91Ó°ÊÓ

A first-order linear differential equation is a type of differential equation that involves the first derivative of a function. The standard form is given by \( y' + p(t)y = g(t) \). Here, \( y' \) represents the derivative of \( y \) with respect to an independent variable \( t \), while \( p(t) \) and \( g(t) \) are given functions of \( t \). In the original problem, the equation \( t^2 y' + y = 0 \) is a first-order linear differential equation. By dividing every term by \( t^2 \), it can be rewritten in the standard form: \( y' + \frac{1}{t^2} y = 0 \). This reformulation allows us to systematically solve the equation using techniques specific to first-order linear differential equations.

The integrating factor is a crucial technique for solving first-order linear differential equations. It simplifies the equation into an easily integrable form. To find the integrating factor, you use the expression \( \mu(t) = e^{\int p(t) \, dt} \), where \( p(t) \) is taken from the standard form \( y' + p(t)y = g(t) \). In the given exercise, \( p(t) = \frac{1}{t^2} \). Calculating the integral, we find\[ \mu(t) = e^{\int \frac{1}{t^2} \, dt} = e^{-1/t} \].This integrating factor \( \mu(t) \) is then multiplied throughout the differential equation, which allows the equation to be rewritten in terms of a total derivative.

The constant of integration, often denoted as \( C \), emerges when integrating a function. In the context of differential equations, it represents the family of solutions that satisfy the homogeneous part of the equation. When integrating the equation after applying the integrating factor, we encounter a formula like \( e^{-1/t}y = C \). This step introduces the constant \( C \). The value of \( C \) is subsequently determined using initial or boundary conditions provided in the problem. In this exercise, the initial condition \( y(1) = -2 \) is applied. This condition enables the determination of \( C \) through substitution, which is essential for finding the particular solution of the differential equation.

A particular solution to a differential equation is a specific solution that satisfies both the differential equation and any given initial conditions. Generally, after finding a general solution that includes the constant of integration \( C \), we use initial conditions to solve for \( C \). In the given problem, after finding \( y = Ce^{1/t} \), we apply the initial condition \( y(1) = -2 \). Substituting \( t = 1 \) and \( y = -2 \) entails that \( -2 = Ce^{1} \), giving us \( C = -2e^{-1} \). Finally, substituting \( C \) back into the expression yields the particular solution: \[ y = -2e^{1/t - 1} \]. This solution is unique to the original initial value problem and describes the behavior of \( y(t) \) for the specific initial conditions given.