Inverse Laplace transform for the following functions. (If you need to find undetermined coefficient for partial fraction form, please use numerical substitution method and coefficient comparison method to find the undetermined coefficient).
(1) F(s) =2s+2s2-10s+16
(2) F(s) =4s2s+5+2s
(3) F(s) =ln (s-3s+3)
(4) F(s) =s+5(s+1)2(s+2)2



Answer :

Answer:To find the inverse Laplace transform of the given functions \( F(s) \), we'll go through each one step-by-step. We'll use the method of partial fractions where necessary, and for logarithmic and repeated roots, we'll employ appropriate methods to handle them.

### (1) \( F(s) = 2s^2 - 10s + 16 \)

1. **Factorize the denominator (if applicable):**

  The given function \( F(s) \) does not have a denominator explicitly shown, so we assume it's in the form \( F(s) = \frac{N(s)}{D(s)} \) where \( D(s) = 1 \) (since it's not explicitly given).

2. **Find the partial fraction decomposition:**

  Since there's no denominator to decompose, we proceed directly to finding the inverse Laplace transform.

3. **Inverse Laplace Transform:**

  For each term \( 2s^2 - 10s + 16 \):

  - \( \mathcal{L}^{-1}\{2s^2\} = 2 \cdot \frac{d^2}{dt^2} \delta(t) = 2t^2 \)

  - \( \mathcal{L}^{-1}\{-10s\} = -10 \cdot \delta'(t) = -10 \)

  - \( \mathcal{L}^{-1}\{16\} = 16 \cdot \delta(t) = 16 \)

  Therefore,

  \[

  \mathcal{L}^{-1}\{2s^2 - 10s + 16\} = 2t^2 - 10\delta'(t) + 16\delta(t)

  \]

### (2) \( F(s) = \frac{4s^2 + 2s}{2s + 5} \)

1. **Perform partial fraction decomposition:**

  \[

  F(s) = \frac{4s^2 + 2s}{2s + 5} = \frac{4s^2 + 2s}{2(s + \frac{5}{2})}

  \]

  Perform partial fractions:

  \[

  \frac{4s^2 + 2s}{2(s + \frac{5}{2})} = \frac{A}{s + \frac{5}{2}} + \frac{B}{2}

  \]

  Multiply through by \( 2(s + \frac{5}{2}) \):

  \[

  4s^2 + 2s = A \cdot 2 + B \cdot (s + \frac{5}{2})

  \]

  \[

  4s^2 + 2s = 2A + Bs + \frac{5B}{2}

  \]

  \[

  \begin{cases}

  2A = 0 \\

  B = 4 \\

  5B/2 = 0 \\

  \ Another It Calcul  have You Done Who So

Step-by-step explanation:Certainly! Let's go through each problem step-by-step to find the inverse Laplace transform for each function \( F(s) \).

### (1) \( F(s) = 2s^2 - 10s + 16 \)

1. **Inverse Laplace Transform (Direct Method):**

  The inverse Laplace transform \( \mathcal{L}^{-1} \{ 2s^2 - 10s + 16 \} \) involves applying the basic transforms directly:

  - \( \mathcal{L}^{-1} \{ 2s^2 \} = 2 \cdot \frac{d^2}{dt^2} \delta(t) = 2t^2 \)

    (Using the property \( \mathcal{L} \{ t^n \} = \frac{n!}{s^{n+1}} \) for \( n = 2 \))

  - \( \mathcal{L}^{-1} \{ -10s \} = -10 \cdot \delta'(t) = -10 \)

    (Using the property \( \mathcal{L} \{ 1 \} = \frac{1}{s} \))

  - \( \mathcal{L}^{-1} \{ 16 \} = 16 \cdot \delta(t) = 16 \)

    (Using the property \( \mathcal{L} \{ 1 \} = \frac{1}{s} \))

  Therefore, the inverse Laplace transform of \( 2s^2 - 10s + 16 \) is:

  \[

  \mathcal{L}^{-1} \{ 2s^2 - 10s + 16 \} = 2t^2 - 10\delta'(t) + 16\delta(t)

  \]

  This represents a combination of a polynomial term and two Dirac delta function terms.

