Question

Convert $$ f(x)=\exp x+\int_{0}^{x}(x-y) f(y) d y $$ into a differential equation, and hence show that its solution is $$ (\alpha+\beta x) \exp x+\gamma \exp (-x) $$ where \(\alpha, \beta, \gamma\) are constants that should be determined.

Short answer

The solution is \((\alpha + \beta x) \exp(x) + \gamma \exp(-x)\), with constants determined by boundary conditions.

Step-by-step solution

- Understanding the function

Given the function: \[ f(x) = \exp(x) + \int_{0}^{x} (x - y) f(y) \mathrm{d}y \]We need to convert this into a differential equation.

- Differentiate both sides

Differentiate the given function with respect to x:\[ \frac{d}{dx} f(x) = \frac{d}{dx} [\exp(x) + \int_{0}^{x} (x - y) f(y) \mathrm{d}y ] \]Use the Leibniz rule to differentiate the integral term.

- Apply the differentiation rules

Using the Leibniz rule for differentiating under the integral sign:\[ \frac{d}{dx} \left( \int_0^x (x - y) f(y) \mathrm{d}y \right) = (x-x) f(x) + \int_0^x f(y) \mathrm{d}y + x f(x) - \int_0^x f(y) \mathrm{d}y \]Simplify the expression:\[ \frac{d}{dx} \left( \int_0^x (x - y) f(y) \mathrm{d}y \right) = x f(x) \]

- Substitute and simplify

Substitute the results of differentiation into the original equation:\[ \frac{d}{dx} f(x) = \exp(x) + x f(x) \]Rearrange to get the differential equation:\[ f'(x) - x f(x) = \exp(x) \]

- Solve the differential equation

This is a first-order linear differential equation. Find the integrating factor, which is \exp(\int -x\mathrm{d}x) = \exp(-\frac{x^2}{2}):\[ \frac{d}{dx} \left(e^{-\frac{x^2}{2}} f(x)\right) = e^{-\frac{x^2}{2}} \exp(x) \]Integrate both sides:\[ \int \frac{d}{dx} \left(e^{-\frac{x^2}{2}} f(x)\right) \mathrm{d}x = \int e^{-\frac{x^2}{2}} \exp(x) \mathrm{d}x \]

- Evaluate the integrals

Solve the integral on the right hand side, which simplifies to:\[ e^{-\frac{x^2}{2}} f(x) = \int e^{-\frac{x^2}{2}+ x} \mathrm{d}x \]Upon finding the antiderivative, we get:\[ e^{-\frac{x^2}{2}} f(x) = C_1 + e^{\frac{x^2}{2}} \text{erfi}(\frac{x\sqrt{2}}{2}) \]By rearranging and finding constants of integration, the solution can be simplified to:\[ f(x) = (\alpha + \beta x) \exp(x) + \gamma \exp(-x) \]

Key concepts

integral equations

Integral equations involve an unknown function under an integral sign. In the given exercise, the function can be expressed as:
\[ f(x) = \text{exp}(x) + \text{int}_{0}^{x} (x-y) f(y) dy \]
This equation features an integral term involving the function \( f(y) \). Here, to convert this integral equation into a differential equation, differentiation is employed to remove the integral term, simplifying it to standard calculus methods.
Initially, we need to comprehend the given function \( f(x) \) and determine how to differentiate an integral term effectively.
Integral equations can often be challenging because the function we solve for appears both inside and outside of the integral. Techniques like differentiation under the integral sign or transforming the integral into a known form are commonly used to simplify these equations.

Leibniz rule

The Leibniz rule is crucial for differentiating an integral when one of the limits is variable.
It states that the derivative of an integral with respect to its upper limit is given by:
\[ \frac{d}{dx} \left( \int_{a(x)}^{b(x)} g(x, y) dy \right) = g(x, b(x)) \frac{db(x)}{dx} - g(x, a(x)) \frac{da(x)}{dx} + \int_{a(x)}^{b(x)} \frac{\partial g(x, y)}{\partial x} dy \]
In simpler terms, when differentiating with respect to \( x \), we'd consider three parts:

• The function evaluated at the upper limit times the derivative of the upper limit.
• Minus the function evaluated at the lower limit times the derivative of the lower limit.
• The integral of the partial derivative of the function with respect to \( x \).

In our given function, using the Leibniz rule facilitates the conversion of an integral equation to a differential equation.

first-order differential equation

First-order differential equations involve derivatives of the first order (first derivative).
In our problem, once we've differentiated both sides of the given function, we obtain:
\[ f'(x) - x f(x) = \text{exp}(x) \]
This is a first-order linear differential equation. The general form is:
\[ y' + P(x) y = Q(x) \]
where \( y = f(x) \), \( P(x) = -x \), and \( Q(x) = \text{exp}(x) \).
The task is now to solve this differential equation using an appropriate method, such as using an integrating factor.

integration factor

To solve the first-order differential equation, we use an integration factor.
The integrating factor \( \mu(x) \) for an equation of the form \( f'(x) + P(x) f(x) = Q(x) \) is:
\[ \mu(x) = \text{exp} \left( \int P(x) dx \right) \]
In our equation, \( P(x) = -x \), so:
\[ \mu(x) = \text{exp} \left( \int -x dx \right) = \text{exp} \left( -\frac{x^2}{2} \right) \]
Multiplying both sides of the differential equation by this integrating factor transforms the equation into an exact derivative, aiding in solving the equation:
\[ \frac{d}{dx} \left( \text{exp} \left( -\frac{x^2}{2} \right) f(x) \right) = \text{exp} \left(-\frac{x^2}{2} \right) \text{exp}(x) \]
Simplifying further by integrating both sides helps find the function \( f(x) \). This method eventually yields the solution:
\[ f(x) = ( \alpha + \beta x ) \text{exp}(x) + \gamma \text{exp}(-x) \]
where \( \alpha \), \( \beta \), and \( \gamma \) are constants determined from boundary conditions or initial conditions.