/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Consider the equation $$ y^{... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the equation $$ y^{\prime \prime}+y=b(x) $$ where \(b\) is a continuous function on \(1 \leqq x<\infty\) satisfying $$ \int_{1}^{\infty}|b(t)| d t<\infty $$ (a) Show that a particular solution \(\psi_{p}\) is given by $$ \psi_{p}(x)=\int_{1}^{x} \sin (x-t) b(t) d t $$ (b) Show that any solution is bounded on \(1 \leqq x<\infty\).

Short Answer

Expert verified
(a) Verify derivatives; (b) show boundedness using Integrability and structure of the solution.

Step by step solution

01

Substitute Particular Solution

We need to verify if \( \psi_p(x) = \int_{1}^{x} \sin(x-t) b(t) \, dt \) is a particular solution of the differential equation \( y'' + y = b(x) \). Substitute \( \psi_p(x) \) into the left-hand side of the equation.
02

Compute Derivatives

Calculate the first derivative \( \psi_p'(x) \) using the Leibniz rule: \( \psi_p'(x) = \int_{1}^{x} \cos(x-t) b(t) \, dt \). Next, compute the second derivative, \( \psi_p''(x) = \int_{1}^{x} -\sin(x-t) b(t) \, dt + \sin(x-x) b(x)\).
03

Verify the Differential Equation

Substitute \( \psi_p''(x) \) and \( \psi_p(x) \) into the differential equation: \( \psi_p''(x) + \psi_p(x) \). Substitute these into the equation and show that it equals \( b(x) \), confirming \( \psi_p(x) \) is a solution.
04

Establish Boundedness

Now show that any solution \( y(x) = C_1 e^x + C_2 e^{-x} + \psi_p(x) \) is bounded on \( 1 \leq x < \infty \). Note that terms involving \( e^x \) must be zero for boundedness, hence focus on \( \psi_p(x) \).
05

Boundedness of Particular Solution

Use the given condition \( \int_{1}^{\infty} |b(t)| \, dt < \infty \) to demonstrate that \( \psi_p(x) \) remains bounded. Show \( \psi_p(x) \) remains finite as \( x \rightarrow \infty \), using the absolute integrability of \( b(t) \).

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Key Concepts

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

Particular Solution
Finding a particular solution to a differential equation involves identifying a specific solution that fits the non-homogeneous equation. In this context, the differential equation is given by \( y'' + y = b(x) \). The particular solution \( \psi_{p}(x) \) is proposed as \( \int_{1}^{x} \sin(x-t) b(t) \, dt \). To verify this, one must substitute \( \psi_{p}(x) \) back into the equation. This involves computing the necessary derivatives and ensuring that the left-hand side becomes \( b(x) \) when the particular solution and its derivatives are substituted. This specific function \( \psi_{p} \) effectively "neutralizes" the non-homogeneous part \( b(x) \), showing it to be a valid particular solution.
Boundedness of Solutions
In the context of differential equations, a solution is said to be bounded if it does not become infinite as \( x \) approaches infinity. For the equation \( y'' + y = b(x) \), the task is to show that all solutions are bounded on the interval \( 1 \leq x < \infty \). In solving a differential equation, the general solution typically includes a homogeneous solution part plus any particular solution. The homogeneous solution is of the form \( C_1 e^x + C_2 e^{-x} \). For boundedness, any terms that could lead to infinity, such as \( e^x \), must be ruled out by setting their coefficients to zero. The solution thus simplifies, and focus shifts to proving that the particular solution \( \psi_{p}(x) \) remains bounded. Given the condition that \( \int_{1}^{\infty} |b(t)| \, dt < \infty \), it implies that the area under the curve of \( b(t) \) is finite, supporting the conclusion that \( \psi_{p}(x) \) remains finite as \( x \to \infty \).
Differential Equation Verification
Verification involves checking that a proposed solution satisfies the given differential equation. This is done by substituting the solution and its derivatives into the equation to see if they simplify correctly back to the function on the right-hand side, \( b(x) \) in this case. For \( \psi_{p}(x) = \int_{1}^{x} \sin(x-t) b(t) \, dt \), the verification process involves a few key steps:
  • Calculate the first and second derivatives of \( \psi_{p}(x) \).
  • Substitute these derivatives into the left-hand side of the differential equation \( y'' + y \).
  • Confirm that the left-hand side equals \( b(x) \).
Once this equality is confirmed, it verifies that \( \psi_{p}(x) \) is a legitimate particular solution of the equation.
Continuous Functions
Continuous functions play a fundamental role in differential equations, especially when dealing with integrals and the behavior of solutions over specific intervals. In this exercise, the function \( b(x) \) must be continuous on the interval \( 1 \leqq x < \infty \). Continuity ensures that the behavior of \( b(x) \) is smooth, with no jumps or discontinuities causing issues in integration. Another important aspect of continuity here is that:
  • The given condition \( \int_{1}^{\infty} |b(t)| \, dt < \infty \) relies on \( b(x) \) being continuous, guaranteeing a well-defined integral.
  • This condition also implies the function is absolutely integrable over the given interval.
These properties are crucial in showing that a solution to the differential equation remains bounded, as it allows the particular solution based on \( b(x) \) to be finite for all \( x \) in the specified domain.

