91Ó°ÊÓ

The concept of an integrating factor is a key technique in solving first-order linear differential equations like the one given in the problem. Consider the equation \( y'(t) + a(t) y(t) = f(t) \). An integrating factor, typically denoted as \( p(t) \), is a function used to simplify and solve these equations. It achieves this by transforming the left side of the equation into an exact derivative. Without this transformation, solving the equation analytically is often quite challenging.

Here's how you calculate the integrating factor:

• First, identify the function \( a(t) \) from the equation. This is the coefficient of \( y(t) \).
• Next, compute \( p(t) \) using the formula \( p(t) = \exp(\int a(t) \, dt) \). In our example, \( a(t) = 2t \), and the integrating factor becomes \( p(t) = \exp(t^2) \) after integrating.

The key property of the integrating factor is that when you multiply the entire differential equation by \( p(t) \), the left side forms an exact derivative. This means it can now be expressed as \( \frac{d}{dt}(p(t) y(t)) \), which greatly simplifies solving the equation when integrated.

In many differential equations like the one given, you're not only asked to find a general solution but also a particular solution that satisfies a given condition. This type of problem is known as an initial value problem. It combines the process of solving the differential equation with applying given specific conditions, such as \( y(0) = 1 \) in this case.

To solve an initial value problem, follow these steps:

• First, find the general solution of the differential equation using techniques such as integrating factors.
• Once you have the solution, apply the initial condition to find any unknown constants in your solution. In our example, after solving for \( y(t) \), you substitute \( y(0) = 1 \) into your equation to determine the constant \( C \).

Applying the initial condition is a crucial step, as it narrows down the solution to the one specific to the given problem. It ensures that your solution fits the particular scenario rather than any possible scenario that the general solution could cover.

Exact derivatives are important to understand as they can simplify solving differential equations. In the context of our problem, the integrating factor transforms the left side of our equation into an exact derivative. Let's discuss what this means and why it helps.

An exact derivative essentially means the expression can be written as the derivative of a product, such as \( \frac{d}{dt}(p(t) y(t)) \). Converting the equation into this form is beneficial because:

• It allows us to integrate directly, bypassing the complexity of solving the differential equation manually. Integration of derivatives resolves into the original function plus a constant, making the process straightforward.
• This transformation leverages the Fundamental Theorem of Calculus, simplifying the problem by reducing it to evaluating integrals.

In our example, multiplying by the integrating factor transforms the left side into \( \frac{d}{dt}(\exp(t^2) y(t)) \), which can then be integrated to find \( y(t) \). Recognizing and utilizing exact derivatives can therefore be a powerful tool in the differential equation toolbox.