The image contains a list of Laplace transform problems labeled "Selesai transformasi Laplace berikut ini", which means "Solve the following Laplace transforms" in Indonesian. The problems are as follows:

1. $\mathcal{L} \{ t^3 e^{-7t} \}$
2. $\mathcal{L} \{ t e^{10t} \}$
3. $\mathcal{L} \{ \sin(5t) e^{-7t} \}$
4. $\mathcal{L} \{ e^{-4t} \sin(2t) \}$
5. $\mathcal{L} \left\{ t \left(e^t + e^{2t}\right)^2 \right\}$
6. $\mathcal{L} \left\{ \left(1 - e^t + 2 e^{-3t}\right) \cos(6t) \right\}$

I'll solve each of these transforms one by one.

1. Problem: $\mathcal{L} \{ t^3 e^{-7t} \}$

   The Laplace transform of $t^n e^{-at}$ is: $\mathcal{L} \{ t^n e^{-at} \} = \frac{n!}{(s + a)^{n+1}}$ For $t^3 e^{-7t}$: $\mathcal{L} \{ t^3 e^{-7t} \} = \frac{3!}{(s + 7)^4} = \frac{6}{(s + 7)^4}$

2. Problem: $\mathcal{L} \{ t e^{10t} \}$

   Using the general formula $\mathcal{L} \{ t^n e^{at} \} = \frac{n!}{(s - a)^{n+1}}$ for $t e^{at}$: $\mathcal{L} \{ t e^{10t} \} = \frac{1!}{(s - 10)^2} = \frac{1}{(s - 10)^2}$

3. Problem: $\mathcal{L} \{ \sin(5t) e^{-7t} \}$

   The Laplace transform of $\sin(kt) e^{-at}$ is: $\mathcal{L} \{ \sin(kt) e^{-at} \} = \frac{k}{(s + a)^2 + k^2}$ For $\sin(5t) e^{-7t}$: $\mathcal{L} \{ \sin(5t) e^{-7t} \} = \frac{5}{(s + 7)^2 + 25}$ Simplifying the denominator: $\frac{5}{(s + 7)^2 + 25}$

4. Problem: $\mathcal{L} \{ e^{-4t} \sin(2t) \}$

   Again, using the same formula for $\sin(kt) e^{-at}$: $\mathcal{L} \{ e^{-4t} \sin(2t) \} = \frac{2}{(s + 4)^2 + 4}$ Simplifying: $\frac{2}{(s + 4)^2 + 4}$

5. Problem: $\mathcal{L} \left\{ t \left(e^t + e^{2t}\right)^2 \right\}$

   First, expand $\left(e^t + e^{2t}\right)^2$: $\left(e^t + e^{2t}\right)^2 = e^{2t} + 2e^{3t} + e^{4t}$ Now, find the Laplace transform of $t \left( e^{2t} + 2e^{3t} + e^{4t} \right)$: $\mathcal{L} \left\{ t e^{2t} \right\} = \frac{1}{(s - 2)^2}, \quad \mathcal{L} \left\{ t e^{3t} \right\} = \frac{1}{(s - 3)^2}, \quad \mathcal{L} \left\{ t e^{4t} \right\} = \frac{1}{(s - 4)^2}$ So: $\mathcal{L} \left\{ t \left(e^t + e^{2t}\right)^2 \right\} = \frac{1}{(s - 2)^2} + 2 \frac{1}{(s - 3)^2} + \frac{1}{(s - 4)^2}$

6. Problem: $\mathcal{L} \left\{ \left(1 - e^t + 2 e^{-3t}\right) \cos(6t) \right\}$

   Break it into three parts: $\mathcal{L} \{ \cos(6t) \}, \quad \mathcal{L} \{ e^t \cos(6t) \}, \quad \mathcal{L} \{ e^{-3t} \cos(6t) \}$ Using the Laplace transform of $\cos(kt) e^{at}$: $\mathcal{L} \{ \cos(kt) \} = \frac{s}{s^2 + k^2}, \quad \mathcal{L} \{ e^t \cos(6t) \} = \frac{s - 1}{(s - 1)^2 + 36}, \quad \mathcal{L} \{ e^{-3t} \cos(6t) \} = \frac{s + 3}{(s + 3)^2 + 36}$ Thus: $\mathcal{L} \left\{ \left(1 - e^t + 2 e^{-3t}\right) \cos(6t) \right\} = \frac{s}{s^2 + 36} - \frac{s - 1}{(s - 1)^2 + 36} + 2 \frac{s + 3}{(s + 3)^2 + 36}$

Let me know if you'd like further details or clarifications. Here are five related questions:

1. How is the Laplace transform used to solve differential equations?
2. What is the significance of shifting in the Laplace domain?
3. How do inverse Laplace transforms work for solving initial value problems?
4. What are some properties of the Laplace transform that simplify