1) Give the Interval of existance for the Solution to the initial valve problem. (+²-16)y" + Cos(4+) y = CSc(+) 4(S) = 4'(s) = 4₁"(s) =0)
1) Give the Interval of existance for the Solution to the initial valve problem. (+²-16)y" + Cos(4+) y = CSc(+) 4(S) = 4'(s) = 4₁"(s) =0)
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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![### Interval of Existence for Initial Value Problems
**Problem Statement:**
1) Give the interval of existence for the solution to the initial value problem.
\[ (t^2 - 16)y^{(iii)} + \cos(4t)y' = \csc(t) \]
With the initial conditions:
\[ y(5) = y'(5) = y^{(iii)}(5) = 0 \]
**Explanation:**
To find the interval of existence for the solution to the given initial value problem, you need to consider the points where the coefficients in the differential equation become undefined or singular.
1. **Coefficient Analysis:**
- \( t^2 - 16 \): This coefficient becomes zero at \( t = \pm 4 \). Hence, \( t = 4 \) and \( t = -4 \) are potential singular points.
- \( \cos(4t) \): This coefficient becomes zero at \( t = \frac{(2n+1)\pi}{8} \) for \( n = 0, \pm1, \pm2, \ldots \).
- \( \csc(t) = \frac{1}{\sin(t)} \): This function becomes undefined at \( t = n\pi \) for \( n = 0, \pm1, \pm2, \ldots \).
2. **Initial Conditions:**
- The initial conditions are given at \( t = 5 \).
Based on the singular points derived from the coefficients and the initial condition, determine the largest interval around \( t = 5 \) within which the coefficients remain well-defined and continuous.
The interval of existence should be:
\[ 4 < t < 8 \]
Therefore, the interval of existence for the solution to the given initial value problem is \( 4 < t < 8 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0d3e718-edd3-44ac-adb9-a3206bb6736d%2F0893ea06-e94f-49ab-b745-cff31ba60a0e%2Ft9ya5qkp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Interval of Existence for Initial Value Problems
**Problem Statement:**
1) Give the interval of existence for the solution to the initial value problem.
\[ (t^2 - 16)y^{(iii)} + \cos(4t)y' = \csc(t) \]
With the initial conditions:
\[ y(5) = y'(5) = y^{(iii)}(5) = 0 \]
**Explanation:**
To find the interval of existence for the solution to the given initial value problem, you need to consider the points where the coefficients in the differential equation become undefined or singular.
1. **Coefficient Analysis:**
- \( t^2 - 16 \): This coefficient becomes zero at \( t = \pm 4 \). Hence, \( t = 4 \) and \( t = -4 \) are potential singular points.
- \( \cos(4t) \): This coefficient becomes zero at \( t = \frac{(2n+1)\pi}{8} \) for \( n = 0, \pm1, \pm2, \ldots \).
- \( \csc(t) = \frac{1}{\sin(t)} \): This function becomes undefined at \( t = n\pi \) for \( n = 0, \pm1, \pm2, \ldots \).
2. **Initial Conditions:**
- The initial conditions are given at \( t = 5 \).
Based on the singular points derived from the coefficients and the initial condition, determine the largest interval around \( t = 5 \) within which the coefficients remain well-defined and continuous.
The interval of existence should be:
\[ 4 < t < 8 \]
Therefore, the interval of existence for the solution to the given initial value problem is \( 4 < t < 8 \).
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