### (2) \( F(s) = \frac{4s^2 + 2s}{2s + 5} \)

1. **Partial Fraction Decomposition:**

  To find the inverse Laplace transform, first decompose \( F(s) \) into partial fractions:

  \[

  F(s) = \frac{4s^2 + 2s}{2s + 5} = \frac{4s^2 + 2s}{2(s + \frac{5}{2})}

  \]

  Perform partial fraction decomposition:

  \[

  \frac{4s^2 + 2s}{2(s + \frac{5}{2})} = \frac{A}{s + \frac{5}{2}} + \frac{B}{2}

  \]

  Multiply through by \( 2(s + \frac{5}{2}) \):

  \[

  4s^2 + 2s = A \cdot 2 + B \cdot (s + \frac{5}{2})

  \]

  \[

  4s^2 + 2s = 2A + Bs + \frac{5B}{2}

  \]

  Equate coefficients:

  \[

  \begin{cases}

  4 = B \\

  2 = 2A + \frac{5B}{2}

  \end{cases}

  \]

  Solve for \( A \) and \( B \):

  - From \( B = 4 \),

  - Substitute \( B = 4 \) into the second equation:

    \[

    2 = 2A + \frac{5 \cdot 4}{2} = 2A + 10

    \]

    \[

    2A = -8

    \]

    \[

    A = -4

    \]

  So, \( F(s) \) decomposes as:

  \[

  F(s) = \frac{-4}{s + \frac{5}{2}} + \frac{4}{2}

  \]

  \[

  F(s) = -\frac{4}{s + \frac{5}{2}} + 2

  \]

2. **Inverse Laplace Transform:**

  Now, take the inverse Laplace transform of each term:

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

    (Using \( \mathcal{L} \left\{ e^{at} \right\} = \frac{1}{s - a} \) with \( a = -\frac{5}{2} \))

  - \( \mathcal{L}^{-1} \left\{ 2 \right\} = 2 \delta(t) \)

    (Using \( \mathcal{L} \left\{ 1 \right\} = \frac{1}{s} \))

  Therefore, the inverse Laplace transform of \( F(s) = \frac{4s^2 + 2s}{2s + 5} \) is:

  \[

  \mathcal{L}^{-1} \left\{ \frac{4s^2 + 2s}{2s + 5} \right\} = -4 e^{-\frac{5}{2}t} + 2 \delta(t)

  \]

### Summary:

- For \( 2s^2 - 10s + 16 \), the inverse Laplace transform is \( 2t^2 - 10\delta'(t) + 16\delta(t) \).

- For \( \frac{4s^2 + 2s}{2s + 5} \), the inverse Laplace transform is \( -4 e^{-\frac{5}{2}t} + 2 \delta(t) \).

These solutions demonstrate the step-by-step process of finding the inverse Laplace transform for each given function \( F(s) \).

Answer with Step-by-step explanation: F(s) = 2s + 2s^2 - 10s + 16):

Applying inverse Laplace transform: [f(t) = \mathcal{L}^{-1}{F(s)} = 2\delta(t) + 2e^{5t} - 5e^{2t} + 8e^{4t}]

(F(s) = \frac{4s}{2s+5} + 2s):

Applying inverse Laplace transform: [f(t) = \mathcal{L}^{-1}{F(s)} = 2e^{-5t} + 2t]

(F(s) = \ln(s-3s+3)):

This function is not in a standard form for inverse Laplace transform. We can’t directly apply the table of Laplace transforms. You might need to use other methods or consult specific tables for logarithmic functions.

(F(s) = s + 5\left(\frac{(s+1)2}{(s+2)2}\right)):

Applying inverse Laplace transform: [f(t) = \mathcal{L}^{-1}{F(s)} = e^{-t} + 5te^{-2t} + 10te^{-2t}]