Answer :
Let's solve the problem of finding the inverse Laplace transforms of the given functions.
### Problem 1: Find the Inverse Laplace Transform of [tex]\(\frac{7s + 15}{s^2 + 2}\)[/tex]
The given function in the Laplace domain is [tex]\( \frac{7s + 15}{s^2 + 2} \)[/tex].
#### Step-by-Step Solution:
1. Identify parts of the numerator that correspond to known Laplace transforms:
- Notice that [tex]\(\frac{7s}{s^2 + 2}\)[/tex] and [tex]\(\frac{15}{s^2 + 2}\)[/tex] can be treated separately.
- The term [tex]\(\frac{7s}{s^2 + 2}\)[/tex] is in the form of the Laplace transform of [tex]\(\cos(\sqrt{2}t)\)[/tex] which transforms to [tex]\(\frac{s}{s^2 + (\sqrt{2})^2}\)[/tex].
- The term [tex]\(\frac{15}{s^2 + 2}\)[/tex] is in the form of the Laplace transform of [tex]\(\sin(\sqrt{2}t)\)[/tex] which transforms to [tex]\(\frac{\sqrt{2}}{s^2 + (\sqrt{2})^2}\)[/tex]. However, our coefficient is 15 instead of [tex]\(\sqrt{2}\)[/tex], hence it requires some adjustment.
2. Compute the inverse Laplace transform separately:
- For [tex]\(\frac{7s}{s^2 + 2}\)[/tex]:
[tex]\[ \mathcal{L}^{-1}\left\{\frac{7s}{s^2 + 2}\right\} = 7\cos(\sqrt{2}t) \][/tex]
- For [tex]\(\frac{15}{s^2 + 2}\)[/tex]:
[tex]\[ \mathcal{L}^{-1}\left\{\frac{15}{s^2 + 2}\right\} = \frac{15}{\sqrt{2}}\sin(\sqrt{2}t) = \frac{15\sqrt{2}}{2}\sin(\sqrt{2}t) \][/tex]
3. Combine results:
- The overall inverse Laplace transform is:
[tex]\[ \mathcal{L}^{-1}\left\{\frac{7s + 15}{s^2 + 2}\right\} = 7\cos(\sqrt{2}t) + \frac{15\sqrt{2}}{2}\sin(\sqrt{2}t) \][/tex]
Therefore, the inverse Laplace transform of [tex]\(\frac{7s + 15}{s^2 + 2}\)[/tex] is:
[tex]\[ (15\sqrt{2}/2 \sin(\sqrt{2}t) + 7\cos(\sqrt{2}t))\Heaviside(t) \][/tex]
### Problem 2: Find the Inverse Laplace Transform of [tex]\(\frac{2}{(s+1)(s+2)^2}\)[/tex]
The given function in the Laplace domain is [tex]\(\frac{2}{(s + 1)(s + 2)^2}\)[/tex].
#### Step-by-Step Solution:
1. Decompose the function using partial fraction decomposition:
- We express [tex]\(\frac{2}{(s + 1)(s + 2)^2}\)[/tex] in terms of its partial fractions.
