First-Order Linear Differential Equations

Precise definition

A first-order equation is linear when it can be written $y'+P(x)y=Q(x)$. On an interval where $P$ and $Q$ are continuous, the integrating factor $\mu(x)=e^{\int P(x)dx}$ converts the left side to $(\mu y)'$, giving $\mu y=\int\mu Q\,dx+C$.

Notation and mathematical language

The coefficient of $y'$ must first be normalized to 1. Multiplying $\mu$ by a nonzero constant changes neither the method nor final family. The homogeneous solution is $Ce^{-\int Pdx}$; one particular solution accounts for $Q$.

Conceptual picture

The integrating factor is designed from the product rule: $(\mu y)'=\mu y'+\mu'y$. Requiring $\mu'=P\mu$ makes this exactly $\mu(y'+Py)$. The procedure therefore derives from a differential identity rather than a formula to memorize.

Conditions and key results

Continuity of normalized $P,Q$ on an interval ensures a unique solution through each initial point there. Dividing by a leading coefficient can expose excluded points, and a solution constant applies only on one connected interval that avoids them.

A reliable strategy

1. Move terms into $a(x)y'+b(x)y=g(x)$ and divide by $a(x)$ on an interval where $a\ne0$.
2. Compute $P=b/a$ and $\mu=e^{\int Pdx}$.
3. Multiply every term by $\mu$, recognize $(\mu y)'$, and integrate both sides.
4. Solve for $y$, apply the initial condition, state the interval, and substitute into the normalized equation.

Fully worked example

Interpretation and application

Linear equations model decay with input, circuits, mixing, and first-order control. Superposition applies to the homogeneous operator, while the forcing $Q$ adds a particular response. Estimated coefficients make the model approximate even when its differential equation is solved exactly.

Common mistakes

Verification and reasonableness

• Differentiate the final expression and evaluate $y'+Py-Q$.
• Check the product-rule identity $(\mu y)'=\mu(y'+Py)$ explicitly.
• Substitute the initial point and inspect the coefficient domain for the maximal interval.

Practice

1. Find an integrating factor for $y'+4y=x$.
2. Solve the homogeneous equation $y'+4y=0$.
3. Why divide by the leading coefficient first?

Further deduction

The solution can be written in definite-integral form without introducing an ambiguous antiderivative constant: $y(x)=e^{-A(x)}[y(x_0)+\int_{x_0}^x e^{A(s)}Q(s)ds]$, where $A(x)=\int_{x_0}^xP(s)ds$. This form builds in the initial value and is useful for proving bounds, continuity, and dependence on forcing.

If two solutions satisfy the same forcing, their difference solves the homogeneous equation. Therefore two solutions meeting the same initial value have difference $z'= -Pz$, $z(x_0)=0$, whose integrating-factor solution is identically zero. This gives a direct uniqueness proof on the continuous-coefficient interval and explains why the initial condition fixes exactly one value of $C$.

Superposition has an affine form here: the difference between any two forced solutions is homogeneous, while adding two forced solutions doubles the forcing. This distinction explains why the full solution set is a translate of the one-dimensional homogeneous solution space rather than a vector space itself.

Related topics

• Bernoulli Differential Equations
• Exact Differential Equations
• Mixing Problems