91Ó°ÊÓ

First-order linear differential equations are a foundational concept in calculus and differential equations. These equations are called 'first-order' because they involve the first derivative of the unknown function, and 'linear' because the function and its derivative appear to the first power. An example of a first-order linear differential equation is given by:

• \( \frac{d y}{d x} = y + 3 \)

This specific form represents a relationship between the derivative \( \frac{d y}{d x} \), which is the rate of change of the function \( y \) with respect to \( x \), and the function itself \( y \). Linear differential equations often have solutions that can be calculated using integration techniques, once they are simplified through algebraic manipulation or separation of variables.

Separation of variables is a helpful technique used to solve differential equations, especially first-order linear ones like the given example. The main strategy is to rearrange the equation so that all the terms involving \( y \) are on one side, and all the terms involving \( x \) are on the other side. Here's how the separation is typically done:

• Original equation: \( \frac{d y}{d x} = y + 3 \)
• Recast it as: \( \frac{d y}{y + 3} = dx \)

This manipulation allows us to perform integration on each side individually. Each variable is isolated which makes it easier to handle and ultimately solve the equation. By integrating both sides separately, we progress towards finding a function \( y(x) \) that satisfies the original differential equation.

Initial conditions are specific values that allow us to find a unique solution to a differential equation. In this problem, the initial condition is given as \( y(0) = 2 \). The significance of this condition is that it lets us determine the integration constant \( C \) that appears after integrating a differential equation. When applying the initial condition to the equation \( \ln|y + 3| = x + C \), first substitute \( x = 0 \) and \( y = 2 \):

• Calculate \( \ln|2 + 3| = 0 + C \)
• This results in \( C = \ln 5 \)

With \( C \) known, we can fully specify our solution, ensuring it not only fits the equation generally but also starts at the correct point in its path, which in this case, is the value 2 when \( x = 0 \).

The integration constant \( C \) is an essential component in solving differential equations. It arises during the integration process of separating variables or other solution techniques. After integrating, you get an equation with an undetermined constant \( C \), such as:

• \( \ln|y + 3| = x + C \)

After solving for \( C \) using the initial condition, substitute it back into the equation to get:

• \( \ln|y + 3| = x + \ln 5 \)

Exponentiating both sides helps solve for \( y \), leading to:

• \( |y + 3| = 5e^x \), thus \( y = 5e^x - 3 \)

The integration constant helps shape solutions so they fit specific conditions stated by the problem, transforming a general solution like \( |y + 3| = e^{x+C} \) into the unique solution demanded by the initial condition.