Math Problem Statement

solve using inverse laplace

Solve using inverse Laplace for the given equation F(s) = e^(-s) / (s^2 + 2s).

Solution

The equation in the image you provided represents a Laplace transform:

F(s)=ess2+2sF(s) = \frac{e^{-s}}{s^2 + 2s}

To solve for the inverse Laplace transform of this function, we will first attempt to factorize the denominator, and then we can use known inverse Laplace transform techniques.

Step 1: Factorizing the Denominator

We begin by factoring the quadratic expression s2+2ss^2 + 2s:

s2+2s=s(s+2)s^2 + 2s = s(s + 2)

Thus, the function becomes:

F(s)=ess(s+2)F(s) = \frac{e^{-s}}{s(s + 2)}

Step 2: Applying the Inverse Laplace Transform

The presence of ese^{-s} in the numerator indicates a time shift property in the Laplace transform. Using the inverse Laplace transform properties and the known inverse transforms, we can start by looking at the function 1s(s+2)\frac{1}{s(s + 2)}, which can be decomposed into partial fractions.

We express:

1s(s+2)=As+Bs+2\frac{1}{s(s + 2)} = \frac{A}{s} + \frac{B}{s + 2}

Multiplying through by s(s+2)s(s + 2), we get:

1=A(s+2)+Bs1 = A(s + 2) + B s

By solving for AA and BB, we find:

  • Set s=0s = 0: 1=2AA=121 = 2A \Rightarrow A = \frac{1}{2}
  • Set s=2s = -2: 1=2BB=121 = -2B \Rightarrow B = -\frac{1}{2}

Thus, the partial fraction decomposition is:

1s(s+2)=12(1s1s+2)\frac{1}{s(s + 2)} = \frac{1}{2} \left( \frac{1}{s} - \frac{1}{s + 2} \right)

Step 3: Inverse Laplace Transform

Now, using the inverse Laplace transform properties:

L1(1s)=1\mathcal{L}^{-1}\left( \frac{1}{s} \right) = 1 L1(1s+2)=e2t\mathcal{L}^{-1}\left( \frac{1}{s + 2} \right) = e^{-2t}

Therefore, we have:

L1(1s(s+2))=12(1e2t)\mathcal{L}^{-1}\left( \frac{1}{s(s + 2)} \right) = \frac{1}{2} \left( 1 - e^{-2t} \right)

Step 4: Time Shifting Property

The ese^{-s} term in the numerator of the original Laplace transform indicates a time shift of 1 unit. Using the time-shifting property of the inverse Laplace transform:

L1(esF(s))=u(t1)f(t1)\mathcal{L}^{-1} \left( e^{-s} F(s) \right) = u(t-1) f(t-1)

where u(t1)u(t-1) is the unit step function shifted to t=1t = 1, and f(t)f(t) is the inverse Laplace transform of the function F(s)F(s) without the ese^{-s} term.

Thus, the solution becomes:

f(t)=u(t1)12(1e2(t1))f(t) = u(t-1) \cdot \frac{1}{2} \left( 1 - e^{-2(t-1)} \right)

This is the inverse Laplace transform of the given function.

Would you like further details or have any questions about this process?

Here are some follow-up questions for your exploration:

  1. How do you apply the inverse Laplace transform to a function with higher-order polynomials in the denominator?
  2. What is the significance of the time-shifting property in Laplace transforms?
  3. Can you apply partial fraction decomposition to other types of rational functions, and how does the complexity change?
  4. How would this process change if the Laplace transform had higher powers of ss?
  5. What are other common Laplace transform pairs and their applications?

Tip: In inverse Laplace transforms, always look for possible simplifications like factoring denominators or recognizing time shifts to streamline your work.

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Math Problem Analysis

Mathematical Concepts

Inverse Laplace Transform
Partial Fractions
Time Shifting Property

Formulas

F(s) = e^(-s) / (s(s + 2))
Time-shifting property: L^(-1)[e^(-as)F(s)] = u(t-a)f(t-a)

Theorems

Time Shifting Theorem
Partial Fraction Decomposition

Suitable Grade Level

Undergraduate Level