Math Problem Statement

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Solution

To solve the given problem, we need to find the inverse Laplace transform of the function:

L1{1s2(s2+3s1)}\mathcal{L}^{-1} \left\{ \frac{1}{s^2 (s^2 + 3s - 1)} \right\}

Here’s a step-by-step solution to this problem:

  1. Factor the denominator if possible: The denominator s2(s2+3s1)s^2 (s^2 + 3s - 1) does not easily factor into simpler components. We need to use partial fraction decomposition.

  2. Partial fraction decomposition: We express the given fraction as a sum of simpler fractions. Let:

1s2(s2+3s1)=As+Bs2+Cs+Ds2+3s1\frac{1}{s^2 (s^2 + 3s - 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 3s - 1}

Multiplying through by the common denominator s2(s2+3s1)s^2 (s^2 + 3s - 1), we get:

1=As(s2+3s1)+B(s2+3s1)+(Cs+D)s21 = A s (s^2 + 3s - 1) + B (s^2 + 3s - 1) + (Cs + D) s^2

Expanding and collecting like terms:

1=(A+C)s3+(3A+B)s2+(AB)sB1 = (A + C)s^3 + (3A + B)s^2 + (A - B)s - B

By equating coefficients of like powers of ss, we obtain:

  • For s3s^3: A+C=0A + C = 0
  • For s2s^2: 3A+B=03A + B = 0
  • For ss: AB=0A - B = 0
  • Constant term: B=1-B = 1

Solving these equations step by step:

  • From B=1-B = 1, we get B=1B = -1.
  • Substituting B=1B = -1 into AB=0A - B = 0, we get A=1A = -1.
  • Substituting A=1A = -1 and B=1B = -1 into 3A+B=03A + B = 0, we get 3(1)+(1)=403(-1) + (-1) = -4 \neq 0. There seems to be an inconsistency in the initial assumption of partial fractions. Let's re-evaluate the structure.

Due to complexity, it may be beneficial to refer to partial fraction techniques tailored for repeated and irreducible quadratic factors.

  1. Correct Partial Fraction Decomposition: Since s2+3s1s^2 + 3s - 1 has no simple roots, let's refine the decomposition form:

1s2(s2+3s1)=As+Bs2+Cs+Ds2+3s1\frac{1}{s^2 (s^2 + 3s - 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 3s - 1}

Solve the coefficients A,B,C,DA, B, C, D:

1=(A+C)s3+(3A+B+D)s2+(AB+3C)s+(D)1 = (A + C)s^3 + (3A + B + D)s^2 + (A - B + 3C)s + (-D)

Equate coefficients: A+C=0A + C = 0 3A+B+D=03A + B + D = 0 AB+3C=0A - B + 3C = 0 D=1    D=1-D = 1 \implies D = -1

Then solving: 3A+B1=0    3A+B=13A + B - 1 = 0 \implies 3A + B = 1 A+3C=0A + 3C = 0

Considering system:

By substitution: B=3A1,C=AB = -3A - 1, \quad C = -A A(3A1)3A=0    5A+1=0    A=15,C=15,B=25A - (-3A - 1) - 3A = 0 \implies 5A + 1 = 0 \implies A = -\frac{1}{5}, \quad C = \frac{1}{5}, \quad B = \frac{2}{5}

Partial fraction: 1s2(s2+3s1)=15s+25s2+15s1s2+3s1\frac{1}{s^2 (s^2 + 3s - 1)} = -\frac{1}{5s} + \frac{2}{5s^2} + \frac{\frac{1}{5}s - 1}{s^2 + 3s - 1}

  1. Inverse Laplace Transform:

Using known transforms: L1(As)=A\mathcal{L}^{-1}\left(\frac{A}{s}\right) = A L1(Bs2)=Bt\mathcal{L}^{-1}\left(\frac{B}{s^2}\right) = Bt L1(Cs+Ds2+3s1)=use convolution and shifting\mathcal{L}^{-1}\left(\frac{Cs + D}{s^2 + 3s - 1}\right) = \text{use convolution and shifting}

Result: 15+25t+L1(15s1s2+3s1)-\frac{1}{5} + \frac{2}{5}t + \mathcal{L}^{-1}\left(\frac{\frac{1}{5}s - 1}{s^2 + 3s - 1}\right)

Ask if user needs more details, also list more questions:

  1. How to solve for Laplace Transform?
  2. Solving quadratic equations?
  3. How convolution integral is used in Laplace?
  4. Can you solve other complex integrals?
  5. What are special properties of Laplace transform?

Tip: Always verify coefficients properly in partial fraction.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Partial Fraction Decomposition
Inverse Laplace Transform
Convolution

Formulas

Inverse Laplace Transform formula

Theorems

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Suitable Grade Level

Advanced Undergraduate