Let's focus on the problems you specified: **2, 6, 12, 18, 20**. I'll use Laplace transform properties and, for 18 & 20, invert the function using formula (6.7) (which is probably the Laplace **inverse table** or partial fraction technique). --- ## **2.** \(\mathcal{L}\left\{e^{-2t} \sin(3t)\right\}\) Recall: \(\mathcal{L}\{ e^{at} f(t)\} = F(s-a)\) \(\mathcal{L}\{\sin(bt)\} = \frac{b}{s^2 + b^2}\) So, \[ \mathcal{L}\left\{e^{-2t} \sin(3t)\right\} = \frac{3}{(s+2)^2 + 9} \] --- ## **6.** \(\mathcal{L}\left\{2t^2 - t + 1\right\}\) Recall: \(\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}\) So, - \(\mathcal{L}\{t^2\} = \frac{2}{s^3}\) - \(\mathcal{L}\{t\} = \frac{1}{s^2}\) - \(\mathcal{L}\{1\} = \frac{1}{s}\) So, \[ \mathcal{L}\left\{2t^2 - t + 1\right\} = 2\left(\frac{2}{s^3}\right) - \frac{1}{s^2} + \frac{1}{s} = \frac{4}{s^3} - \frac{1}{s^2} + \frac{1}{s} \] --- ## **12.** Not directly visible in your images, but based on context, let's skip this unless you clarify! --- ## **18.** Given: \( G(s) = \frac{2s+3}{s^2 + 2s + 26} \) **Find the inverse Laplace transform using (6.7).** First, rewrite the denominator: \( s^2 + 2s + 26 = (s+1)^2 + 25 \) So, \[ G(s) = \frac{2s+3}{(s+1)^2 + 5^2} \] Express numerator in terms of \(s+1\): \[ 2s + 3 = 2(s+1) + 1 \] So, \[ G(s) = \frac{2(s+1) + 1}{(s+1)^2 + 5^2} = \frac{2(s+1)}{(s+1)^2 + 25} + \frac{1}{(s+1)^2 + 25} \] Recall: - \(\mathcal{L}\{ e^{-at} \cos(\omega t) \} = \frac{s+a}{(s+a)^2 + \omega^2}\) - \(\mathcal{L}\{ e^{-at} \sin(\omega t) \} = \frac{\omega}{(s+a)^2 + \omega^2}\) So, \[ \mathcal{L}^{-1}\{G(s)\} = 2 e^{-t} \cos(5t) + \frac{1}{5} e^{-t} \sin(5t) \] --- ## **20.** Given: \( x'' + 2x' + 26x = 0, \quad x(0)=2, \quad x'(0)=-1 \) Take Laplace transform: \[ \mathcal{L}\{x''\} + 2\mathcal{L}\{x'\} + 26\mathcal{L}\{x\} = 0 \] \[ s^2 X(s) - s x(0) - x'(0) + 2[s X(s) - x(0)] + 26 X(s) = 0 \] Plug in initial conditions: \(x(0) = 2,\, x'(0) = -1\) \[ s^2 X(s) - 2s + 1 + 2[s X(s) - 2] + 26 X(s) = 0 \] \[ (s^2 + 2s + 26) X(s) - 2s + 1 - 4 = 0 \] \[ (s^2 + 2s + 26) X(s) = 2s + 3 \] \[ X(s) = \frac{2s+3}{s^2 + 2s + 26} \] But that's exactly the \(G(s)\) from 18! So, the solution is: \[ x(t) = 2 e^{-t} \cos(5t) + \frac{1}{5} e^{-t} \sin(5t) \] --- **If you need number 12 or more details, let me know!**

Certainly! Here are the final answers for the specified problems: --- ### **2. Laplace Transform of \( e^{-2t} \sin(3t) \):** \[ \boxed{ \mathcal{L}\left\{ e^{-2t} \sin(3t) \right\} = \frac{3}{(s+2)^2 + 9} } \] --- ### **6. Laplace Transform of \( 2t^2 - t + 1 \):** \[ \boxed{ \mathcal{L}\left\{ 2t^2 - t + 1 \right\} = \frac{4}{s^3} - \frac{1}{s^2} + \frac{1}{s} } \] --- ### **12.** *Please specify the function or context for problem 12 if you need its transform or solution.* --- ### **18. Inverse Laplace Transform of \( G(s) = \frac{2s+3}{(s+1)^2 + 25} \):** \[ \boxed{ \mathcal{L}^{-1}\left\{ G(s) \right\} = 2 e^{-t} \cos(5t) + \frac{1}{5} e^{-t} \sin(5t) } \] --- ### **20. Solution to \( x'' + 2x' + 26x = 0, \quad x(0)=2, \quad x'(0)=-1 \):** The Laplace transform method yields: \[ X(s) = \frac{2s+3}{s^2 + 2s + 26} \] Inverse transforming gives: \[ \boxed{ x(t) = 2 e^{-t} \cos(5t) + \frac{1}{5} e^{-t} \sin(5t) } \] --- If you need any further clarification or details on other problems, feel free to ask!