Integrating Factor¶

Method¶

Consider the linear first-order differential equation

with $p,q$ continuous on an interval $I$ and $x'(t)=\dfrac{\mathrm{d}x}{\mathrm{d}t}(t)$$\frac{\mathrm{d}x}{\mathrm{d}t}(t)$.

1. Choose an antiderivative $P$ of $p$ and set $\mu(t)=\mathrm{e}^{P(t)}$${\mathrm{e}}^{P(t)}$, the integrating factor.
2. Multiply the equation by $\mu(t)$. Since $\mu'(t)=p(t)\mu(t)$$p(t)\mu (t)$, the left-hand side becomes an exact derivative: $\mu x'+\mu p x=(\mu x)'$$(\mu x{)}^{\prime }$.
3. Integrate $(\mu x)'=\mu q$$\mu q$: $\ \mu(t)x(t)=\int\mu(t)q(t)\,\mathrm{d}t+C$$C$.
4. Divide by $\mu(t)>0$$0$: $\ x(t)=\mathrm{e}^{-P(t)}\Big(\int \mathrm{e}^{P(t)}q(t)\,\mathrm{d}t+C\Big)$$C)$, and fix $C$ from an initial condition.

With $x(t_0)=x_0$ and $P(t_0)=0$ this gives the closed form

Interpretation¶

The left-hand side $x'+px$ is not a derivative of anything simple; multiplying by $\mu$ turns it into the derivative of the product $\mu x$. Step 2 uses the product rule from differentiation rules and the chain rule for $\frac{\mathrm{d}}{\mathrm{d}t}\mathrm{e}^{P(t)}=P'(t)\mathrm{e}^{P(t)}$$P^{\prime }(t){\mathrm{e}}^{P(t)}$.

The two terms of the closed form have a meaning: $\mathrm{e}^{-P(t)}x_0$ solves the homogeneous equation $x'+px=0$$0$, and the integral is a particular solution that starts at $0$, so the solution is homogeneous part plus particular part, as in the general solution of a linear ODE. For constant $p=-a$ the formula reads $x(t)=\mathrm{e}^{a(t-t_0)}x_0+\int_{t_0}^t \mathrm{e}^{a(t-s)}q(s)\,\mathrm{d}s$${\int }_{t_0}^t{\mathrm{e}}^{a(t-s)}q(s)\mathrm{d}s$, the scalar form of the variation-of-constants formula that returns in Week 4 with the matrix exponential.

Common slips: forgetting to write the equation with coefficient $1$ in front of $x'$ before reading off $p$; using $\mathrm{e}^{-P}$ instead of $\mathrm{e}^{P}$; adding the constant $C$ after dividing by $\mu$ instead of before.

Key Results¶

• $\mu(t)=\mathrm{e}^{P(t)}$ is never zero, so multiplying by it neither loses nor creates solutions.
• Any antiderivative of $p$ can be used: replacing $P$ by $P+k$ multiplies $\mu$ by $\mathrm{e}^{k}$, which cancels.
• If $p,q$ are continuous on $I$, the method yields the unique solution of each initial value problem on all of $I$.
• For constant $p\neq0$ and constant $q$ the solution tends to $q/p$ when $p>0$$0$; this limit is the equilibrium.

Examples¶

Constant coefficients. $x'+2x=1$$1$, $x(0)=0$$0$. Take $\mu=\mathrm{e}^{2t}$${\mathrm{e}}^{2t}$: $(\mathrm{e}^{2t}x)'=\mathrm{e}^{2t}$${\mathrm{e}}^{2t}$, so $\mathrm{e}^{2t}x=\tfrac{1}{2}\mathrm{e}^{2t}+C$ and $x=\tfrac{1}{2}+C\mathrm{e}^{-2t}$$C{\mathrm{e}}^{-2t}$. From $x(0)=0$$0$, $C=-\tfrac{1}{2}$$-\frac{1}{2}$: $x(t)=\tfrac{1}{2}(1-\mathrm{e}^{-2t})$${\mathrm{e}}^{-2t})$.

The figure plots the general solution $x=\tfrac{1}{2}+C\mathrm{e}^{-2t}$ for $C=1$$1$, $C=-\tfrac{1}{2}$ and $C=-1$$-1$. The constant $C$ only shifts the decaying homogeneous part $C\mathrm{e}^{-2t}$, so every curve approaches the equilibrium $x=\tfrac{1}{2}=q/p$$q/p$; the initial condition $x(0)=0$ picks out the highlighted curve with $C=-\tfrac{1}{2}$$-\frac{1}{2}$.

Variable coefficient. $x'+\frac{1}{t} x=t$ on $t>0$$0$. Here $P(t)=\ln t$ and $\mu=t$$t$: $(tx)'=t^2$$t^2$, so $tx=\tfrac{t^3}{3}+C$ and $x(t)=\tfrac{t^2}{3}+\tfrac{C}{t}$$\frac{C}{t}$. Check: $x'+\frac{x}{t}=\tfrac{2t}{3}-\tfrac{C}{t^2}+\tfrac{t}{3}+\tfrac{C}{t^2}=t$$t$.

With integration by parts. $x'-x=t$$t$. Here $\mu=\mathrm{e}^{-t}$ and $(\mathrm{e}^{-t}x)'=t\mathrm{e}^{-t}$$t{\mathrm{e}}^{-t}$. Integration by parts gives $\int t\mathrm{e}^{-t}\,\mathrm{d}t=-t\mathrm{e}^{-t}-\mathrm{e}^{-t}+C$$C$, so $x(t)=-t-1+C\mathrm{e}^{t}$$C{\mathrm{e}}^t$. Check: $x'=-1+C\mathrm{e}^t$ and $x+t=-1+C\mathrm{e}^t$$C{\mathrm{e}}^t$.

Prerequisites¶

• First-Order Differential Equation
• Antiderivative
• Differentiation Rules
• Chain Rule

Related Concepts¶

• Particular Solution of an Inhomogeneous Linear ODE
• General Solution of a Linear ODE
• Equilibrium of a Differential Equation
• Matrix Exponential

Sources¶

• Math V: Advanced Linear Algebra · Week 3

Aliases: integrating factor method, method of integrating factors Keywords: mu(t)=e^{P(t)}, x'+p(t)x=q(t), linear first-order equation, variation of constants, product rule Symbols: element of, integral, mu

Last updated: 22 October 2026 19:04