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Chapter 3: Problem 8

determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution. $$ (t-1) y^{\prime \prime}-3 t y^{\prime}+4 y=\sin t, \quad y(-2)=2, \quad y^{\prime}(-2)=1 $$

Short Answer

Expert verified
Answer: The longest interval is \((-\infty, \infty)\).

Step by step solution

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01

Analyze the coefficients

Since the problem asks not to find the solution to the initial value problem, we focus on the characteristics of the differential equation. The given differential equation is a second-order linear nonhomogeneous equation: $$(t-1)y^{\prime\prime} - 3ty^{\prime} + 4y = \sin(t)$$ When analyzing the coefficients, we want to make sure they are continuous to guarantee a unique and twice differentiable solution. In this case, the coefficients are: 1. \((t-1)\) 2. \((-3t)\) 3. \((4)\) All of these functions are continuous for all real numbers, which means we do not have any singular points.
02

Analyze the forcing function

The forcing function, \(\sin(t)\), is also continuous for all real numbers. Since both the coefficients and forcing function are continuous, the differential equation qualifies for the existence and uniqueness theorem.
03

Determine the interval for unique, twice differentiable solution

Based on the analysis in Steps 1 and 2, we can conclude that the initial value problem has a unique twice differentiable solution for all real numbers \((-\infty, \infty)\).
04

Verify the initial conditions

We are given the initial conditions: \(y(-2) = 2\) and \(y^{\prime}(-2) = 1\). Since they are provided, we can conclude that the given point \((-2, 2)\) with initial slope \(1\) is within the interval where the solution exists.
05

State the conclusion

The longest interval in which the given initial value problem is certain to have a unique twice differentiable solution is \((-\infty, \infty)\). The differential equation and the forcing function are continuous, making the solution unique and twice differentiable in this entire interval.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Existence and Uniqueness Theorem
The existence and uniqueness theorem for differential equations is a foundational principle guaranteeing that, under certain conditions, there is one and only one solution to a differential equation with given initial conditions. A key aspect of the theorem is it requires the functions involved to be continuous. In the context of our initial value problem, consisting of a second-order nonhomogeneous differential equation with specified initial values, this theorem assures us that the solution is not just a possibility but a certainty, provided that the equation meets the theorem's criteria. Specifically, for a second-order differential equation of the form \(ay'' + by' + cy = f(t)\),with initial conditions provided at a point \(y(t_0) = y_0\) and \(y'(t_0) = y_0'\),the existence and uniqueness theorem holds if the coefficients \(a, b,\) and \(c\) are continuous on an open interval containing \(t_0\), and \(f(t)\) is continuous on the same interval. In our exercise, the continuity of the coefficients and the forcing function \(\text{sin}(t)\) throughout all real numbers implies this theorem applies, ensuring the existence and the uniqueness of the solution across an infinite interval.
Second-Order Differential Equation
A second-order differential equation involves the second derivative of a function and can often describe phenomena such as motion in physics or acceleration in mechanics. The given initial value problem \((t-1) y'' - 3ty' + 4y = \text{sin}(t)\), is a second-order differential equation because it involves the term \(y''\), which represents the second derivative of \(y\) with respect to \(t\). The solution to such an equation is dependent on both the form of the equation itself and the initial conditions supplied. These equations can be particularly challenging when trying to find explicit solutions. Our focus, therefore, is not on solving the equation but on utilizing the driven continuity of coefficients to assert the existence of a unique solution through the specified range.
Twice Differentiable Solution
The term 'twice differentiable solution' refers to a function that is smooth enough to have a well-defined first and second derivative. In the context of our exercise, the second derivative is not just an abstract concept but a crucial part of the differential equation's structure. The solutions of the differential equation are expected to be twice differentiable to satisfy both the equations themselves and initial value conditions. When the theorem and conditions align, as they do in the given problem, it ensures the solution will possess these smooth characteristics over the interval in which it is defined. In practice, this means if we were to graph the solution, it would be a smooth curve without any breaks, corners, or cusps, which is significant in applications where continuity and smoothness are required.
Nonhomogeneous Differential Equation
A nonhomogeneous differential equation, in contrast to a homogeneous equation, includes a term that is not a function of the unknown function \(y\) and its derivatives. In simpler terms, nonhomogeneous equations are those that equal something other than zero on one side of the equation. For the initial value problem we are analyzing, the nonhomogeneous term is \(\text{sin}(t)\), making the equation nonhomogeneous. This distinction is crucial because the presence of a nonhomogeneous term often signifies that the equation models a system subject to external forces or influences, which is common in real-world applications, such as in oscillatory systems like springs with external driving forces. Despite the additional complexity introduced by such terms, the existence and uniqueness theorem still helps to ascertain the predictability and behavior of solutions provided all conditions of the theorem are met.

