Math Problem Statement

please solve problem number 6 which great, insightful and step-by-step explanation!

Solution

The problem is asking to find the inverse Laplace transform of the function:

F(s)=1s(1s2+1)F(s) = \frac{1}{s} \left(\frac{1}{s^2 + 1}\right)

Let's solve this step-by-step.

Step 1: Understand the given function

The function F(s)=1s1s2+1F(s) = \frac{1}{s} \cdot \frac{1}{s^2 + 1} can be written as:

F(s)=1s(s2+1)F(s) = \frac{1}{s(s^2 + 1)}

We need to find the inverse Laplace transform of F(s)F(s), i.e., we want to find f(t)f(t) such that:

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

Step 2: Decompose the function using Partial Fraction Decomposition

To make the inverse Laplace transform easier, we'll use partial fraction decomposition on F(s)F(s).

Let’s express F(s)F(s) as:

1s(s2+1)=As+Bs+Cs2+1\frac{1}{s(s^2 + 1)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 1}

To find AA, BB, and CC, multiply both sides by s(s2+1)s(s^2 + 1):

1=A(s2+1)+(Bs+C)s1 = A(s^2 + 1) + (Bs + C)s

This expands to:

1=A(s2+1)+Bs2+Cs1 = A(s^2 + 1) + Bs^2 + Cs

1=(A+B)s2+Cs+A1 = (A + B)s^2 + Cs + A

Now, equate the coefficients of s2s^2, ss, and the constant term from both sides of the equation:

  • Coefficient of s2s^2: A+B=0A + B = 0
  • Coefficient of ss: C=0C = 0
  • Constant term: A=1A = 1

Step 3: Solve for AA, BB, and CC

From A=1A = 1 and A+B=0A + B = 0, we find:

A=1,B=1,C=0A = 1, \quad B = -1, \quad C = 0

So the partial fraction decomposition is:

1s(s2+1)=1sss2+1\frac{1}{s(s^2 + 1)} = \frac{1}{s} - \frac{s}{s^2 + 1}

Step 4: Find the inverse Laplace transforms

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

  1. Inverse Laplace transform of 1s\frac{1}{s}:

L1{1s}=1\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1

  1. Inverse Laplace transform of ss2+1\frac{s}{s^2 + 1}:

L1{ss2+1}=cos(t)\mathcal{L}^{-1}\left\{\frac{s}{s^2 + 1}\right\} = \cos(t)

Step 5: Combine the results

Combine the results from the inverse Laplace transforms:

f(t)=1cos(t)f(t) = 1 - \cos(t)

Final Answer:

The inverse Laplace transform of F(s)=1s(s2+1)F(s) = \frac{1}{s(s^2 + 1)} is:

f(t)=1cos(t)f(t) = 1 - \cos(t)

This function f(t)=1cos(t)f(t) = 1 - \cos(t) represents the solution in the time domain.

Would you like to see more details or have any questions?

Here are some related questions you might find helpful:

  1. How can the convolution theorem be applied to find the inverse Laplace transform of products like 1s(s2+1)\frac{1}{s(s^2 + 1)}?
  2. What are some common Laplace transforms and their corresponding time-domain functions?
  3. How does the inverse Laplace transform change if additional terms like ease^{-as} are involved?
  4. How can partial fraction decomposition be used in solving differential equations with Laplace transforms?
  5. What are the applications of inverse Laplace transforms in solving real-world problems?

Tip: When working with Laplace transforms, keep a table of common transforms handy to quickly identify and apply the inverse transform.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Laplace Transform
Inverse Laplace Transform
Partial Fraction Decomposition

Formulas

-

Theorems

-

Suitable Grade Level

Advanced