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Most popular questions from this chapter

Consider the equation $$L(y)=y^{\prime \prime \prime}+a_{1}(x) y^{\prime \prime}+a_{2}(x) y^{\prime}+a_{3}(x) y=0$$ Suppose \(\phi_{1}, \phi_{2}\) are given linearly independent solutions of \(L(y)=0\). (a) Let \(\phi=u \phi_{1}\), and compute the equation of order two satisfied by \(u^{\prime}\) in order that \(L(\phi)=0\). Show that \(\left(\phi_{2} / \phi_{1}\right)^{\prime}\) is a solution of this equation of order two. (b) Use the fact that \(\left(\phi_{2} / \phi_{1}\right)^{\prime}\) satisfies the equation of order two to reduce the order of this equation by one.

Let \(\phi_{1}, \cdots, \phi_{n}\) be \(n\) continuous functions on the interval \(a \leqq x \leqq b\). Let $$ \alpha_{i j}=\int_{a}^{b} \overline{\phi_{i}(x)} \phi_{j}(x) d x, \quad(i, j=1, \cdots, n) $$ and let \(\Delta\) denote the determinant $$ \Delta=\left|\begin{array}{llll} \alpha_{11} & \alpha_{12} & \cdots & \alpha_{1 n} \\ \alpha_{21} & \alpha_{22} & \cdots & \alpha_{2 n} \\ \cdot & & & \cdot \\ \cdot & & & \cdot \\ \cdot & & & \cdot \\ \alpha_{n 1} & \alpha_{n 2} & \cdots & \alpha_{n n} \end{array}\right| $$ Prove that \(\phi_{1}, \cdots, \phi_{n}\) are linearly independent on \(a \leqq x \leqq b\) if, and only if, \(\Delta \neq 0\). (Hint: Suppose $$ \Delta \neq 0, \text { and } c_{1} \phi_{1}+\cdots+c_{n} \phi_{n}=0 . $$ Multiply this equation in turn by \(\overline{\phi_{1}}, \overline{\phi_{2}}, \cdots, \overline{\phi_{n}}\) and integrate to obtain $$ \begin{array}{l} c_{1} \alpha_{11}+c_{2} \alpha_{12}+\cdots+c_{n} \alpha_{1 n}=0 \\ \cdot \\ \cdot \\ \cdot \\ c_{1} \alpha_{n 1}+c_{2} \alpha_{n 2}+\cdots+c_{n} \alpha_{n n}=0 \end{array} $$ The only solution of these is \(c_{1}=c_{2}=\cdots=c_{n}=0\), Conversely, if \(\phi_{1}, \cdots, \phi_{n}\) are linearly independent and \(\Delta=0\), then there are \(c_{1}, \cdots, c_{n}\) satisfying \(\left(^{*}\right)\) not all zero. Multiply the first equation by \(\overline{c_{1}}\), the second by \(\overline{c_{2}}\), etc., to obtain $$ 0=\sum_{j=1}^{n} \sum_{i=1}^{n} \bar{c}_{i} \alpha_{i j} c_{j}=\int_{a}^{b}\left|\sum_{i=1}^{n} c_{i} \phi_{i}(x)\right|^{2} d x $$ Show that this implies \(c_{1} \phi_{1}+\cdots+c_{n} \phi_{n}=0\) The determinant \(\Delta\) is called the Gramian of \(\phi_{1}, \cdots, \phi_{n}\). Note that \(\alpha_{i j}=\overline{\alpha_{i i}}\).)

The equation $$y^{\prime \prime}+e^{x} y=0$$ has a solution \(\phi\) of the form $$\phi(x)=\sum_{k=0}^{\infty} c_{k} x^{k}$$ which satisfies \(\phi(0)=1, \phi^{\prime}(0)=0\). Compute \(c_{0}, c_{1}, c_{2}, c_{3}, c_{4}, c_{5} .\) (Hint: \(c_{k}=\phi^{(k)}(0) / k !\) and \(\phi^{\prime \prime}\) \(\left.(x)=-e^{x} \phi(x) .\right)\)

Consider the equation $$y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{2}(x) y=0$$ where \(a_{1}, a_{2}\) are continuous on some interval \(I\) containing \(x_{0}\). Suppose \(\left(\phi_{1}\right.\) is a solution such that \(\phi_{1}(x) \neq 0\) for all \(x\) in \(I\). (a) Show that there is a second solution \(\phi_{2}\) on \(I\) such that $$W\left(\phi_{1}, \phi_{2}\right)\left(x_{0}\right)=1$$ (b) Compute such a \(\phi_{2}\) in terms of \(\phi_{1}\), by solving the first order equation $$\phi_{1}(x) \phi_{2}^{\prime}(x)-\phi_{1}^{\prime}(x) \phi_{2}(x)=\exp \left[-\int_{x_{0}}^{x} a_{1}(t) d t\right]$$ for \(\phi_{2}\).

The equation $$y^{\prime \prime}-2 x y^{\prime}+2 \alpha y=0$$ where \(\alpha\) is a constant, is called the Hermite equation. (a) Find two linearly independent solutions on \(-\infty

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