Determine the first three nonzero terms in the Taylor polynomial approximation for the given initial value problem. 3x, y' = 8 siny +4 e ³x; y(0) = 0 +.... The Taylor approximation to three nonzero terms is y(x) = -
Determine the first three nonzero terms in the Taylor polynomial approximation for the given initial value problem. 3x, y' = 8 siny +4 e ³x; y(0) = 0 +.... The Taylor approximation to three nonzero terms is y(x) = -
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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![### Taylor Series Approximation in Differential Equations
To solve differential equations, one particularly useful method is the Taylor polynomial approximation. Here, we will determine the first three nonzero terms in the Taylor polynomial approximation for the given initial value problem:
\[ y' = 8 \sin y + 4e^{3x}, \quad y(0) = 0 \]
This involves finding an approximation for the function \(y(x)\) around the point \(x = 0\).
Essentially, in this example, we aim to construct a Taylor polynomial that approximates the solution to the differential equation. Taylor series can simplify the representation of functions by using polynomials.
The general form of the Taylor series for a function \(y(x)\) about \(x = 0\) is given by:
\[ y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \cdots \]
For this problem:
\[ y(x) = a_0 + a_1x + a_2x^2 + \cdots \]
#### Steps to determine the Taylor polynomial:
1. **Initial Value:**
\[ y(0) = 0 \]
2. **First Derivative:**
Evaluate \( y' \) at \( x = 0 \):
\[ y' = 8 \sin y + 4e^{3x} \]
Substituting \( y(0) \) and \( x = 0 \), we get:
\[ y'(0) = 8 \sin(0) + 4e^{0} = 4 \]
Therefore, the first term, \( a_1 \), is 4.
3. **Second Derivative:**
Differentiate \( y' \):
\[ y'' = 8 \cos y \cdot y' + 12 e^{3x} \]
Substituting \( y(0) \), \( x = 0 \), and \( y'(0) \):
\[ y''(0) = 8 \cos(0) \cdot 4 + 12 e^{0} = 8 \cdot 4 + 12 = 44 \]
Therefore, the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbba0909-7b4f-4e80-8e24-06dfda2d061e%2F2ae9385e-5dd7-4512-887e-cb645b801f6e%2Fuw4gjhj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Taylor Series Approximation in Differential Equations
To solve differential equations, one particularly useful method is the Taylor polynomial approximation. Here, we will determine the first three nonzero terms in the Taylor polynomial approximation for the given initial value problem:
\[ y' = 8 \sin y + 4e^{3x}, \quad y(0) = 0 \]
This involves finding an approximation for the function \(y(x)\) around the point \(x = 0\).
Essentially, in this example, we aim to construct a Taylor polynomial that approximates the solution to the differential equation. Taylor series can simplify the representation of functions by using polynomials.
The general form of the Taylor series for a function \(y(x)\) about \(x = 0\) is given by:
\[ y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \cdots \]
For this problem:
\[ y(x) = a_0 + a_1x + a_2x^2 + \cdots \]
#### Steps to determine the Taylor polynomial:
1. **Initial Value:**
\[ y(0) = 0 \]
2. **First Derivative:**
Evaluate \( y' \) at \( x = 0 \):
\[ y' = 8 \sin y + 4e^{3x} \]
Substituting \( y(0) \) and \( x = 0 \), we get:
\[ y'(0) = 8 \sin(0) + 4e^{0} = 4 \]
Therefore, the first term, \( a_1 \), is 4.
3. **Second Derivative:**
Differentiate \( y' \):
\[ y'' = 8 \cos y \cdot y' + 12 e^{3x} \]
Substituting \( y(0) \), \( x = 0 \), and \( y'(0) \):
\[ y''(0) = 8 \cos(0) \cdot 4 + 12 e^{0} = 8 \cdot 4 + 12 = 44 \]
Therefore, the
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