First Order Linear Differential Equations

1. Idea of a First Order Linear Differential Equation

A first order linear differential equation connects a function and its first derivative in a linear manner. The equation involves no powers or products of the dependent variable and its derivative. The structure is simple and allows a systematic method of solution using an integrating factor.

The idea is that even though the function itself may appear with powers or other expressions, the derivative appears only to the first power, and the dependent variable does not multiply its own derivative.

1.1. Recognising linear form

A differential equation is linear in \(y\) if \(y\) and \(\frac{dy}{dx}\) appear only to the first power, and are not multiplied together. If this holds, the equation can be rearranged into a standard form that is easy to solve.

2. Standard Form of a First Order Linear Differential Equation

The standard form of a first order linear differential equation is:

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

Here, \(P(x)\) and \(Q(x)\) are known functions of the independent variable \(x\). Once the equation is in this form, the integrating factor method can be applied directly.

2.1. Key components

\(P(x)\): coefficient of \(y\)
\(Q(x)\): the non-homogeneous term
\(\frac{dy}{dx}\): appears with coefficient 1 in standard form

3. Integrating Factor (IF)

The integrating factor is a multiplier that converts the left-hand side of the standard form into the derivative of a product. This simplifies the solving process drastically.

The integrating factor is defined as:

\(\text{IF} = e^{\int P(x)\, dx}.\)

3.1. Why the integrating factor works

Multiplying the entire equation by the integrating factor makes the left-hand side become:

\(\frac{d}{dx}\big(\text{IF} \cdot y\big).\)

This turns the linear differential equation into an easily integrable expression. The method works because it reverses the product rule for differentiation.

4. Method of Solution

Once the equation is in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\), apply the following steps:

4.1. Step-by-step procedure

Identify \(P(x)\) and \(Q(x)\).
Find the integrating factor (IF):
\(\text{IF} = e^{\int P(x)\, dx}.\)
Multiply the entire differential equation by the IF.
Recognise the left-hand side as the derivative of a product:
\(\frac{d}{dx}(y\,\text{IF}).\)
Integrate both sides with respect to \(x\).
Solve for \(y(x)\).

5. General Solution Formula

The general solution to the equation

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

is given by:

\(y\, e^{\int P(x) dx} = \int Q(x)\, e^{\int P(x) dx}\, dx + C.\)

Solving for \(y\) gives the explicit solution.

6. Worked Examples

The following examples demonstrate the integrating factor method clearly.

6.1. Example 1: Simple linear equation

Solve:

\(\frac{dy}{dx} + y = e^x.\)

Here, \(P(x) = 1\) and \(Q(x) = e^x\).

Integrating factor:

\(\text{IF} = e^{\int 1 \, dx} = e^x.\)

Multiply through by \(e^x\):

\(e^x \frac{dy}{dx} + e^x y = e^{2x}.\)

Left-hand side becomes:

\(\frac{d}{dx}(y e^x).\)

Integrate both sides:

\(y e^x = \int e^{2x} dx = \frac{e^{2x}}{2} + C.\)

Thus:

\(y = \frac{e^x}{2} + Ce^{-x}.\)

6.2. Example 2: Coefficient depending on x

Solve:

\(\frac{dy}{dx} + 2x y = x.\)

Here, \(P(x) = 2x\), \(Q(x) = x\).

Integrating factor:

\(\text{IF} = e^{\int 2x dx} = e^{x^2}.\)

Multiply through by the integrating factor:

\(e^{x^2}\frac{dy}{dx} + 2x e^{x^2} y = x e^{x^2}.\)

Left-hand side becomes:

\(\frac{d}{dx}(y e^{x^2}).\)

Integrate:

\(y e^{x^2} = \int x e^{x^2} dx.\)

Substitute \(t = x^2\), \(dt = 2x dx\):

\(\int x e^{x^2} dx = \frac{1}{2} e^{x^2} + C.\)