- Let [tex]\(\frac{2}{(s + 1)(s + 2)^2} = \frac{A}{s + 1} + \frac{B}{s + 2} + \frac{C}{(s + 2)^2}\)[/tex]
2. Solve for the constants [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
- Multiply both sides by [tex]\((s + 1)(s + 2)^2\)[/tex] to get:
[tex]\[ 2 = A(s + 2)^2 + B(s + 1)(s + 2) + C(s + 1) \][/tex]
- Substitute [tex]\(s = -1\)[/tex]: [tex]\(2 = C \cdot 1 \Rightarrow C = 2\)[/tex]
- Substitute [tex]\(s = -2\)[/tex]: [tex]\(2 = A \cdot 1 \Rightarrow A = 2\)[/tex]
- Differentiate both sides:
[tex]\[ 0 = 2A(s + 2) + B(s + 2 + s + 1) + C \Rightarrow 0 = 2A(s + 2) + B(2s + 3) + C \][/tex]
- Solve for [tex]\(B\)[/tex]:
[tex]\[ \Rightarrow B = -6 \][/tex]
3. Construct the decomposed form:
- The decomposed form is:
[tex]\[ \frac{2}{(s+1)(s+2)^2} = \frac{2}{s+1} - \frac{6}{(s+2)^2} + \frac{2}{s + 2 } \][/tex]
4. Compute the inverse Laplace transform of each term:
- For [tex]\(\frac{2}{s+1}\)[/tex]:
[tex]\[ \mathcal{L}^{-1} \left\{\frac{2}{s+1}\right\} = 2e^{-t} \][/tex]
- For [tex]\(\frac{2}{s+2}\)[/tex]:
[tex]\[ \mathcal{L}^{-1} \left\{\frac{2}{s+2}\right\} = 2e^{-2t} \][/tex]
- For [tex]\(\frac{-6}{(s+2)^2}\)[/tex]:
[tex]\[ \mathcal{L}^{-1} \left\{\frac{-6}{(s+2)^2}\right\} = -6te^{-2t} \][/tex]
5. Combine results:
- The overall inverse Laplace transform is:
[tex]\[ 2e^{-t} + 2e^{-2t} - 6te^{-2t} \][/tex]
Therefore, the inverse Laplace transform of [tex]\(\frac{2}{(s+1)(s+2)^2}\)[/tex] is:
[tex]\[ (2e^{-t} + 2e^{-2t} - 6te^{-2t}) \Heaviside(t) \][/tex]
### Final Answer:
1. The inverse Laplace transform of [tex]\( \frac{7s + 15}{s^2 + 2} \)[/tex] is:
[tex]\[ (15\sqrt{2}\sin(\sqrt{2}t)/2 + 7\cos(\sqrt{2}t))\Heaviside(t) \][/tex]
2. The inverse Laplace transform of [tex]\( \frac{2}{(s + 1)(s + 2)^2} \)[/tex] is:
[tex]\[ -2te^{-2t} + 2e^{-t}- 2e^{-2t}) \Heaviside(t) \][/tex]
### Problem 1: Find the Inverse Laplace Transform of [tex]\(\frac{7s + 15}{s^2 + 2}\)[/tex]
The given function in the Laplace domain is [tex]\( \frac{7s + 15}{s^2 + 2} \)[/tex].
#### Step-by-Step Solution:
1. Identify parts of the numerator that correspond to known Laplace transforms:
- Notice that [tex]\(\frac{7s}{s^2 + 2}\)[/tex] and [tex]\(\frac{15}{s^2 + 2}\)[/tex] can be treated separately.
- The term [tex]\(\frac{7s}{s^2 + 2}\)[/tex] is in the form of the Laplace transform of [tex]\(\cos(\sqrt{2}t)\)[/tex] which transforms to [tex]\(\frac{s}{s^2 + (\sqrt{2})^2}\)[/tex].
- The term [tex]\(\frac{15}{s^2 + 2}\)[/tex] is in the form of the Laplace transform of [tex]\(\sin(\sqrt{2}t)\)[/tex] which transforms to [tex]\(\frac{\sqrt{2}}{s^2 + (\sqrt{2})^2}\)[/tex]. However, our coefficient is 15 instead of [tex]\(\sqrt{2}\)[/tex], hence it requires some adjustment.
2. Compute the inverse Laplace transform separately:
- For [tex]\(\frac{7s}{s^2 + 2}\)[/tex]:
[tex]\[ \mathcal{L}^{-1}\left\{\frac{7s}{s^2 + 2}\right\} = 7\cos(\sqrt{2}t) \][/tex]
- For [tex]\(\frac{15}{s^2 + 2}\)[/tex]:
[tex]\[ \mathcal{L}^{-1}\left\{\frac{15}{s^2 + 2}\right\} = \frac{15}{\sqrt{2}}\sin(\sqrt{2}t) = \frac{15\sqrt{2}}{2}\sin(\sqrt{2}t) \][/tex]
3. Combine results:
- The overall inverse Laplace transform is:
[tex]\[ \mathcal{L}^{-1}\left\{\frac{7s + 15}{s^2 + 2}\right\} = 7\cos(\sqrt{2}t) + \frac{15\sqrt{2}}{2}\sin(\sqrt{2}t) \][/tex]
Therefore, the inverse Laplace transform of [tex]\(\frac{7s + 15}{s^2 + 2}\)[/tex] is:
[tex]\[ (15\sqrt{2}/2 \sin(\sqrt{2}t) + 7\cos(\sqrt{2}t))\Heaviside(t) \][/tex]
### Problem 2: Find the Inverse Laplace Transform of [tex]\(\frac{2}{(s+1)(s+2)^2}\)[/tex]
The given function in the Laplace domain is [tex]\(\frac{2}{(s + 1)(s + 2)^2}\)[/tex].
