Give step-by-step solution with explanation and final answer:QUESTION 3
Determine the following:
1 Lfcteon2e].
32 L158),
33 LA)

Question

Give step-by-step solution with explanation and final answer:QUESTION 3
Determine the following:
1 Lfcteon2e].
32 L158),
33 LA)

VIPSolutions · Accepted Answer

Let's tackle each part step by step:

---

### **3.1 $ \mathcal{L}\{e^{4t}\cos(2t)\} $**

**Laplace Transform Formula:**
$$
\mathcal{L}\{e^{at}f(t)\} = F(s-a)
$$
where $ F(s) = \mathcal{L}\{f(t)\} $.

The Laplace transform of $ \cos(\omega t) $ is:
$$
\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}
$$

So, for $ f(t) = \cos(2t) $:
$$
F(s) = \frac{s}{s^2 + 4}
$$

Applying the shifting property with $ a = 4 $:
$$
\mathcal{L}\{e^{4t}\cos(2t)\} = F(s-4) = \frac{s-4}{(s-4)^2 + 4}
$$

**Final Answer:**
$$
\boxed{\mathcal{L}\{e^{4t}\cos(2t)\} = \frac{s-4}{(s-4)^2 + 4}}
$$

---

### **3.2 $ \mathcal{L}^{-1}\left\{\frac{e^{-4s}}{s+1}\right\} $**

**Inverse Laplace Transform Formula:**
$$
\mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\} = e^{-at}
$$

The presence of $ e^{-4s} $ indicates a time delay (shifting property):
$$
\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)
$$

Here,
$$
F(s) = \frac{1}{s+1} \implies f(t) = e^{-t}
$$

So,
$$
\mathcal{L}^{-1}\left\{\frac{e^{-4s}}{s+1}\right\} = e^{-(t-4)}u(t-4)
$$

**Final Answer:**
$$
\boxed{\mathcal{L}^{-1}\left\{\frac{e^{-4s}}{s+1}\right\} = e^{-(t-4)}u(t-4)}
$$

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### **3.3 $ \mathcal{L}^{-1}\left\{ \frac{2(s+1)}{s^2 + 2s + 10} \right\} $**

First, rewrite the denominator:
$$
s^2 + 2s + 10 = (s+1)^2 + 3^2
$$

So,
$$
\frac{2(s+1)}{(s+1)^2 + 3^2}
$$

**Laplace Transform Formula:**
$$
\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}
$$
$$
\mathcal{L}\{e^{-at}f(t)\} = F(s+a)
$$

Let’s compare:
$$
\frac{2(s+1)}{(s+1)^2 + 3^2} = 2 \cdot \frac{s+1}{(s+1)^2 + 3^2}
$$

This matches the Laplace transform of $ e^{-t}\cos(3t) $:

$$
\mathcal{L}\{e^{-t}\cos(3t)\} = \frac{s+1}{(s+1)^2 + 9}
$$

Multiply by 2:

$$
\mathcal{L}^{-1}\left\{ \frac{2(s+1)}{(s+1)^2 + 9} \right\} = 2e^{-t}\cos(3t)
$$

**Final Answer:**
$$
\boxed{\mathcal{L}^{-1}\left\{ \frac{2(s+1)}{s^2 + 2s + 10} \right\} = 2e^{-t}\cos(3t)}
$$

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## **Summary of Answers:**

1. $ \boxed{\frac{s-4}{(s-4)^2 + 4}} $
2. $ \boxed{e^{-(t-4)}u(t-4)} $
3. $ \boxed{2e^{-t}\cos(3t)} $

Let me know if you need further explanation for any step!

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