Let's solve **Problem 1** step by step: ### 1. Given: \[ F(s) = \frac{2s^2 + 5}{s^2 + 3s + 6} \] #### **Step 1: Factor the denominator if possible** The denominator is \( s^2 + 3s + 6 \). This does **not factor nicely** with real roots. Let's complete the square: \[ s^2 + 3s + 6 = (s^2 + 3s + \frac{9}{4}) + 6 - \frac{9}{4} = (s + \frac{3}{2})^2 + \left(6 - \frac{9}{4}\right) = (s + \frac{3}{2})^2 + \frac{15}{4} \] #### **Step 2: Write numerator in terms of the denominator's completed square form** Let \( x = s + \frac{3}{2} \), then \( s = x - \frac{3}{2} \): \[ 2s^2 + 5 = 2(s^2) + 5 \] But let's express it as \( A(s^2 + 3s + 6) + B s + C \): Assume \[ 2s^2 + 5 = A(s^2 + 3s + 6) + Bs + C \] Expand: \[ 2s^2 + 5 = A s^2 + 3A s + 6A + B s + C \] \[ = A s^2 + (3A + B)s + (6A + C) \] Set coefficients equal: - \( s^2 \): \( 2 = A \) ⇒ \( A = 2 \) - \( s \): \( 0 = 3A + B \) ⇒ \( 0 = 3 \cdot 2 + B \Rightarrow B = -6 \) - constant: \( 5 = 6A + C \Rightarrow 5 = 12 + C \Rightarrow C = -7 \) So, \[ 2s^2 + 5 = 2(s^2 + 3s + 6) - 6s - 7 \] Therefore, \[ F(s) = \frac{2s^2 + 5}{s^2 + 3s + 6} = 2 - \frac{6s + 7}{s^2 + 3s + 6} \] #### **Step 3: Inverse Laplace Transform** Recall: - \( \mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\} = e^{-a t} \) - \( \mathcal{L}^{-1}\left\{\frac{s+a}{(s+a)^2 + b^2}\right\} = e^{-a t} \cos(bt) \) - \( \mathcal{L}^{-1}\left\{\frac{b}{(s+a)^2 + b^2}\right\} = e^{-a t} \sin(bt) \) Let’s write denominator as: \[ s^2 + 3s + 6 = (s + \frac{3}{2})^2 + \frac{15}{4} \] Let’s express numerator: \[ -6s - 7 = -6(s + \frac{3}{2}) + (2) \] So, \[ F(s) = 2 - \frac{-6(s + \frac{3}{2}) + 2}{(s + \frac{3}{2})^2 + (\sqrt{15}/2)^2} \] \[ = 2 + \frac{6(s + \frac{3}{2}) - 2}{(s + \frac{3}{2})^2 + (\sqrt{15}/2)^2} \] Split numerator: \[ = 2 + \frac{6(s + \frac{3}{2})}{(s + \frac{3}{2})^2 + (\sqrt{15}/2)^2} - \frac{2}{(s + \frac{3}{2})^2 + (\sqrt{15}/2)^2} \] #### **Step 4: Write the inverse Laplace** - \( \mathcal{L}^{-1}\left\{ \frac{s+a}{(s+a)^2 + b^2} \right\} = e^{-a t} \cos(bt) \) - \( \mathcal{L}^{-1}\left\{ \frac{b}{(s+a)^2 + b^2} \right\} = e^{-a t} \sin(bt) \) Here, \( a = \frac{3}{2} \), \( b = \frac{\sqrt{15}}{2} \): So, - \( 6(s + \frac{3}{2}) \): coefficient for cosine - \( -2 \): for sine (since \( b \) is in denominator, write as \( 2 = (\sqrt{15}) \cdot \left( \frac{2}{\sqrt{15}} \right) \)) So, \[ \frac{2}{(s + \frac{3}{2})^2 + (\sqrt{15}/2)^2} = \frac{2}{(\sqrt{15}/2)} \cdot \frac{\sqrt{15}/2}{(s + \frac{3}{2})^2 + (\sqrt{15}/2)^2} \] \[ = \frac{4}{\sqrt{15}} \cdot \frac{\sqrt{15}/2}{(s + \frac{3}{2})^2 + (\sqrt{15}/2)^2} \] So: ### **Final Answer** \[ f(t) = 2 \delta(t) + 6 e^{-\frac{3}{2} t} \cos\left( \frac{\sqrt{15}}{2} t \right) - \frac{4}{\sqrt{15}} e^{-\frac{3}{2} t} \sin\left( \frac{\sqrt{15}}{2} t \right) \] **Or, without the delta function if you ignore the 2 as 2 times the Laplace of 1:** \[ f(t) = 2 + 6 e^{-\frac{3}{2} t} \cos\left( \frac{\sqrt{15}}{2} t \right) - \frac{4}{\sqrt{15}} e^{-\frac{3}{2} t} \sin\left( \frac{\sqrt{15}}{2} t \right) \] --- **If you want the step-by-step solution for other items, let me know which one!**

Certainly! Let's proceed step-by-step to find the inverse Laplace transforms for all the provided functions. --- ### **2. \( F(s) = \frac{5}{(s+1)(s+2)^2} \)** #### **Step 1: Partial fraction decomposition** Express: \[ \frac{5}{(s+1)(s+2)^2} = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{(s+2)^2} \] Multiply both sides by denominator: \[ 5 = A(s+2)^2 + B(s+1)(s+2) + C(s+1) \] Set \( s = -1 \): \[ 5 = A(1)^2 + B(0) + C(0) \Rightarrow A = 5 \] Set \( s = -2 \): \[ 5 = A(0) + B(-1)(0) + C(-1) \Rightarrow 5 = -C \Rightarrow C = -5 \] Now, pick any \( s \), say \( s=0 \): \[ 5 = A(2)^2 + B(1)(2) + C(1) \Rightarrow 5 = 4A + 2B + C \] Plugging \( A=5 \), \( C=-5 \): \[ 5 = 4 \times 5 + 2B - 5 \Rightarrow 5 = 20 + 2B - 5 \Rightarrow 5 = 15 + 2B \] \[ 2B = -10 \Rightarrow B = -5 \] #### **Step 2: Write partial fractions** \[ F(s) = \frac{5}{s+1} - \frac{5}{s+2} - \frac{5}{(s+2)^2} \] #### **Step 3: Inverse Laplace transforms** - \( \mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\} = e^{-a t} \) - \( \mathcal{L}^{-1}\left\{\frac{1}{(s+a)^2}\right\} = t e^{-a t} \) Thus: \[ f(t) = 5 e^{-t} - 5 e^{-2 t} - 5 t e^{-2 t} \] --- ### **3. \( F(s) = \frac{3}{(s+1)(s+2)^2} \)** This is similar to problem 2, with numerator 3 instead of 5. **Partial fractions:** \[ \frac{3}{(s+1)(s+2)^2} = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{(s+2)^2} \] Multiply through: \[ 3 = A(s+2)^2 + B(s+1)(s+2) + C(s+1) \] Using previous calculations: - At \( s=-1 \): \[ 3 = A(1)^2 \Rightarrow A=3 \] - At \( s=-2 \): \[ 3 = - C \Rightarrow C= -3 \] - At \( s=0 \): \[ 3= 4A + 2B + C = 4 \times 3 + 2B -3 = 12 + 2B -3 \] \[ 3= 9 + 2B \Rightarrow 2B= -6 \Rightarrow B= -3 \] **Partial fractions:** \[ F(s) = \frac{3}{s+1} - \frac{3}{s+2} - \frac{3}{(s+2)^2} \] **Inverse Laplace:** \[ f(t) = 3 e^{-t} - 3 e^{-2 t} - 3 t e^{-2 t} \] --- ### **4. \( F(s) = \frac{5}{s^2 + 5s + 4} \)** #### **Step 1: Factor denominator** \[ s^2 + 5s + 4 = (s+4)(s+1) \] Partial fractions: \[ \frac{5}{(s+4)(s+1)} = \frac{A}{s+4} + \frac{B}{s+1} \] Multiply