Linear First-Order Differential Equations

A linear first-order differential equation is an equation of the form dy/dx + P(x)y = Q(x), where P(x) and Q(x) are continuous functions of x alone. These equations appear throughout science and engineering — modelling radioactive decay, population growth, mixing problems, and electric circuits. Unlike nonlinear equations, they can always be solved by a systematic algebraic method, making them one of the most accessible and powerful tools in calculus. The key insight behind solving these equations is the concept of an integrating factor, a cleverly chosen function μ(x) that transforms the left-hand side into an exact derivative. Once the equation is rewritten as d/dx[μ(x)y] = μ(x)Q(x), both sides can be integrated directly to find y. This technique turns what looks like a complicated relationship between y and its derivative into a straightforward integration problem. Mastering linear first-order DEs builds the foundation for studying higher-order equations, systems of DEs, and applied modelling. The method is entirely algorithmic: identify P(x) and Q(x), compute the integrating factor, multiply through, integrate, and solve for y. With practice, the process becomes fast and reliable, and the solutions reveal how quantities change continuously over time or space.

Key takeaways
  • Write the equation in standard form
  • Compute the integrating factor μ(x)
  • Multiply both sides by μ(x)
  • Integrate both sides
  • Solve for y

Write the equation in standard form

Rearrange the equation so it reads dy/dx + P(x)y = Q(x). Every term involving y or dy/dx must be on the left, and everything else on the right. For example, dy/dx - 3y = e^{2x} is already in standard form with P(x) = -3 and Q(x) = e^{2x}.

Compute the integrating factor μ(x)

The integrating factor is μ(x) = e^{\int P(x)\,dx}. You do not need the constant of integration at this stage — any antiderivative of P(x) works. For P(x) = -3, we get μ(x) = e^{\int -3\,dx} = e^{-3x}.

Multiply both sides by μ(x)

Multiply every term in the standard-form equation by μ(x). The left-hand side must now equal d/dx[μ(x)y] — this is the whole point of the integrating factor. You can verify this using the product rule: d/dx[μy] = μ(dy/dx) + μ'y = μ(dy/dx) + μPy, which matches μ times the left side.

Integrate both sides

Since the left side is d/dx[μ(x)y], integrating gives μ(x)y = \int μ(x)Q(x)\,dx + C. Be sure to include the constant of integration C, as it generates the family of all solutions.

Solve for y

Divide both sides by μ(x) to isolate y: y = \frac{1}{μ(x)}\left[\int μ(x)Q(x)\,dx + C\right]. This is the general solution. If an initial condition y(x_0) = y_0 is given, substitute it to find the specific value of C and obtain the particular solution.

The Key Equation

dydx+P(x) y=Q(x),μ(x)=e∫P(x) dx,y=1μ(x)∫μ(x) Q(x) dx+Cμ(x)\frac{dy}{dx} + P(x)\,y = Q(x), \qquad \mu(x) = e^{\int P(x)\,dx}, \qquad y = \frac{1}{\mu(x)}\int \mu(x)\,Q(x)\,dx + \frac{C}{\mu(x)}
  • y — Dependent variable (varies): The unknown function of x being solved for.
  • P(x) — Coefficient function (dimensionless): A continuous function of x that multiplies y; determines the integrating factor.
  • Q(x) — Forcing function (varies): The right-hand side function; represents an external input or source term.
  • \mu(x) — Integrating factor (dimensionless): The function e^{\int P(x)dx} used to make the left side an exact derivative.
  • C — Constant of integration (varies): Arbitrary constant that produces the full family of solutions; fixed by an initial condition.

Worked Example

Problem. Solve the initial value problem: dy/dx + (2/x)y = 4x, with y(1) = 3.

y=x2+2x2y = x^2 + \dfrac{2}{x^2}

Answer. y = x² + 2/x²

Figure
Diagram
Solution Curves: dy/dx + (2/x)y = 4xxy012342612y(1)=3 (C=2)y(1)=2 (C=1)y(1)=4 (C=3)Each curve y = x² + C/x² passes through a different initial point
Origins · 1 minGottfried Wilhelm Leibniz and how this was discovered
GW
Gottfried Wilhelm Leibniz
1646–1716 · German

Leibniz co-invented calculus independently of Newton and developed much of the notation still used today, including dy/dx for derivatives and the integral sign ∫. He was among the first to study differential equations systematically and introduced techniques for separating variables and finding integrating factors, laying the groundwork for the theory of linear first-order ODEs. Leibniz designed one of the first mechanical calculators capable of multiplication and division — the Stepped Reckoner — and spent years trying to build a reliable working model.

  • Separable Differential Equations — Separable equations are another type of first-order ODE solved by a different technique; comparing the two methods deepens understanding of when each applies.
  • Integration Techniques — Solving a linear first-order ODE requires integrating both P(x) to find the integrating factor and μ(x)Q(x) for the solution; fluency with integration is essential.
  • Initial Value Problems — An initial condition y(x₀) = y₀ selects one particular solution from the family; understanding IVPs is key to applying DEs to real-world situations.
  • Exponential Functions and the Natural Logarithm — The integrating factor e^{∫P(x)dx} almost always involves exponentials; recognising and manipulating e^x and ln(x) is crucial throughout this topic.