When y = -cos(2t) - sin(2t) - e, dy = dt d²y dt² = Thus, in terms of t, d²x dt² 4y - et = and d²y dt² 4x + et = - 4(-cos(2t) - sin(2t) — -—-et) - et - (cos(2t) + sin(2t) + ¹-et) + et
When y = -cos(2t) - sin(2t) - e, dy = dt d²y dt² = Thus, in terms of t, d²x dt² 4y - et = and d²y dt² 4x + et = - 4(-cos(2t) - sin(2t) — -—-et) - et - (cos(2t) + sin(2t) + ¹-et) + et
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![## Calculus Derivatives and Equations
### Derivatives:
\[ \frac{dx}{dt} = \]
\[ \frac{d^2x}{dt^2} = \]
### Given Function:
When \( y = -\cos(2t) - \sin(2t) - \frac{1}{5} e^t \),
\[ \frac{dy}{dt} = \]
\[ \frac{d^2y}{dt^2} = \]
### Equations in Terms of \( t \):
#### Equation 1:
\[ \frac{d^2x}{dt^2} - 4y - e^t = \]
\[ = -4(-\cos(2t) - \sin(2t) - \frac{1}{5} e^t) - e^t \]
\[ = \]
#### Equation 2:
\[ \frac{d^2y}{dt^2} - 4x + e^t = \]
\[ = -4 (\cos(2t) + \sin(2t) + \frac{1}{5} e^t) + e^t \]
\[ = \]
These equations and derivatives are part of an exercise in understanding differential calculus, focusing on derivatives and equations involving trigonometric and exponential functions. The placeholders are meant for computations based on given initial functions and conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F379782e3-36e2-4b06-9a33-e2601af30423%2F49eaa2f7-e1ad-4004-afb0-a91ae595e5a9%2Fc7ci7ma_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Calculus Derivatives and Equations
### Derivatives:
\[ \frac{dx}{dt} = \]
\[ \frac{d^2x}{dt^2} = \]
### Given Function:
When \( y = -\cos(2t) - \sin(2t) - \frac{1}{5} e^t \),
\[ \frac{dy}{dt} = \]
\[ \frac{d^2y}{dt^2} = \]
### Equations in Terms of \( t \):
#### Equation 1:
\[ \frac{d^2x}{dt^2} - 4y - e^t = \]
\[ = -4(-\cos(2t) - \sin(2t) - \frac{1}{5} e^t) - e^t \]
\[ = \]
#### Equation 2:
\[ \frac{d^2y}{dt^2} - 4x + e^t = \]
\[ = -4 (\cos(2t) + \sin(2t) + \frac{1}{5} e^t) + e^t \]
\[ = \]
These equations and derivatives are part of an exercise in understanding differential calculus, focusing on derivatives and equations involving trigonometric and exponential functions. The placeholders are meant for computations based on given initial functions and conditions.
![**Euler's Method for Approximations**
Use Euler's method to obtain a four-decimal approximation of the indicated value. First, use \( h = 0.1 \) and then use \( h = 0.05 \). Find an explicit solution for the initial-value problem and then fill in the following tables. (Round your answers to four decimal places. Percentages may be rounded to two decimal places.)
Given:
\[ y' = 2xy, \quad y(1) = 1; \quad y(1.5) \]
Explicit solution: \( y(x) = \_\_\_\_\_ \)
### Table 1: \( h = 0.1 \)
| \( x_n \) | \( y_n \) | Actual Value | Absolute Error | % Rel. Error |
|-----------|-----------|--------------|----------------|-------------|
| 1.00 | 1.0000 | 1.0000 | 0.0000 | 0.00 |
| 1.10 | | 1.2337 | | |
| 1.20 | | 1.5527 | | |
| 1.30 | | 1.9937 | | |
| 1.40 | | 2.6117 | | |
| 1.50 | | 3.4903 | | |
### Table 2: \( h = 0.05 \)
| \( x_n \) | \( y_n \) | Actual Value | Absolute Error | % Rel. Error |
|-----------|-----------|--------------|----------------|-------------|
| 1.00 | 1.0000 | 1.0000 | 0.0000 | 0.00 |
| 1.05 | 1.1079 | 1.1079 | 0.0000 | 0.01 |
| 1.10 | 1.2155 | 1.2337 | 0.0182 | 1.48 |
| 1.15 | | | | |
| 1.20 | | 1.5527 | 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F379782e3-36e2-4b06-9a33-e2601af30423%2F49eaa2f7-e1ad-4004-afb0-a91ae595e5a9%2F4sbnkkq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Euler's Method for Approximations**
Use Euler's method to obtain a four-decimal approximation of the indicated value. First, use \( h = 0.1 \) and then use \( h = 0.05 \). Find an explicit solution for the initial-value problem and then fill in the following tables. (Round your answers to four decimal places. Percentages may be rounded to two decimal places.)
Given:
\[ y' = 2xy, \quad y(1) = 1; \quad y(1.5) \]
Explicit solution: \( y(x) = \_\_\_\_\_ \)
### Table 1: \( h = 0.1 \)
| \( x_n \) | \( y_n \) | Actual Value | Absolute Error | % Rel. Error |
|-----------|-----------|--------------|----------------|-------------|
| 1.00 | 1.0000 | 1.0000 | 0.0000 | 0.00 |
| 1.10 | | 1.2337 | | |
| 1.20 | | 1.5527 | | |
| 1.30 | | 1.9937 | | |
| 1.40 | | 2.6117 | | |
| 1.50 | | 3.4903 | | |
### Table 2: \( h = 0.05 \)
| \( x_n \) | \( y_n \) | Actual Value | Absolute Error | % Rel. Error |
|-----------|-----------|--------------|----------------|-------------|
| 1.00 | 1.0000 | 1.0000 | 0.0000 | 0.00 |
| 1.05 | 1.1079 | 1.1079 | 0.0000 | 0.01 |
| 1.10 | 1.2155 | 1.2337 | 0.0182 | 1.48 |
| 1.15 | | | | |
| 1.20 | | 1.5527 | 0.
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