#### Step-by-Step Solution:
1. Decompose the function using partial fraction decomposition:
- We express [tex]\(\frac{2}{(s + 1)(s + 2)^2}\)[/tex] in terms of its partial fractions.
- Let [tex]\(\frac{2}{(s + 1)(s + 2)^2} = \frac{A}{s + 1} + \frac{B}{s + 2} + \frac{C}{(s + 2)^2}\)[/tex]
2. Solve for the constants [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
- Multiply both sides by [tex]\((s + 1)(s + 2)^2\)[/tex] to get:
[tex]\[ 2 = A(s + 2)^2 + B(s + 1)(s + 2) + C(s + 1) \][/tex]
- Substitute [tex]\(s = -1\)[/tex]: [tex]\(2 = C \cdot 1 \Rightarrow C = 2\)[/tex]
- Substitute [tex]\(s = -2\)[/tex]: [tex]\(2 = A \cdot 1 \Rightarrow A = 2\)[/tex]
- Differentiate both sides:
[tex]\[ 0 = 2A(s + 2) + B(s + 2 + s + 1) + C \Rightarrow 0 = 2A(s + 2) + B(2s + 3) + C \][/tex]
- Solve for [tex]\(B\)[/tex]:
[tex]\[ \Rightarrow B = -6 \][/tex]
3. Construct the decomposed form:
- The decomposed form is:
[tex]\[ \frac{2}{(s+1)(s+2)^2} = \frac{2}{s+1} - \frac{6}{(s+2)^2} + \frac{2}{s + 2 } \][/tex]
4. Compute the inverse Laplace transform of each term:
- For [tex]\(\frac{2}{s+1}\)[/tex]:
[tex]\[ \mathcal{L}^{-1} \left\{\frac{2}{s+1}\right\} = 2e^{-t} \][/tex]
- For [tex]\(\frac{2}{s+2}\)[/tex]:
[tex]\[ \mathcal{L}^{-1} \left\{\frac{2}{s+2}\right\} = 2e^{-2t} \][/tex]
- For [tex]\(\frac{-6}{(s+2)^2}\)[/tex]:
[tex]\[ \mathcal{L}^{-1} \left\{\frac{-6}{(s+2)^2}\right\} = -6te^{-2t} \][/tex]
5. Combine results:
- The overall inverse Laplace transform is:
[tex]\[ 2e^{-t} + 2e^{-2t} - 6te^{-2t} \][/tex]
Therefore, the inverse Laplace transform of [tex]\(\frac{2}{(s+1)(s+2)^2}\)[/tex] is:
[tex]\[ (2e^{-t} + 2e^{-2t} - 6te^{-2t}) \Heaviside(t) \][/tex]
### Final Answer:
1. The inverse Laplace transform of [tex]\( \frac{7s + 15}{s^2 + 2} \)[/tex] is:
[tex]\[ (15\sqrt{2}\sin(\sqrt{2}t)/2 + 7\cos(\sqrt{2}t))\Heaviside(t) \][/tex]
2. The inverse Laplace transform of [tex]\( \frac{2}{(s + 1)(s + 2)^2} \)[/tex] is:
[tex]\[ -2te^{-2t} + 2e^{-t}- 2e^{-2t}) \Heaviside(t) \][/tex]
Answer:
### 1. Find the inverse Laplace transform (ILT) of \(\frac{7}{s^4}\)
To find the inverse Laplace transform of \(\frac{7}{s^4}\), use the formula for the inverse Laplace transform of \(\frac{1}{s^n}\), which is \(\frac{t^{n-1}}{(n-1)!}\).
