Question

Show that \(y=\int_{0}^{x} f(u) \sin (x-u) d u\) satisfies \(y^{\prime \prime}+y=f(x).\)

Short answer

It is shown that the function \( y \) satisfies the equation \( y'' + y = f(x) \).

Step-by-step solution

Define the Given Function

The given function is: \[ y = \int_{0}^{x} f(u) \sin (x-u) \; du \].

Compute the First Derivative

Use the Leibniz rule for differentiating an integral with variable limits: \[ y' = \frac{d}{dx} \left( \int_{0}^{x} f(u) \sin (x-u) \; du \right) = f(x)\sin(x-x) + \int_{0}^{x} f(u) \frac{\partial}{\partial x} \sin(x-u) \; du \]. Simplify the expression: \[ y' = f(x) \cdot 0 + \int_{0}^{x} f(u) \cos(x-u) \; du = \int_{0}^{x} f(u) \cos(x-u) \; du \].

Compute the Second Derivative

Differentiate the first derivative using the Leibniz rule again: \[ y'' = \frac{d}{dx} \left( \int_{0}^{x} f(u) \cos (x-u) \; du \right) = f(x)\cos(x-x) + \int_{0}^{x} f(u) \frac{\partial}{\partial x} \cos(x-u) \; du \]. Simplify the expression: \[ y'' = f(x)\cdot 1 + \int_{0}^{x} f(u) (-\sin(x-u)) \; du = f(x) - \int_{0}^{x} f(u) \sin(x-u) \; du \].

Substitute Back and Conclude

Recognize that the expression \( \int_{0}^{x} f(u) \sin(x-u) \; du \) is the original function \( y \): \[ y'' + y = f(x) - \int_{0}^{x} f(u)\sin(x-u) \; du + \int_{0}^{x} f(u)\sin(x-u) \; du = f(x). \] Thus, it is shown that \( y'' + y = f(x) \).

Key concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They describe a wide variety of phenomena in science and engineering. For example, the given problem involves a second-order linear differential equation. Understanding how to solve these types of equations is crucial for modeling real-world systems.
There are different types of differential equations, such as ordinary differential equations (ODEs) and partial differential equations (PDEs). This exercise focuses on an ODE.

Leibniz Rule

The Leibniz rule is a formula for differentiating an integral whose limits are functions of the variable of differentiation. In simpler terms, it tells you how to differentiate an integral when the integrand or the limits of integration depend on the variable with respect to which you are differentiating.
The general form of the Leibniz rule for an integral from \(\alpha(x)\) to \(\beta(x)\) is:
\[ \frac{d}{dx} \left( \int_{\alpha(x)}^{\beta(x)} f(x, t) \; dt \right) = f(x, \beta(x)) \cdot \beta'(x) - f(x, \alpha(x)) \cdot \alpha' (x) + \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(x, t)}{\partial x} \; dt \].
In this exercise, the Leibniz rule is used twice to find the first and second derivatives of the given function.

Integral Calculus

Integral calculus deals with integrals and their properties. Integrals are used to calculate areas, volumes, central points, among other things. Definite integrals, like the one in this exercise, compute the accumulated quantity over an interval.
The given function is formulated as a definite integral:
\[ y = \int_{0}^{x} f(u) \sin (x-u) \; du \].
Understanding how to work with integrals and their properties, such as the limits of integration and the integrand, is essential in solving problems like these.

Second Derivative

The second derivative measures the rate at which the first derivative of a function is changing. It provides information about the concavity of the original function.
In this problem, you first find the first derivative of the given integral function, and then you differentiate the result again to get the second derivative. The process involves the application of the Leibniz rule twice.
The steps are:
1. Compute the first derivative: \[ y' = \int_{0}^{x} f(u) \cos(x-u) \; du \].
2. Compute the second derivative: \[ y'' = f(x) - \int_{0}^{x} f(u) \sin(x-u) \; du \].
The second step shows how changes in the original function affect the second derivative, which is central to verifying the given differential equation.