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Most popular questions from this chapter

The differential equation $$ x y^{\prime \prime}-(x+N) y^{\prime}+N y=0 $$ where \(N\) is a nonnegative integer, has been discussed by several authors. 6 One reason it is interesting is that it has an exponential solution and a polynomial solution. (a) Verify that one solution is \(y_{1}(x)=e^{x}\). (b) Show that a second solution has the form \(y_{2}(x)=c e^{x} \int x^{N} e^{-x} d x\). Calculate \(y_{2 (x)\) for \(N=1\) and \(N=2 ;\) convince yourself that, with \(c=-1 / N !\) $$ y_{2}(x)=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\cdots+\frac{x^{N}}{N !} $$ Note that \(y_{2}(x)\) is exactly the first \(N+1\) terms in the Taylor series about \(x=0\) for \(e^{x},\) that is, for \(y_{1}(x) .\)

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ (1-x) y^{\prime \prime}+x y^{\prime}-y=g(x), \quad 0

By combining the results of Problems 24 through \(26,\) show that the solution of the initial value problem $$ L[y]=\left(a D^{2}+b D+c\right) y=g(t), \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ where \(a, b,\) and \(c\) are constants, has the form $$ y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s $$ The function \(K\) depends only on the solutions \(y_{1}\) and \(y_{2}\) of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once \(K\) is determined, all nonhomogeneous problems involving the same differential operator \(L\) are reduced to the evaluation of an integral. Note also that although \(K\) depends on both \(t\) and \(s,\) only the combination \(t-s\) appears, so \(K\) is actually a function of a single variable. Thinking of \(g(t)\) as the input to the problem and \(\phi(t)\) as the output, it follows from Eq. (i) that the output depends on the input over the entire interval from the initial point \(t_{0}\) to the current value \(t .\) The integral in Eq. (i) is called the convolution of \(K\) and \(g,\) and \(K\) is referred to as the kernel.

Show that the solution of the initial value problem $$ m u^{\prime \prime}+\gamma u^{\prime}+k u=0, \quad u\left(t_{0}\right)=u_{0}, \quad u^{\prime}\left(t_{0}\right)=u_{0}^{\prime} $$ can be expressed as the sum \(u=v+w,\) where \(v\) satisfies the initial conditions \(v\left(t_{0}\right)=\) \(u_{0}, v^{\prime}\left(t_{0}\right)=0, w\) satisfies the initial conditions \(w\left(t_{0}\right)=0, w^{\prime}\left(t_{0}\right)=u_{0}^{\prime},\) and both \(v\) and \(w\) satisfy the same differential equation as \(u\). This is another instance of superposing solutions of simpler problems to obtain the solution of a more general problem.

Find the solution of the initial value problem $$ u^{\prime \prime}+u=F(t), \quad u(0)=0, \quad u^{\prime}(0)=0 $$ where $$ F(t)=\left\\{\begin{array}{ll}{F_{0}(2 \pi-t),} & {0 \leq t \leq \pi} \\ {-0} & {(2 \pi-t),} & {\pi
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