/*! 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 12 By considering functions of the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 12

By considering functions of the form \(h(x)=\int_{0}^{x}(x-y) f(y) d y\), show that the solution \(f(x)\) of the integral equation $$ f(x)=x+\frac{1}{2} \int_{0}^{1}|x-y| f(y) d y $$ satisfies the equation \(f^{\prime \prime}(x)=f(x)\).

Short Answer

Expert verified
The solution function satisfies \( f''(x) = f(x) \).

Step by step solution

01

Define the given function

Consider the function provided: \[ h(x)=\int_{0}^{x}(x-y) f(y) d y \]
02

Differentiate h(x) with respect to x

To find the first derivative of \( h(x) \) with respect to \( x \), use the Leibniz rule for differentiating an integral: \[ h'(x) = \frac{d}{dx} \left( \int_{0}^{x} (x-y) f(y) \, dy \right) = f(x) \int_{0}^{x} (1) f(y) \, dy \] Simplify to get: \[ h'(x) = \int_{0}^{x} f(y) \, dy \]
03

Differentiate h'(x) with respect to x to find h''(x)

Differentiate \( h'(x) \) again to find \( h''(x) \): \[ h''(x) = \frac{d}{dx} \left( \int_{0}^{x} f(y) \, dy \right) = f(x) \]
04

Relate h''(x) to f(x)

We have shown that \( f(x) = h''(x) \). Now substitute this into the original integral equation to verify: \[ f(x)=x+\frac{1}{2} \int_{0}^{1} |x-y| f(y) \, dy \]
05

Verify the integral equation

Calculate the integral equation using the substitution \( h''(x) = f(x) \): \[ f(x) = x + \frac{1}{2} \int_{0}^{1} |x-y| h''(y) \, dy \]
06

Conclude that f''(x) = f(x)

Since \( h(x) = \int_{0}^{x} (x-y) f(y) \, dy \) and \( h''(x) = f(x) \), it follows that \( f''(x) = f(x) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

differentiation under the integral sign
Differentiation under the integral sign is a powerful technique in calculus. It allows us to differentiate an integral with respect to a parameter.
This method helps simplify complex problems and is often called the Leibniz rule for differentiation.
In the context of the exercise, we consider a function involving an integral: \ \( h(x) = \int_0^x (x - y) f(y) dy \)
We differentiate this integral with respect to x.
According to the Leibniz rule: \ \[ \frac{d}{dx} \left( \int_0^x g(x, y) dy \right) = g(x, x) + \int_0^x \frac{\partial g(x, y)}{\partial x} dy \] In our exercise: \ \( g(x, y) = (x - y) f(y) \).
Differentiating, we get: \ \[ h'(x) = (x - x) f(x) + \int_0^x \frac{\partial}{\partial x} (x - y) f(y) dy \]
Notice that the term \ \( (x - x) f(x)\ \) is zero, leaving us with: \ \[ h'(x) = \int_0^x f(y) dy \]
This shows how differentiating under the integral sign helps simplify the problem.
second-order differential equations
A second-order differential equation involves the second derivative of a function.
These equations commonly appear in physics and engineering.
They describe systems like simple harmonic oscillators and wave equations.
In this exercise, we show that the given function \ \( f(x) \) satisfies: \ \[ f^{\prime \prime}(x) = f(x) \]
First, we start by differentiating the integral form \ \( h(x) = \int_0^x (x-y) f(y) dy \), as discussed before.
Taking the second derivative: \ \[ h''(x) = \frac{d}{dx} \left( \int_0^x f(y) dy \right) = f(x) \] confirms that \ \( h''(x) = f(x) \).
Since we have established this, we can substitute back into the integral equation and confirm that it also holds true.
These steps help us show that the function \ \( f(x) \) must satisfy the second-order differential equation.
Leibniz rule
The Leibniz rule provides a method for differentiating an integral.
It's fundamental for problems involving parameter-dependent integrals.
It tells us how to handle derivatives of integrals with variable limits.
In our context, the Leibniz rule helps differentiate: \ \[ h(x) = \int_0^x (x-y) f(y) dy \] According to the Leibniz rule: \ \[ \frac{de}{dx} \left( \int_0^x g(x, y) dy \right) = g(x, x) + \int_0^x \frac{d}{dx} g(x, y) dy \] For our function \ \( g(x, y) = (x-y) f(y) \), differentiating gives: \ \[ h'(x) = (x-x) f(x) + \int_0^x \frac{d}{dx}(x-y) f(y) dy \]
The first term \ \( (x-x) f(x) \) vanishes, simplifying the expression to: \ \[ h'(x) = \int_0^x f(y) dy \] The Leibniz rule thus aids in simplifying our results and forms key steps in solving the problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the integral equation $$ \int_{0}^{\infty} \cos (x v) y(v) d v=\exp \left(-x^{2} / 2\right) $$ for the function \(y=y(x)\) for \(x>0 .\) Note that for \(x<0, y(x)\) can be chosen as is most convenient.

The kernel of the integral equation $$ \psi(x)=\lambda \int_{a}^{b} K(x, y) \psi(y) d y $$ has the form $$ K(x, y)=\sum_{n=0}^{\infty} h_{n}(x) g_{n}(y) $$ where the \(h_{n}(x)\) form a complete orthonormal set of functions over the interval \([a, b]\) (a) Show that the eigenvalues \(\lambda_{i}\) are given by $$ \left|\mathrm{M}-\lambda^{-1}\right| \mid=0 $$ where \(M\) is the matrix with elements $$ M_{k j}=\int_{a}^{b} g_{k}(u) h_{j}(u) d u $$ If the corresponding solutions are \(\psi^{(i)}(x)=\sum_{n=0}^{\infty} a_{n}^{(i)} h_{n}(x)\), find an expression for \(a_{n}^{(i)}\). (b) Obtain the eigenvalues and eigenfunctions over the interval \([0,2 \pi]\) if $$ K(x, y)=\sum_{n=1}^{\infty} \frac{1}{n} \cos n x \cos n y $$

For \(f(t)=\exp \left(-t^{2} / 2\right)\), use the relationships of the Fourier transforms of \(f^{\prime}(t)\) and \(t f(t)\) to that of \(f(t)\) itself to find a simple differential equation satisfied by \(\tilde{f}(\omega)\), the Fourier transform of \(f(t)\), and hence determine \(\tilde{f}(\omega)\) to within a constant. Use this result to solve the integral equation $$ \int_{-\infty}^{\infty} e^{-t(t-2 x) / 2} h(t) d t=e^{3 x^{2} / 8} $$ for \(h(t)\).

Use Fredholm theory to show that, for the kernel $$ K(x, z)=(x+z) \exp (x-z) $$ over the interval \([0,1]\), the resolvent kernel is $$ R(x, z ; \lambda)=\frac{\exp (x-z)\left[(x+z)-\lambda\left(\frac{1}{2} x+\frac{1}{2} z-x z-\frac{1}{3}\right)\right]}{1-\lambda-\frac{1}{12} \lambda^{2}} $$ and hence solve $$ y(x)=x^{2}+2 \int_{0}^{1}(x+z) \exp (x-z) y(z) d z $$ expressing your answer in terms of \(I_{n}\), where \(I_{n}=\int_{0}^{1} u^{n} \exp (-u) d u\).

The operator \(\mathscr{M}\) is defined by $$ \mathscr{M} f(x) \equiv \int_{-\infty}^{\infty} K(x, y) f(y) d y $$ where \(K(x, y)=1\) inside the square \(|x|
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Wave Optics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.