Give step-by-step solution with explanation and final answer:
Find the solution
t
o
the following differential equation using the Laplace transform method,
i
f
the input
i
s
a unit step:
d
3

Question

Give step-by-step solution with explanation and final answer: 
Find the solution 
t
o
 ﻿the following differential equation using the Laplace transform method, 
i
f
 ﻿the input
i
s
 ﻿a unit step:
d
3
y
(
t
)
d
t
3
+
3
d
2
y
(
t
)
d
t
2
+
4
d
y
(
t
)
d
t
+
y
(
t
)
=
6
x
(
t
)
Assume zero initial conditions.
(
b
)
 ﻿Find the impulse response 
o
f
 ﻿the system 
h
(
t
)
(
c
)
 ﻿Find the step response for the system 
δ
(
t
)
(
d
)
 ﻿Use the final value theorem 
t
o
 ﻿find the final value 
o
f
 
y
(
t
)
.

VIPSolutions · Accepted Answer

Let's solve the problem step by step:

## Given Differential Equation

$$
\frac{d^3y(t)}{dt^3} + 3\frac{d^2y(t)}{dt^2} + 4\frac{dy(t)}{dt} + y(t) = 6x(t)
$$

Assume **zero initial conditions**.

---

## (a) **Find the solution using the Laplace transform method if the input is a unit step**

### Step 1: Take Laplace Transforms

Let $ Y(s) = \mathcal{L}\{y(t)\} $, $ X(s) = \mathcal{L}\{x(t)\} $.

For zero initial conditions:
- $\mathcal{L}\{y'(t)\} = sY(s)$
- $\mathcal{L}\{y''(t)\} = s^2Y(s)$
- $\mathcal{L}\{y'''(t)\} = s^3Y(s)$

So,
$$
s^3Y(s) + 3s^2Y(s) + 4sY(s) + Y(s) = 6X(s)
$$
$$
Y(s) [s^3 + 3s^2 + 4s + 1] = 6 X(s)
$$
$$
Y(s) = \frac{6 X(s)}{s^3 + 3s^2 + 4s + 1}
$$

### Step 2: For Unit Step Input ($x(t) = u(t)$), $X(s) = \frac{1}{s}$:

$$
Y(s) = \frac{6}{s[s^3 + 3s^2 + 4s + 1]}
$$

---

## (b) **Impulse response $ h(t) $ of the system**

The **impulse response** is the output when the input is a delta ($ \delta(t) $), whose Laplace transform is 1:

$$
H(s) = \frac{Y(s)}{X(s)} = \frac{6}{s^3 + 3s^2 + 4s + 1}
$$

So,
$$
h(t) = \mathcal{L}^{-1}\left\{ \frac{6}{s^3 + 3s^2 + 4s + 1} \right\}
$$

---

## (c) **Step response of the system**

The step response is the output when the input is a **unit step** ($ x(t) = u(t) $), already found above:
$$
Y(s) = \frac{6}{s[s^3 + 3s^2 + 4s + 1]}
$$

So,
$$
y_{\text{step}}(t) = \mathcal{L}^{-1}\left\{ \frac{6}{s[s^3 + 3s^2 + 4s + 1]} \right\}
$$

---

## (d) **Final value theorem to find the final value of $ y(t) $**

The **Final Value Theorem**:
$$
\lim_{t \to \infty} y(t) = \lim_{s \to 0} s Y(s)
$$

Recall:
$$
Y(s) = \frac{6}{s[s^3 + 3s^2 + 4s + 1]}
$$
So,
$$
\lim_{t \to \infty} y(t) = \lim_{s \to 0} s \cdot \frac{6}{s[s^3 + 3s^2 + 4s + 1]} = \lim_{s \to 0} \frac{6}{s^3 + 3s^2 + 4s + 1}
$$

Plug $ s = 0 $:
$$
= \frac{6}{1} = 6
$$

---

## **Summary of Answers**

### (a) **Laplace Transform Solution**
$$
Y(s) = \frac{6}{s[s^3 + 3s^2 + 4s + 1]}
$$

### (b) **Impulse Response**
$$
h(t) = \mathcal{L}^{-1}\left\{ \frac{6}{s^3 + 3s^2 + 4s + 1} \right\}
$$

### (c) **Step Response**
$$
y_{\text{step}}(t) = \mathcal{L}^{-1}\left\{ \frac{6}{s[s^3 + 3s^2 + 4s + 1]} \right\}
$$

### (d) **Final Value**
$$
\boxed{6}
$$

---

If you want to explicitly compute the inverse Laplace for (b) or (c), or want to factor the denominator, let me know!

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