(s) = \frac{4s + 6}{s^2(s+1)}, we need to express it in a form where we can easily apply the inverse Laplace transform. The general form of partial fractions for a function of this type is:

F(s)=4s+6s2(s+1)=As+Bs2+Cs+1F(s) = \frac{4s + 6}{s^2(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1}

Here, we will find the constants AA, BB, and CC.

Step 1: Set up the equation

4s+6s2(s+1)=As+Bs2+Cs+1\frac{4s + 6}{s^2(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1}

Multiply both sides by s2(s+1)s^2(s+1) to eliminate the denominators:

4s+6=As(s+1)+B(s+1)+Cs24s + 6 = A s(s+1) + B(s+1) + C s^2

Step 2: Expand the terms

Expand both sides of the equation:

4s+6=As2+As+Bs+B+Cs24s + 6 = A s^2 + A s + B s + B + C s^2

Group like terms:

4s+6=(A+C)s2+(A+B)s+B4s + 6 = (A + C) s^2 + (A + B) s + B

Step 3: Equate coefficients

Now, equate the coefficients of the powers of ss on both sides:

  • For s2s^2: A+C=0A + C = 0
  • For ss: A+B=4A + B = 4
  • For the constant term: B=6B = 6

Step 4: Solve the system of equations

From B=6B = 6, substitute into the second equation:

A+6=4    A=2A + 6 = 4 \implies A = -2

Now substitute A=2A = -2 into the first equation:

2+C=0    C=2-2 + C = 0 \implies C = 2

Step 5: Write the partial fraction decomposition

Substitute the values of AA, BB, and CC into the partial fraction decomposition:

4s+6s2(s+1)=2s+6s2+2s+1\frac{4s + 6}{s^2(s+1)} = \frac{-2}{s} + \frac{6}{s^2} + \frac{2}{s+1}

Thus, the partial fraction decomposition is:

F(s)=2s+6s2+2s+1F(s) = \frac{-2}{s} + \frac{6}{s^2} + \frac{2}{s+1}

Step 6: Inverse Laplace Transform

To find the inverse Laplace transform, apply the standard transforms:

  • L1{2s}=2\mathcal{L}^{-1} \left\{ \frac{-2}{s} \right\} = -2
  • L1{6s2}=6t\mathcal{L}^{-1} \left\{ \frac{6}{s^2} \right\} = 6t
  • L1{2s+1}=2et\mathcal{L}^{-1} \left\{ \frac{2}{s+1} \right\} = 2e^{-t}

Thus, the inverse Laplace transform is:

f(t)=2+6t+2etf(t) = -2 + 6t + 2e^{-t}

Would you like further details or have any questions?

Related Questions:

  1. How do you handle repeated factors in partial fraction decomposition?
  2. What is the Laplace transform of eate^{-at}?
  3. How would the partial fraction decomposition change if the denominator was s2(s+2)s^2(s+2)?
  4. Can you apply partial fractions to solve differential equations with Laplace transforms?
  5. What is the significance of the terms in the inverse Laplace transform?

Tip:

When solving for constants in partial fraction decomposition, equate powers of ss carefully to avoid errors in coefficients!

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Math Problem Analysis

Mathematical Concepts

Partial Fraction Decomposition
Laplace Transform
Inverse Laplace Transform

Formulas

F(s) = \frac{4s + 6}{s^2(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1}
\mathcal{L}^{-1} \left\{ \frac{A}{s} \right\} = A
\mathcal{L}^{-1} \left\{ \frac{B}{s^2} \right\} = Bt
\mathcal{L}^{-1} \left\{ \frac{C}{s+1} \right\} = Ce^{-t}

Theorems

Laplace Transform Theorem
Inverse Laplace Transform Theorem

Suitable Grade Level

University (Calculus or Engineering level)