Chapter 1 · First-Order ODEs

First-Order Linear Differential Equations

The Integrating Factor Method

Dr. Mohamed Mabrok · Qatar University

What is a Linear First-Order ODE?

A linear equation is one where $y$ and $y'$ appear only to the first power — no $y^2$, $\sin(y)$, or $y \cdot y'$ terms.

Standard Form

      A first-order linear ODE has the form:
    
      $$\frac{dy}{dx} + P(x)y = Q(x)$$

where $P(x)$ and $Q(x)$ are functions of $x$ alone (not $y$).

What Makes It Linear?

      Both $y$ and $\frac{dy}{dx}$ appear to the first power only. The equation is linear in both the dependent variable and its derivative.

The Integrating Factor Method

The key insight: multiply the equation by a special function $\mu(x)$ called the integrating factor.

Integrating Factor Formula

      $$\mu(x) = e^{\int P(x)\,dx}$$
    
      Multiply both sides of the ODE by $\mu(x)$.

General Solution Formula

      After multiplying by $\mu$ and integrating:
    
      $$y(x) = \frac{1}{\mu(x)}\left[\int \mu(x)\,Q(x)\,dx + C\right]$$

Why the Integrating Factor Works

STEP 1 Start with the standard form

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

STEP 2 Multiply both sides by $\mu(x) = e^{\int P(x)\,dx}$

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

STEP 3 The left side is the product rule!

\frac{d}{dx}[\mu(x) \cdot y] = \mu(x)Q(x)

STEP 4 Integrate both sides and solve for $y$

The 4-Step Solution Method

STEP 1
        Write in standard form

        Divide by coefficient of $y'$ if needed: $\frac{dy}{dx} + P(x)y = Q(x)$
      

STEP 2
        Compute $\mu(x)$

        $\mu(x) = e^{\int P(x)\,dx}$ (ignore the $+C$ in the exponent)
      

STEP 3
        Apply the formula

        $y = \frac{1}{\mu}\left[\int \mu Q\,dx + C\right]$
      

STEP 4
        Apply initial conditions

        Use $y(x_0) = y_0$ to find the constant $C$
      

Worked Example

Solve: $y' + 2y = 6$, $y(0)=1$

STEPS 1–2 Identify $P(x)$ and compute $\mu(x)$

P(x) = 2, \quad Q(x) = 6 \quad \Longrightarrow \quad \mu(x) = e^{\int 2\,dx} = e^{2x}

STEP 3 Apply the formula: $y = \frac{1}{e^{2x}}\left[\int 6e^{2x}\,dx + C\right]$

\int 6e^{2x}\,dx = 3e^{2x} + C_1 \quad \Longrightarrow \quad y = e^{-2x}(3e^{2x} + C) = 3 + Ce^{-2x}

General solution: $\boxed{y = 3 + Ce^{-2x}}$

Worked Example (continued)

Apply Initial Condition: $y(0) = 1$

STEP 4 Substitute $x=0$, $y=1$

1 = 3 + Ce^{0} = 3 + C \quad \Longrightarrow \quad C = -2

Particular solution:

\boxed{y = 3 - 2e^{-2x}}

Key Observation

      As $x \to \infty$, $e^{-2x} \to 0$, so $y \to 3$. The value $y = 3$ is called the equilibrium.

      All solutions (regardless of initial condition) converge to this equilibrium!

Variable Coefficients Example

The integrating factor method also works when $P(x)$ varies with $x$.

Example Solve $xy' + y = x^2$

Step 1: Divide by $x$ to get standard form:

y' + \frac{1}{x}y = x

Step 2: Compute $\mu(x)$:

\mu(x) = e^{\int (1/x)\,dx} = e^{\ln|x|} = x

Step 3: General solution:

y = \frac{1}{x}\left[\int x \cdot x\,dx + C\right] = \frac{1}{x}\left[\frac{x^3}{3} + C\right] = \frac{x^2}{3} + \frac{C}{x}

Common Pitfalls

Pitfall 1: Coefficient of $y'$ is not 1

      Always divide first! If your equation is $2y' + 4y = 10$, divide by 2 to get $y' + 2y = 5$ before finding $\mu(x)$.

Pitfall 2: Forgetting the integration constant

      When you compute $\mu(x) = e^{\int P(x)\,dx}$, ignore the $+C$ in the integral. But don't forget $C$ in the final formula!

Pitfall 3: Negative $P(x)$ leads to exponential growth

      If $P(x) = -3$ (negative), then $\mu(x) = e^{-3x}$ causes growth as $x$ increases. This is correct — solutions grow!

Pitfall 4: Not all integrals have closed forms

      If $\int P(x)\,dx$ or $\int \mu Q\,dx$ cannot be evaluated, your solution may be left in integral form (implicit).

Key Takeaways

📄

        Standard form is essential

        Always write as $\frac{dy}{dx} + P(x)y = Q(x)$ with coefficient of $y'$ equal to 1.

✓️

        $\mu$ transforms via product rule

        Multiplying by $\mu = e^{\int P\,dx}$ makes the left side $\frac{d}{dx}[\mu y]$.

➡

        Constant coefficients give equilibrium

        For $y' + ay = b$ (constant), the equilibrium is $y_{eq} = \frac{b}{a}$. All solutions converge to it.

🌌

        Real-world applications abound

        Mixing tanks, RC circuits, Newton's cooling law, pharmacokinetics, and population dynamics all use linear ODEs.

\text{Standard form} \;\Rightarrow\; \mu(x) = e^{\int P\,dx} \;\Rightarrow\; \text{Formula} \;\Rightarrow\; \text{Apply IC}