Math Problem Statement

Inverse Laplace in partial fraction

Inverse Laplace in partial fraction

Solution

It looks like you've uploaded an image of a mathematical expression that you want to solve using the inverse Laplace transform in partial fractions. Let's first break down the expression in the image:

The expression shown is: 1(s2)(s+3)\frac{1}{(s-2)(s+3)}

Steps to solve using Partial Fraction Decomposition:

  1. Express the function as a sum of simpler fractions: 1(s2)(s+3)=As2+Bs+3\frac{1}{(s-2)(s+3)} = \frac{A}{s-2} + \frac{B}{s+3} where AA and BB are constants to be determined.

  2. Multiply both sides by (s2)(s+3)(s-2)(s+3) to get rid of the denominator: 1=A(s+3)+B(s2)1 = A(s + 3) + B(s - 2) Simplify and group like terms: 1=(A+B)s+(3A2B)1 = (A + B)s + (3A - 2B)

  3. Set up a system of equations by equating coefficients:

    • Coefficient of ss: A+B=0A + B = 0
    • Constant term: 3A2B=13A - 2B = 1
  4. Solve the system of equations:

    • From A+B=0A + B = 0, we have B=AB = -A.
    • Substitute into the second equation: 3A2(A)=13A - 2(-A) = 1, which simplifies to 3A+2A=13A + 2A = 1, so 5A=15A = 1, hence A=15A = \frac{1}{5}.
    • Since B=AB = -A, B=15B = -\frac{1}{5}.

    So, the partial fraction decomposition is: 1(s2)(s+3)=15s215s+3\frac{1}{(s-2)(s+3)} = \frac{\frac{1}{5}}{s-2} - \frac{\frac{1}{5}}{s+3}

  5. Apply the inverse Laplace transform:

    • The inverse Laplace transform of 1sa\frac{1}{s-a} is eate^{at}.
    • Therefore: L1(1(s2)(s+3))=15e2t15e3t\mathcal{L}^{-1}\left(\frac{1}{(s-2)(s+3)}\right) = \frac{1}{5}e^{2t} - \frac{1}{5}e^{-3t}

Final Answer:

L1(1(s2)(s+3))=15e2t15e3t\mathcal{L}^{-1}\left(\frac{1}{(s-2)(s+3)}\right) = \frac{1}{5}e^{2t} - \frac{1}{5}e^{-3t}

Would you like any further details or have any questions on this solution?

Here are some related questions you might consider:

  1. How does the Laplace transform of 1s2+a2\frac{1}{s^2 + a^2} compare to this case?
  2. What happens if the denominator has repeated roots?
  3. How can partial fraction decomposition be applied to higher-degree polynomials?
  4. What is the significance of each term in the inverse Laplace transform?
  5. How would the procedure change if ss had a complex root?

Tip: When solving inverse Laplace problems, always make sure your decomposition is correct by re-multiplying to check your work before applying the inverse transform.

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Math Problem Analysis

Mathematical Concepts

Inverse Laplace Transform
Partial Fraction Decomposition

Formulas

Inverse Laplace of 1/(s-a) is e^{at}
Partial Fraction: 1/(s-2)(s+3) = A/(s-2) + B/(s+3)

Theorems

Laplace Transform Theorem
Partial Fraction Theorem

Suitable Grade Level

Undergraduate