91Ó°ÊÓ

In the world of mathematics, a _first-order differential equation_ is an equation that involves the function and its first derivative. This means it essentially describes how a change in one variable can affect another through the rate of change of one of those variables. The equation has the form \( P(t) \frac{d y}{d t} + Q(t) y = g(t) \), where \( P(t) \), \( Q(t) \), and \( g(t) \) are functions of the independent variable, usually considered as time \( t \).

In our specific problem, the equation \( t \frac{d y}{d t} + 2y = t^3 \) can be seen fitting to this type because it contains the first derivative of \( y \) with respect to \( t \), which aligns it with the first-order category. Here's how each part lines up:

• \( P(t) = t \)
• \( Q(t) = 2 \)
• \( g(t) = t^3 \)

Breaking down the components of the equation helps us understand what factors influence the function's change over time. Recognizing the type of differential equation is crucial as it guides us on how to approach finding the solution.

The _integrating factor method_ is a strategy used to solve linear first-order differential equations. It revolves around simplifying and transforming the equation to make it easier to integrate. This method is particularly useful when handling equations of the form \( \frac{d y}{d t} + P(t) y = g(t) \).

To apply this method, you first need to identify the integrating factor, denoted as \( \mu(t) \). This factor is determined by:
\[\mu(t) = e^{\int P(t)} \thinspace dt\]

In the given exercise, after rewriting the equation in the standard form, we find \( P(t) = \frac{2}{t} \). Thus, the integrating factor becomes:
\[\mu(t) = e^{\int \frac{2}{t} \, dt} = e^{2 \ln|t|} = t^2\]

Multiplying the entire differential equation by this integrating factor transforms it into a form that can be easily integrated on both sides. This clever manipulation allows us to combine the terms with \( y \) and its derivative into a single derivative, \( \frac{d}{d t} (t^2 y) = t^4 \), paving the way for straightforward integration.

An _initial value problem_ (IVP) occurs when a differential equation is accompanied by a specific condition that the solution must fulfill, known as an initial condition. This adds a starting point to the problem, enabling the determination of a unique solution fitting specific criteria.

Here, the initial condition given in the problem is \( y(2) = 1 \). This provides a concrete value of the function at the point \( t = 2 \), which is essential for finding the particular solution. After integrating the equation and finding the general solution \( y(t) = \frac{t^3}{5} + \frac{C}{t^2} \), the initial condition is used to solve for the constant \( C \).

Let's break it down:

• Substitute \( t = 2 \) and \( y = 1 \) into the equation: \( 1 = \frac{2^3}{5} + \frac{C}{4} \)
• Solve for \( C \) to obtain \( C = \frac{12}{5} \)

Finally, by substituting the value of \( C \) back, you get the particular solution \( y(t) = \frac{t^3}{5} + \frac{12}{5t^2} \), satisfying both the differential equation and the initial condition. This complete solution is what characterizes the problem as an initial value one.