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

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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.
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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