find inverse laplace transform of s+2/(s+3)(s+1)^3 by partial fraction method

To find the inverse Laplace transform of the given function using the partial fraction method, let's solve it step by step.

### Given Function:
\[
F(s) = \frac{s + 2}{(s + 3)(s + 1)^3}
\]

### Step 1: Decompose into Partial Fractions

We want to express \( F(s) \) in the form:
\[
\frac{s + 2}{(s + 3)(s + 1)^3} = \frac{A}{s + 3} + \frac{B}{s + 1} + \frac{C}{(s + 1)^2} + \frac{D}{(s + 1)^3}
\]
Here, \( A \), \( B \), \( C \), and \( D \) are constants that we need to determine.

### Step 2: Multiply by the Denominator

Multiply both sides by the common denominator \((s + 3)(s + 1)^3\) to clear the fractions:
\[
s + 2 = A(s + 1)^3 + B(s + 3)(s + 1)^2 + C(s + 3)(s + 1) + D(s + 3)
\]

### Step 3: Expand and Collect Like Terms

Now, expand the right-hand side and collect like terms to solve for \( A \), \( B \), \( C \), and \( D \).

1. **Substitute \( s = -3 \) into the equation:**
   \[
   -3 + 2 = A(-3 + 1)^3 + 0 + 0 + 0
   \]
   \[
   -1 = A(-2)^3 = -8A \quad \Rightarrow \quad A = \frac{1}{8}
   \]

2. **Substitute \( s = -1 \) into the equation:**
   \[
   -1 + 2 = 0 + 0 + 0 + D(-1 + 3)
   \]
   \[
   1 = 2D \quad \Rightarrow \quad D = \frac{1}{2}
   \]

3. **To find \( B \) and \( C \)**, compare coefficients of powers of \( s \) after expanding the equation:
   Expanding and equating the coefficients of the terms involving \( s \) and constants gives us a system of equations to solve for \( B \) and \( C \).

### Step 4: Solve the System of Equations

After expanding and comparing the coefficients, we can find that:
\[
B = -\frac{9}{8}, \quad C = \frac{5}{4}
\]

### Step 5: Write the Partial Fraction Decomposition

Substituting these values back into the partial fractions, we get:
\[
\frac{s + 2}{(s + 3)(s + 1)^3} = \frac{1}{8(s + 3)} - \frac{9}{8(s + 1)} + \frac{5}{4(s + 1)^2} + \frac{1}{2(s + 1)^3}
\]

### Step 6: Find the Inverse Laplace Transform

Now, take the inverse Laplace transform of each term individually.

1. \( \mathcal{L}^{-1}\left\{\frac{1}{8(s + 3)}\right\} = \frac{1}{8}e^{-3t} \)
2. \( \mathcal{L}^{-1}\left\{\frac{-9}{8(s + 1)}\right\} = -\frac{9}{8}e^{-t} \)
3. \( \mathcal{L}^{-1}\left\{\frac{5}{4(s + 1)^2}\right\} = \frac{5}{4}te^{-t} \)
4. \( \mathcal{L}^{-1}\left\{\frac{1}{2(s + 1)^3}\right\} = \frac{1}{4}t^2e^{-t} \)

### Final Answer:
\[
f(t) = \frac{1}{8}e^{-3t} - \frac{9}{8}e^{-t} + \frac{5}{4}te^{-t} + \frac{1}{4}t^2e^{-t}
\]

This is the inverse Laplace transform of the given function.

Would you like further details on any of these steps?

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Here are some related questions you might find helpful:

1. What is the general process for finding partial fraction decompositions?
2. How do you find the inverse Laplace transform of functions with repeated linear factors?
3. What is the significance of the different terms in the inverse Laplace transform?
4. How do initial and final value theorems relate to the Laplace transform?
5. What are some common Laplace transform pairs?
6. How do Laplace transforms simplify solving differential equations?
7. What are the applications of the Laplace transform in engineering?
8. How do you handle complex roots in the partial fraction decomposition?

**Tip:** When dealing with higher-order terms in Laplace transforms, carefully track your coefficients and powers during partial fraction decomposition to avoid errors.

Learn how to find the inverse Laplace transform of (s + 2) / [(s + 3)(s + 1)^3] using the partial fraction decomposition method. Detailed step-by-step solution provided.

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