Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. 1(t – 18) y' + y = 0, y(8) = 1 i i
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. 1(t – 18) y' + y = 0, y(8) = 1 i i
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![### Determining Intervals of Existence for Initial Value Problems
**Problem Statement:**
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.
**Equation:**
\[ t(t - 18)y' + y = 0, \quad y(8) = 1 \]
**Interval Specification:**
\[ \boxed{\ } < t < \boxed{\ } \]
**Explanation:**
The given initial value problem involves solving a differential equation of the form \(t(t - 18)y' + y = 0\) with an initial condition \(y(8) = 1\). To determine the interval of t in which a solution is guaranteed to exist, it is essential to identify the values of t where the coefficient of \(y'\) (i.e., \(t(t - 18)\)) might cause singularities or discontinuities.
In this equation, singularities occur where \(t(t - 18) = 0\), that is, at \(t = 0\) and \(t = 18\). The solution is therefore guaranteed to exist in an interval that does not include these points of singularity. Given the initial condition \(y(8) = 1\), we look for an interval around t = 8 which does not include 0 or 18.
Thus, the interval in which the solution is certain to exist will be:
\[ 0 < t < 18 \]
In this interval, the differential equation remains well-behaved and does not hit any points of singularity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77cac1a6-5ad9-4f72-bdd9-21a202e53df4%2Fc00ec273-671d-4b95-bcc3-8448e44ef435%2Fi4lnaf_processed.png&w=3840&q=75)
Transcribed Image Text:### Determining Intervals of Existence for Initial Value Problems
**Problem Statement:**
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.
**Equation:**
\[ t(t - 18)y' + y = 0, \quad y(8) = 1 \]
**Interval Specification:**
\[ \boxed{\ } < t < \boxed{\ } \]
**Explanation:**
The given initial value problem involves solving a differential equation of the form \(t(t - 18)y' + y = 0\) with an initial condition \(y(8) = 1\). To determine the interval of t in which a solution is guaranteed to exist, it is essential to identify the values of t where the coefficient of \(y'\) (i.e., \(t(t - 18)\)) might cause singularities or discontinuities.
In this equation, singularities occur where \(t(t - 18) = 0\), that is, at \(t = 0\) and \(t = 18\). The solution is therefore guaranteed to exist in an interval that does not include these points of singularity. Given the initial condition \(y(8) = 1\), we look for an interval around t = 8 which does not include 0 or 18.
Thus, the interval in which the solution is certain to exist will be:
\[ 0 < t < 18 \]
In this interval, the differential equation remains well-behaved and does not hit any points of singularity.
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