through: \[ 5 = A(s+1) + B(s+4) \] Set \( s=-1 \): \[ 5= A(0) + B(3) \Rightarrow B= \frac{5}{3} \] Set \( s=-4 \): \[ 5= A(-3) + B(0) \Rightarrow A= -\frac{5}{3} \] **Partial fractions:** \[ F(s) = -\frac{5}{3} \frac{1}{s+4} + \frac{5}{3} \frac{1}{s+1} \] **Inverse Laplace:** \[ f(t) = -\frac{5}{3} e^{-4 t} + \frac{5}{3} e^{-t} \] --- ### **5. \( F(s) = \frac{8}{(s+3)^3} \)** Recall: \[ \mathcal{L}^{-1}\left\{\frac{n!}{(s+a)^{n+1}}\right\} = t^{n} e^{-a t} \] Here: \[ F(s) = \frac{8}{(s+3)^3} = \frac{2! \times 4}{(s+3)^3} \] Since \( 2! = 2 \), and numerator is 8: \[ 8 = 2 \times 4 \Rightarrow \text{but better to write as:} \] Alternatively, note: \[ \mathcal{L}^{-1}\left\{\frac{n!}{(s+a)^{n+1}}\right\} = t^{n} e^{-a t} \] So, \[ f(t) = \frac{8}{(s+3)^3} \Rightarrow \text{corresponds to} \quad n=2, \quad \text{since} \quad (s+3)^{3} = (s+3)^{n+1} \] And numerator: \[ n! = 2! = 2 \] But numerator is 8, so: \[ \frac{8}{(s+3)^3} = 4 \times \frac{2}{(s+3)^3} \] Inverse Laplace: \[ f(t) = 4 t^{2} e^{-3 t} \] --- ### **6. \( F(s) = \frac{4}{(s+5)^2} \)** Recall: \[ \mathcal{L}^{-1}\left\{\frac{n!}{(s+a)^{n+1}}\right\} = t^n e^{-a t} \] Here: \[ \frac{4}{(s+5)^2} = 4 \times \frac{1}{(s+5)^2} \] Since \( n=1 \), \( 1! =1 \): \[ f(t) = 4 t e^{-5 t} \] --- ### **7. \( F(s) = \frac{1}{s} \)** Inverse Laplace transform: \[ f(t) = 1 \] --- ### **8. \( F(s) = \frac{5}{s^2} \)** Recall: \[ \mathcal{L}^{-1}\left\{\frac{1}{s^{n+1}}\right\} = \frac{t^{n}}{n!} \] Here: \[ F(s) = 5 \times \frac{1}{s^2} \] Corresponds to \( n=1 \), so: \[ f(t) = 5 \times t \] --- ### **9. \( F(s) = \frac{1}{s-1} \)** Inverse Laplace: \[ f(t) = e^{t} \] --- --- ## **Summary of Final Results:** | Problem | \( F(s) \) | **Inverse Laplace \( f(t) \)** | |---|---|---| | 1 | \( \frac{2s^2 + 5}{s^2 + 3s + 6} \) | \( 2 + 6 e^{-\frac{3}{2} t} \cos\left( \frac{\sqrt{15}}{2} t \right) - \frac{4}{\sqrt{15}} e^{-\frac{3}{2} t} \sin\left( \frac{\sqrt{15}}{2} t \right) \) | | 2 | \( \frac{5}{(s+1)(s+2)^2} \) | \( 5 e^{-t} - 5 e^{-2 t} - 5 t e^{-2 t} \) | | 3 | \( \frac{3}{(s+1)(s+2)^2} \) | \( 3 e^{-t} - 3 e^{-2 t} - 3 t e^{-2 t} \) | | 4 | \( \frac{5}{s^2 + 5s + 4} \) | \( -\frac{5}{3} e^{-4 t} + \frac{5}{3} e^{-t} \) | | 5 | \( \frac{8}{(s+3)^3} \) | \( 4 t^{2} e^{-3 t} \) | | 6 | \( \frac{4}{(s+5)^2} \) | \( 4 t e^{-5 t} \) | | 7 | \( \frac{1}{s} \) | \( 1 \) | | 8 | \( \frac{5}{s^2} \) | \( 5 t \) | | 9 | \( \frac{1}{s-1} \) | \( e^{t} \) | --- If you'd like detailed steps for any specific problem or further clarifications, feel free to ask!