In this case, \(n = 4\), so:
\[
\mathcal{L}^{-1}\left\{\frac{1}{s^4}\right\} = \frac{t^{4-1}}{(4-1)!} = \frac{t^3}{6}
\]
Thus,
\[
\mathcal{L}^{-1}\left\{\frac{7}{s^4}\right\} = 7 \cdot \frac{t^3}{6} = \frac{7t^3}{6}
\]
### 2. Find the inverse Laplace transform (ILT) of \(\frac{7s + 15}{s^2 + 2s + 7}\)
To find the inverse Laplace transform, first simplify the expression by completing the square for the quadratic term in the denominator:
\[
s^2 + 2s + 7 = (s+1)^2 + 6
\]
Thus, the expression becomes:
\[
\frac{7s + 15}{(s+1)^2 + 6}
\]
Separate the numerator into terms that fit the standard Laplace transform forms:
\[
7s + 15 = 7(s+1) + 8
\]
So,
\[
\frac{7s + 15}{(s+1)^2 + 6} = \frac{7(s+1) + 8}{(s+1)^2 + 6} = \frac{7(s+1)}{(s+1)^2 + 6} + \frac{8}{(s+1)^2 + 6}
\]
The inverse Laplace transforms are:
- For \(\frac{7(s+1)}{(s+1)^2 + 6}\), use the fact that \(\mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^2 + b^2}\right\} = e^{at} \cos(bt)\). Here, \(a = -1\) and \(b = \sqrt{6}\). Thus, the inverse transform is \(7e^{-t} \cos(\sqrt{6}t)\).
- For \(\frac{8}{(s+1)^2 + 6}\), use the fact that \(\mathcal{L}^{-1}\left\{\frac{b}{(s-a)^2 + b^2}\right\} = \frac{b}{\sqrt{b^2}} e^{at} \sin(\sqrt{b^2}t)\). Here, \(a = -1\), \(b = \sqrt{6}\), and \(b = 8\). Thus, the inverse transform is \(\frac{8}{\sqrt{6}} e^{-t} \sin(\sqrt{6}t)\).
Combining these, the inverse Laplace transform is:
\[
\mathcal{L}^{-1}\left\{\frac{7s + 15}{s^2 + 2s + 7}\right\} = 7e^{-t} \cos(\sqrt{6}t) + \frac{8}{\sqrt{6}} e^{-t} \sin(\sqrt{6}t)
\]
### 3. Find the inverse Laplace transform (ILT) of \(\frac{2}{(s + 1)(s + 2)^2}\)
To find the inverse Laplace transform, use partial fraction decomposition:
\[
\frac{2}{(s + 1)(s + 2)^2} = \frac{A}{s + 1} + \frac{B}{s + 2} + \frac{C}{(s + 2)^2}
\]
Multiply through by the denominator \((s + 1)(s + 2)^2\) and solve for \(A\), \(B\), and \(C\):
\[
2 = A(s + 2)^2 + B(s + 1)(s + 2) + C(s + 1)
\]
Substitute suitable values for \(s\) to solve for \(A\), \(B\), and \(C\):
1. Let \(s = -1\):
\[
2 = A(1) \implies A = 2
\]
2. Let \(s = -2\):
\[
2 = C(-1) \implies C = -2
\]
3. Expand and compare coefficients to find \(B\):
\[
2 = 2(s^2 + 4s + 4) + B(s^2 - s - 2) - 2(s + 1)
\]
\[
2 = 2s^2 + 8s + 8 + Bs^2 - Bs - 2s - 2
\]
\[
2 = (2 + B)s^2 + (8 - B - 2)s + (8 - 2)
\]
\[
2 = (2 + B)s^2 + (6 - B)s + 6
\]
Equating coefficients with \(2s^2 + 0s + 2\):
\[
2 + B = 0 \implies B = -2
\]
\[
6 - B = 0 \implies B = 6
\]
Correcting calculation errors, \(B = -4\).
Thus, the partial fractions are:
\[
\frac{2}{(s + 1)(s + 2)^2} = \frac{2}{s + 1} - \frac{4}{s + 2} + \frac{-2}{(s + 2)^2}
\]
Take the inverse Laplace transform of each term:
\[
\mathcal{L}^{-1}\left\{\frac{2}{s + 1}\right\} = 2e^{-t}
\]
\[
\mathcal{L}^{-1}\left\{\frac{-4}{s + 2}\right\} = -4e^{-2t}
\]
\[
\mathcal{L}^{-1}\left\{\frac{-2}{(s + 2)^2}\right\} = -2te^{-2t}
\]
Thus, the inverse Laplace transform is:
\[
\mathcal{L}^{-1}\left\{\frac{2}{(s + 1)(s + 2)^2}\right\} = 2e^{-t} - 4e^{-2t} - 2te^{-2t}
\]
