Math Problem Statement

laplace inverse of 4/p(5P+1)

Solution

To compute the inverse Laplace transform of the given expression:

L1(4p(5p+1))\mathcal{L}^{-1}\left(\frac{4}{p(5p + 1)}\right)

Step-by-Step Solution:

  1. Decompose the expression: The expression can be split using partial fraction decomposition. We aim to express the given fraction as a sum of simpler fractions.

    4p(5p+1)=Ap+B5p+1\frac{4}{p(5p + 1)} = \frac{A}{p} + \frac{B}{5p + 1}

    Multiply both sides by p(5p+1)p(5p + 1) to clear the denominator:

    4=A(5p+1)+Bp4 = A(5p + 1) + Bp

    Now, expand the right-hand side:

    4=A5p+A+Bp4 = A \cdot 5p + A + Bp

    Group terms by powers of pp:

    4=(5A+B)p+A4 = (5A + B)p + A

    To solve for AA and BB, compare the coefficients of like powers of pp:

    • For the p0p^0 term (constant term): A=4A = 4
    • For the p1p^1 term: 5A+B=05A + B = 0, so 5×4+B=05 \times 4 + B = 0, which gives B=20B = -20

    Thus, the partial fraction decomposition is:

    4p(5p+1)=4p205p+1\frac{4}{p(5p + 1)} = \frac{4}{p} - \frac{20}{5p + 1}

  2. Find the inverse Laplace transform:

    We now compute the inverse Laplace transform of each term separately:

    • L1(4p)=41=4\mathcal{L}^{-1}\left(\frac{4}{p}\right) = 4 \cdot 1 = 4

    • L1(205p+1)\mathcal{L}^{-1}\left(\frac{20}{5p + 1}\right): This matches the form 1p+a\frac{1}{p + a}, whose inverse Laplace transform is eate^{-at}. Here, a=15a = \frac{1}{5}, so:

      L1(205p+1)=20et5\mathcal{L}^{-1}\left(\frac{20}{5p + 1}\right) = 20e^{-\frac{t}{5}}

  3. Final result:

    Combine the two inverse Laplace transforms:

    L1(4p(5p+1))=420et5\mathcal{L}^{-1}\left(\frac{4}{p(5p + 1)}\right) = 4 - 20e^{-\frac{t}{5}}

Final Answer:

L1(4p(5p+1))=420et5\mathcal{L}^{-1}\left(\frac{4}{p(5p + 1)}\right) = 4 - 20e^{-\frac{t}{5}}


Would you like more details or have any further questions?

5 Related Questions:

  1. What is the inverse Laplace transform of 1p2+a2\frac{1}{p^2 + a^2}?
  2. How do you compute the Laplace transform of tneatt^n e^{at}?
  3. What are the properties of Laplace transforms used in solving differential equations?
  4. Can you explain the convolution theorem in Laplace transforms?
  5. How does partial fraction decomposition help in solving inverse Laplace transforms?

Tip:

When dealing with inverse Laplace transforms, partial fraction decomposition is a powerful technique to simplify complex fractions into forms with known transforms.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Partial Fraction Decomposition
Inverse Laplace Transform

Formulas

Partial Fraction Decomposition: \frac{4}{p(5p+1)} = \frac{A}{p} + \frac{B}{5p+1}
Inverse Laplace Transform of \frac{1}{p+a} = e^{-at}

Theorems

Partial Fraction Decomposition
Inverse Laplace Transform Properties

Suitable Grade Level

Undergraduate level (Calculus/Differential Equations)