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}]
