3. Use Laplace transforms to solve y+3y=1351 24 y(0) = 6
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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![### Solving Differential Equations Using Laplace Transforms
In this tutorial, we will learn how to solve a differential equation using Laplace transforms. Let’s consider the following differential equation and initial condition:
\[ y' + 3y = 13 \sin 2t \]
\[ y(0) = 6 \]
#### Steps to Solve:
1. **Take the Laplace transform of both sides of the differential equation:**
Recall that the Laplace transform of \( y(t) \) is \( Y(s) \).
Using properties of Laplace transforms, we have:
\[ \mathcal{L}\{ y' \} = sY(s) - y(0) \]
\[ \mathcal{L}\{ 3y \} = 3Y(s) \]
\[ \mathcal{L}\{ 13 \sin 2t \} = 13 \cdot \frac{2}{s^2 + 4} \ (since \mathcal{L}\{\sin at\} = \frac{a}{s^2 + a^2}) \]
Now apply the Laplace transform to the given differential equation:
\[ sY(s) - y(0) + 3Y(s) = 13 \cdot \frac{2}{s^2 + 4} \]
2. **Substitute the initial condition \( y(0) \):**
Given \( y(0) = 6 \), substitute this into the equation:
\[ sY(s) - 6 + 3Y(s) = \frac{26}{s^2 + 4} \]
3. **Combine like terms:**
\[ (s + 3)Y(s) - 6 = \frac{26}{s^2 + 4} \]
\[ (s + 3)Y(s) = \frac{26}{s^2 + 4} + 6 \]
4. **Solve for \( Y(s) \):**
\[ Y(s) = \frac{26}{(s^2 + 4)(s + 3)} + \frac{6}{s + 3} \]
5. **Perform partial fraction decomposition as needed:**
Suppose:
\[ \frac{26}{(s^2 +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabc5aa9d-951a-40e1-86a3-f092042c4046%2Ff2f75d66-34c1-4063-a2d7-cf12d6221363%2F5zs68gs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Differential Equations Using Laplace Transforms
In this tutorial, we will learn how to solve a differential equation using Laplace transforms. Let’s consider the following differential equation and initial condition:
\[ y' + 3y = 13 \sin 2t \]
\[ y(0) = 6 \]
#### Steps to Solve:
1. **Take the Laplace transform of both sides of the differential equation:**
Recall that the Laplace transform of \( y(t) \) is \( Y(s) \).
Using properties of Laplace transforms, we have:
\[ \mathcal{L}\{ y' \} = sY(s) - y(0) \]
\[ \mathcal{L}\{ 3y \} = 3Y(s) \]
\[ \mathcal{L}\{ 13 \sin 2t \} = 13 \cdot \frac{2}{s^2 + 4} \ (since \mathcal{L}\{\sin at\} = \frac{a}{s^2 + a^2}) \]
Now apply the Laplace transform to the given differential equation:
\[ sY(s) - y(0) + 3Y(s) = 13 \cdot \frac{2}{s^2 + 4} \]
2. **Substitute the initial condition \( y(0) \):**
Given \( y(0) = 6 \), substitute this into the equation:
\[ sY(s) - 6 + 3Y(s) = \frac{26}{s^2 + 4} \]
3. **Combine like terms:**
\[ (s + 3)Y(s) - 6 = \frac{26}{s^2 + 4} \]
\[ (s + 3)Y(s) = \frac{26}{s^2 + 4} + 6 \]
4. **Solve for \( Y(s) \):**
\[ Y(s) = \frac{26}{(s^2 + 4)(s + 3)} + \frac{6}{s + 3} \]
5. **Perform partial fraction decomposition as needed:**
Suppose:
\[ \frac{26}{(s^2 +
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