Linear Equations and Integrating Factors

A first-order equation $y^{\prime }=f(t,y)$ is linear when $f$ depends on $y$ to the first power only. Collecting the terms in $y^{\prime }$ and $y$ on the left, every such equation can be put in the standard form

where $p$ and $g$ are given functions of the independent variable $t$. When the original equation carries a leading coefficient, $P(t)y^{\prime }+Q(t)y=G(t)$, dividing by $P(t)$ recovers the standard form on any interval where $P(t)\ne 0$, with $p=Q/P$ and $g=G/P$. This is the one first-order class that a single, uniform procedure always solves.¹ An integrating factor does this: a function $\mu (t)$ chosen so that multiplying the equation by it turns the left side into the derivative of a product.

When the left side is already a derivative

Occasionally the left side is a total derivative before any manipulation. For

the product rule gives $\frac{\mathbf{d}}{\mathbf{d}t}[(4+t^2)y]=(4+t^2)y^{\prime }+2ty$, exactly the left side. The equation collapses to

and integrating both sides yields $(4+t^2)y=2t^2+c$, so $y=(2t^2+c)/(4+t^2)$. The method generalizes this case: given an equation whose left side is not a derivative, find a multiplier that makes it one.

Constructing the integrating factor

Take the constant-coefficient case $y^{\prime }+ay=g(t)$ first. Multiply by an undetermined $\mu (t)$:

The left side equals $\frac{\mathbf{d}}{\mathbf{d}t}[\mu (t)y]=\mu y^{\prime }+{\mu }^{\prime }y$ provided the coefficient of $y$ matches, that is ${\mu }^{\prime }(t)=a\mu (t)$. That condition is itself a differential equation, solved by $\mu (t)=e^{at}$. With this factor the equation becomes $\frac{\mathbf{d}}{\mathbf{d}t}\left[e^{at}y\right]=e^{at}g(t)$, and one integration finishes it.

The general variable-coefficient case runs the same way. Multiplying $y^{\prime }+p(t)y=g(t)$ by $\mu (t)$ and demanding that the left side be $\frac{\mathbf{d}}{\mathbf{d}t}[\mu y]$ forces

The arbitrary constant of integration in $\int p\mathbf{d}t$ only rescales $\mu$ by a constant, which cancels out of the final answer, so we drop it and take the simplest factor. Because $\mu$ is an exponential, it is positive for all $t$, which justifies dividing by it later.

Carrying out the last integration and solving for $y$ gives the general solution in closed form.

The formula involves two integrations: one to build $\mu$, one to recover $y$. In practice it is cleaner to reconstruct the steps than to memorize the formula, since the key step is recognizing $\frac{\mathbf{d}}{\mathbf{d}t}[\mu y]$.

Worked initial value problems

The same sinusoidal steady state, now with damping and mass, reappears in the forced vibrations of second-order systems.

Existence and uniqueness for linear equations

The linear case carries a guarantee stronger than anything available for nonlinear equations, a direct consequence of the explicit solution formula.

The interval of existence is read off in advance from the coefficients: a solution runs as far as $p$ and $g$ stay continuous, and singularities can occur only where $p$ or $g$ is discontinuous. In $t{y}^{\prime }+2y=4t^2$ the coefficient $p=2/t$ fails at $t=0$, and indeed the solution through $t_0=1$ lives on $(0,\infty )$ and no further. This advance prediction is special to linear equations; the nonlinear theory guarantees a solution only on some possibly small interval, and even that interval usually cannot be found without solving the equation.

Transient and steady state

When $p$ is a positive constant $a$, the general solution splits into two named pieces. Write $y^{\prime }+ay=g(t)$; the solution is

where $y_p$ is any particular solution driven by $g$. The two parts behave differently as $t$ grows:

• Transient. The term $c{e}^{-at}$ carries all the dependence on the initial condition and decays to zero. It matters early and vanishes late.
• Steady state. The part $y_p(t)$ is what the forcing $g$ sustains. Every solution, regardless of $y_0$, converges to it.

For $y^{\prime }-2y=4-t$ the integrating factor is $e^{-2t}$ and the general solution is $y=-\frac{7}{4}+\frac{1}{2}t+c{e}^{2t}$. Here $a=-2<0$, so the transient $c{e}^{2t}$ actually grows and the roles invert: the exponential dominates and solutions diverge, except for the single curve $c=0$ that follows the line $y=-\frac{7}{4}+\frac{1}{2}t$. The sign of the coefficient decides whether the exponential part is a decaying transient or a runaway.

Variation of parameters

The integrating factor is one route to the solution; variation of parameters is another, and it is the one that generalizes to higher-order linear equations. The homogeneous equation $y^{\prime }+p(t)y=0$ has solution $y=A\exp (-\int p\mathbf{d}t)$ with $A$ constant. To solve the full equation, promote the constant to a function, $y=A(t)\exp (-\int p\mathbf{d}t)$, and substitute. The terms in $A$ cancel by construction, leaving

which integrates directly and reproduces the general solution formula above. The name records the idea: the parameter $A$ that was constant for the homogeneous problem is allowed to vary to absorb the forcing $g$.

Summary

Every first-order linear equation is solved by the same procedure:

• Standardize. Divide by the leading coefficient to reach $y^{\prime }+p(t)y=g(t)$.
• Build the factor. Compute $\mu =\exp (\int p\mathbf{d}t)$.
• Collapse and integrate. Multiply through; the left side is $\frac{\mathbf{d}}{\mathbf{d}t}[\mu y]$, so $\mu y=\int \mu g\mathbf{d}t+c$.
• Apply data. Fix $c$ from $y(t_0)=y_0$; the solution exists wherever $p$ and $g$ are continuous.