Question

In each of Problems I through 6 determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. $$ y^{\prime}+(\tan t) y=\sin t, \quad y(\pi)=0 $$

Short answer

Question: Determine an interval in which the solution of the following initial value problem is certain to exist without actually solving the problem: \(y' + (\tan t)y = \sin t, y(\pi) = 0\). Answer: The solution of the given initial value problem is certain to exist on the interval \((\frac{\pi}{2}, \frac{3\pi}{2})\).

Step-by-step solution

Identify the given ODE and initial condition

We are given the first-order linear ODE: $$ y' + (\tan t)y = \sin t $$ and the initial condition: $$ y(\pi) = 0 $$

Analyze the continuity of the functions p(t) and q(t)

We have the functions \(p(t) = \tan t\) and \(q(t) = \sin t\). Both of these functions have points of discontinuity. In this case, \(\tan t\) is discontinuous at its asymptotes, which occur at \((2n + 1)\frac{\pi}{2}\) for all integers \(n\). \(\sin t\) is continuous for all real t.

Determine the open interval containing the initial value t = π

We need to find an open interval containing the initial value \(t = \pi\) in which both \(p(t)\) and \(q(t)\) are continuous. The discontinuities of the function \(p(t) = \tan t\) are of the form \((2n + 1)\frac{\pi}{2}\). Specifically, to the left of \(t = \pi\), the closest asymptote is \(t = \frac{3\pi}{2}\), and to the right, the closest asymptote is \(t = \frac{\pi}{2}\). Therefore, the open interval \((\frac{\pi}{2}, \frac{3\pi}{2})\) contains the initial value \(t = \pi\) and is a region where both \(p(t)\) and \(q(t)\) are continuous.

Conclusion

We have found an interval, \((\frac{\pi}{2}, \frac{3\pi}{2})\), in which the solution of the given initial value problem is certain to exist. This is because the functions \(p(t) = \tan t\) and \(q(t) = \sin t\) are continuous on this interval, which contains the initial value of \(t = \pi\). The Existence and Uniqueness Theorem guarantees a unique solution on this interval.