91Ó°ÊÓ

An integrating factor is a clever mathematical tool used to simplify solving linear ordinary differential equations, particularly those which are not exact. The integrating factor is a function, often denoted as \( \mu(t) \), which is multiplied across the entire differential equation.
This operation transforms the equation into one that is "exact," meaning we can easily integrate both sides.
Here's how it works:

• First, identify the coefficient of \( y \) in the equation \( y' + P(t)y = Q(t) \). The coefficient is \( P(t) \).
• Compute the integrating factor by evaluating the integral \( \mu(t) = e^{\int P(t) \, dt} \).
• Multiply the entire differential equation by this integrating factor.

This step rewrites the equation in a more integrable form, preparing it for easier solution.

The error function, or \( \operatorname{erf}(t) \), is a special function that appears in probability, statistics, and differential equations.
Its definition revolves around Gaussian probability and is given as:\[\operatorname{erf}(t) = \frac{2}{\sqrt{\pi}} \int_{0}^{t} e^{-s^2} \, ds.\]This function measures the probability that a Gaussian random variable falls within a certain range.
In differential equations, the error function comes up when solving integrals of the form \( e^{-t^2} \), like in our exercise.
Some key properties of \( \operatorname{erf}(t) \) include:

• \( \operatorname{erf}(0) = 0 \)
• It asymptotically approaches 1 as \( t \to \infty \)

Incorporating the error function helps express solutions involving these special integrals concisely and accurately.

An initial value problem (IVP) involves determining a function that satisfies a differential equation and passes through a given point, known as an initial condition.
For a differential equation of the form \( y' = f(t, y) \), the initial value \( y(t_0) = y_0 \) specifies the value of the function at a particular point \( t_0 \).
The given exercise specifies an IVP with \( y(0) = 2 \).
To solve an IVP:

• First, solve the differential equation to find a general solution.
• Next, apply the initial condition to pinpoint the specific solution among the possible functions.

This method ensures the solution is tailored to the problem's conditions, fitting the curve appropriately from the start.

An exact equation in the context of differential equations is one where a single derivative forms the entire equation.
This structure allows for direct integration.
Given a differential equation in form \( M(x, y) + N(x, y) \frac{dy}{dx} = 0 \), if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.
For solving, the goal is to find a potential function \( \Psi(x, y) \) such that:

• \( \Psi_x = M \)
• \( \Psi_y = N \)

In our exercise, the equation \( (y e^{-t^2})' = e^{-t^2} \) arises after applying the integrating factor, simplifying the problem to integrating both sides.
This process helps convert complex linear differential equations into forms we can easily solve through integration.