91Ó°ÊÓ

First-order linear differential equations are one of the simplest types of differential equations you'll encounter. They are called "first-order" because they involve the first derivative of the unknown function, but not higher-order derivatives. The general form of a first-order linear differential equation is given by:\[ y' + P(t) y = Q(t) \]Key components of this equation include:

• \( y' \) is the first derivative of \( y \) with respect to \( t \).
• \( P(t) \) and \( Q(t) \) are given functions of \( t \).

This form allows for specific techniques to be applied for finding the solution. In this exercise, we identified that our equation, \( y' = y + 2 \cos t \), fits this form by rearranging it to \( y' - y = 2 \cos t \), making \( P(t) = -1 \) and \( Q(t) = 2 \cos t \). These differential equations can model real-world phenomena, such as cooling dynamics or population growth, which often depend linearly on the current state of the system.

The integrating factor method is a crucial technique for solving first-order linear differential equations. When you're dealing with an equation of the form \( y' + P(t) y = Q(t) \), an important tool is the integrating factor, \( \mu(t) \). This factor simplifies the equation, making it easier to solve.To find the integrating factor:

• Calculate \( \mu(t) = e^{\int P(t) \, dt} \).
• In many cases, this involves a simple integral, like \( \int -1 \, dt \), leading to \( \mu(t) = e^{-t} \).

Once you have \( \mu(t) \):

• Multiply every term of the differential equation by this factor.
• Transform the left-hand side of the equation into the derivative of a product, making integration straightforward.

For our example, multiplying by \( e^{-t} \) led to:\[ \frac{d}{dt}(e^{-t} y) = 2 e^{-t} \cos t \]This allowed us to integrate both sides easily. Remember, the goal is to make both sides integrable, helping you to find the general solution through integration.

An initial value problem (IVP) is a differential equation paired with specific conditions at a starting point. For our exercise, we began with an equation: \( y' = y + 2 \cos t \), and an initial condition: \( y(0) = -1 \). This initial condition implies that at \( t = 0 \), the function \( y \) equals \(-1\).Solving an IVP involves:

• Finding the general solution to the differential equation using methods like integrating factors.
• Substituting the initial condition to determine any constants in the solution.

In the given problem, after deriving the general solution, \( y = \sin t + \cos t + C e^t \), we applied the initial condition:

• Plug \( t = 0 \) into the equation: \(-1 = \sin 0 + \cos 0 + C e^0 \).
• Simplify to find \( C = -2 \).

Thus, the solution satisfying both the equation and the initial condition is \( y = \sin t + \cos t - 2 e^t \). Initial value problems are ubiquitous in applications, providing solutions that respect specific starting conditions, such as temperature levels or population sizes at time zero.