1.3. First-Order Linear Differential Equations

The general form of a first-order linear differential equation is:

$\frac{dy}{dx} + P(x)y = Q(x)$

In this equation, $y$ is the unknown function that depends on $x$, and it can be thought of as a dependent variable. The function $P(x)$ is a given function that multiplies with $y$, and $Q(x)$ is another given function that can be seen as a source term. This equation essentially describes how the rate of change of the variable $y$ relates to its current state and the influences represented by $P(x)$ and $Q(x)$.

Imagine a tank that is being filled with water. The amount of water in the tank at any time $y$ is influenced by both the speed at which water flows in (represented by $Q(x)$) and the existing pressure or flow conditions inside the tank (represented by $P(x)$).

To solve the equation, the standard method involves the following steps:

1. Multiply both sides by an integrating factor: $\mu(x) = e^{\int P(x)dx}$
2. This transforms the equation into: $\frac{d}{dx}[\mu(x)y] = \mu(x)Q(x)$
3. Integrate both sides: $y = \frac{1}{\mu(x)}\int \mu(x)Q(x)dx + C$

The integrating factor $\mu(x)$ is calculated based on the function $P(x)$. This integrating factor serves to simplify the left side of the equation so that it can be expressed as the derivative of a product. After multiplying by $\mu(x)$, the equation can be integrated, yielding a solution for $y$ in terms of an integral involving $Q(x)$ and the integrating factor itself, along with an arbitrary constant $C$. This constant represents the general solution's family of curves.

If we think of a car's speedometer, the rate of change of speed can depend on various factors such as acceleration and road conditions. By determining an integrating factor (like using additional information such as fuel consumption rates), we can better predict the car's speed over time, leading to a clearer understanding of its behavior under different conditions.

Consider the differential equation:

$\frac{dy}{dx} + 2y = e^{-x}$

Solution:

1. Identify $P(x) = 2$, thus $\mu(x) = e^{\int 2dx} = e^{2x}$.
2. Multiply both sides: $e^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{x}$
3. Rewrite as: $\frac{d}{dx}(e^{2x}y) = e^{x}$
4. Integrate both sides: $e^{2x}y = \int e^{x}dx + C$
5. Solve: $y = e^{-x} + Ce^{-2x}$

In this example, we begin by determining the appropriate integrating factor based on the function $P(x) = 2$. Once we have calculated $\mu(x)$, we multiply the entire differential equation by it, transforming our equation to expose the derivative on the left. After this transformation, we integrate both sides to find the general solution for $y$ with an arbitrary constant.

If we think about cooling a cup of coffee, the rate at which the coffee cools down is affected by the current temperature (which can be analogous to $y$). The outside environment (like air temperature) plays a role too (analogous to $Q(x)$). Using the integrating factor can be thought of as recognizing how these external influences change the cooling rate over time, helping us reach a clearer conclusion about the coffee's temperature changes.

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