Math Problem Statement

L(6/(s^2-1)(s+2))

Solution

To find the Laplace transform L{6(s21)(s+2)}L\left\{\frac{6}{(s^2-1)(s+2)}\right\}, we'll first simplify the expression and then apply partial fraction decomposition.

Step 1: Factor the denominator

The expression is:

6(s21)(s+2)\frac{6}{(s^2-1)(s+2)}

Factor s21s^2 - 1:

s21=(s1)(s+1)s^2 - 1 = (s - 1)(s + 1)

So the expression becomes:

6(s1)(s+1)(s+2)\frac{6}{(s - 1)(s + 1)(s + 2)}

Step 2: Partial fraction decomposition

We express the function as a sum of simpler fractions:

6(s1)(s+1)(s+2)=As1+Bs+1+Cs+2\frac{6}{(s - 1)(s + 1)(s + 2)} = \frac{A}{s - 1} + \frac{B}{s + 1} + \frac{C}{s + 2}

Multiplying both sides by the denominator (s1)(s+1)(s+2)(s - 1)(s + 1)(s + 2) gives:

6=A(s+1)(s+2)+B(s1)(s+2)+C(s1)(s+1)6 = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s - 1)(s + 1)

Step 3: Solve for A, B, and C

To find AA, BB, and CC, we substitute convenient values of ss.

  1. Let s=1s = 1:

6=A(1+1)(1+2)+0+06=6AA=16 = A(1 + 1)(1 + 2) + 0 + 0 \Rightarrow 6 = 6A \Rightarrow A = 1

  1. Let s=1s = -1:

6=0+B(11)(1+2)+06=2BB=36 = 0 + B(-1 - 1)(-1 + 2) + 0 \Rightarrow 6 = 2B \Rightarrow B = 3

  1. Let s=2s = -2:

6=0+0+C(21)(2+1)6=3CC=26 = 0 + 0 + C(-2 - 1)(-2 + 1) \Rightarrow 6 = 3C \Rightarrow C = -2

Thus, the decomposition is:

6(s1)(s+1)(s+2)=1s1+3s+12s+2\frac{6}{(s - 1)(s + 1)(s + 2)} = \frac{1}{s - 1} + \frac{3}{s + 1} - \frac{2}{s + 2}

Step 4: Apply the inverse Laplace transform

Now, we apply the inverse Laplace transform to each term:

L1{1s1}=etL^{-1}\left\{\frac{1}{s - 1}\right\} = e^{t}

L1{3s+1}=3etL^{-1}\left\{\frac{3}{s + 1}\right\} = 3e^{-t}

L1{2s+2}=2e2tL^{-1}\left\{\frac{2}{s + 2}\right\} = 2e^{-2t}

Final Answer:

Combining these, we get the inverse Laplace transform:

L1{6(s1)(s+1)(s+2)}=et+3et2e2tL^{-1}\left\{\frac{6}{(s - 1)(s + 1)(s + 2)}\right\} = e^{t} + 3e^{-t} - 2e^{-2t}

Would you like any further details or have any questions?

Here are 5 related questions you might find interesting:

  1. How do you perform partial fraction decomposition with repeated factors?
  2. What is the Laplace transform of 1s2(s+1)\frac{1}{s^2(s + 1)}?
  3. How does the initial value theorem apply to Laplace transforms?
  4. Can you explain the convolution theorem in the context of Laplace transforms?
  5. What are some applications of Laplace transforms in engineering?

Tip: When performing partial fraction decomposition, check for repeated roots and handle them appropriately to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Partial Fraction Decomposition
Inverse Laplace Transform

Formulas

L{f(t)} = ∫₀^∞ e^(-st)f(t) dt
Partial fraction decomposition: 6/(s-1)(s+1)(s+2) = A/(s-1) + B/(s+1) + C/(s+2)
Inverse Laplace Transforms: L⁻¹{1/(s-a)} = e^(at)

Theorems

Laplace Transform Theorem
Partial Fraction Decomposition

Suitable Grade Level

Grades 11-12 / College