First-Order Linear Differential Equations and Integrating Factors

1Introduction

This lecture derives an integrating factor as a factor that converts the left-hand side of a first-order linear equation into the derivative of a product, rather than presenting it as a formula to memorize.

The standard form of a first-order linear equation is

Because $$y^{\prime }$$y' and $$y$$y occur as separate terms, the left-hand side cannot generally be integrated directly. The objective is to multiply the equation by a function $$\mu \left(x\right)$$\mu(x) so that

becomes the single derivative

2Why an Integrating Factor Is Introduced

The expression $$\mu =e^{\int p\left(x\right)dx}$$\mu=e^{\int p(x)\,dx} follows systematically from the product rule.

An equation of the form

can be integrated directly because its left-hand side is already one derivative. By contrast, in

the term $$p\left(x\right)y$$p(x)y prevents the left-hand side from being integrated directly as the derivative of a single function.

Apply the product rule in reverse:

If multiplication by an unknown function $$\mu \left(x\right)$$\mu(x) transforms the original left-hand side into

then it becomes one derivative. Multiplication of the original equation by $$\mu$$\mu gives

Comparison with the product rule shows that the required condition is

Thus the purpose of the integrating factor is to transform the left-hand side into the derivative of a product and thereby reduce the equation to direct integration.

3First-Order Linear Differential Equations

A first-order linear differential equation is an equation that can be written as

where $$p\left(x\right)$$p(x) and $$q\left(x\right)$$q(x) depend only on $$x$$x.

The term linear means that, for $$L\left[y\right]=y^{\prime }+p\left(x\right)y$$L[y]=y'+p(x)y,

Equivalently, the unknown function and its derivative do not occur in products with each other or inside nonlinear functions. For example,

is first-order linear, whereas

is not linear in $$y$$y because it contains $$y^2$$y^2.

4Proposition 1: Derivation of the Integrating Factor

Suppose that $$p$$p and $$q$$q are continuous on an interval and that

If $$\mu$$\mu is nonzero on the interval and satisfies

then

Moreover, because $$\mu \ne 0$$\mu\ne0, the transformed equation is equivalent to the original equation.

4.1Proof

Multiply the original equation by $$\mu$$\mu:

The assumption $${\mu }^{\prime }=\mu p\left(x\right)$$\mu'=\mu p(x) transforms the left-hand side into

The product rule therefore gives

Conversely, expand $$(\mu y{)}^{\prime }=\mu q(x)$$(\mu y)'=\mu q(x) by the product rule and substitute $${\mu }^{\prime }=\mu p\left(x\right)$$\mu'=\mu p(x). This gives $$\mu y^{\prime }+\mu p\left(x\right)y=\mu q\left(x\right)$$\mu y'+\mu p(x)y=\mu q(x). Since $$\mu$$\mu is nonzero on the interval, division by $$\mu$$\mu recovers the original equation.

Consequently, the essential step is to determine a function $$\mu$$\mu satisfying $${\mu }^{\prime }=\mu p\left(x\right)$$\mu'=\mu p(x).

5Form of the Integrating Factor

On a range where $$\mu \ne 0$$\mu\ne0, the equation $${\mu }^{\prime }=\mu p\left(x\right)$$\mu'=\mu p(x) becomes

Integration gives

One may therefore choose

An integration constant only multiplies $$\mu$$\mu by a nonzero constant, which is absorbed into the arbitrary constant in the final solution.

6Proposition 2: General-Solution Formula

Suppose that $$p$$p and $$q$$q are continuous on an interval $$I$$I. With

the solutions of

are given by

6.1Proof

Proposition 1 gives

Integrating both sides yields

Since $$\mu =e^{\int p\left(x\right)dx}$$\mu=e^{\int p(x)\,dx} is nonzero, division by $$\mu$$\mu gives

Thus the formula results directly from transforming the equation into an integrable product derivative.

7Example 1: $$y^{\prime }+y=x$$y'+y=x

For

we have $$p\left(x\right)=1$$p(x)=1 and $$q\left(x\right)=x$$q(x)=x. An integrating factor is

Multiplication by $$e^x$$e^x gives

The left-hand side is $$(e^{x}y{)}^{\prime }$$(e^x y)', so

Integration yields

and hence

To verify the result, note that $$y^{\prime }=1-C{e}^{-x}$$y'=1-Ce^{-x}. Therefore,

8Relation to Variation of Constants

Variation of constants provides another derivation for the solution of a first-order linear equation. First consider the associated homogeneous equation

The zero function is a solution. On a branch where $$y\ne 0$$y\ne0, separation gives

and hence

Allowing $$C=0$$C=0 includes the zero solution. For the nonhomogeneous equation, replace the constant $$C$$C by a function $$C\left(x\right)$$C(x):

If $$\mu =e^{\int p\left(x\right)dx}$$\mu=e^{\int p(x)\,dx}, then

The integrating-factor equation

is therefore equivalent to

Variation of constants and the integrating-factor method express the same transformation: the constant multiplying a homogeneous solution is varied to account for the nonhomogeneous term.

9Linear-Operator Formulation

Define

Then the first-order linear equation is

This is an equation for a linear operator. The solutions of the homogeneous equation $$L\left[y\right]=0$$L[y]=0 form the kernel $$\ker L$$\ker L, and the general solution of the nonhomogeneous equation has the form

The same operator formulation will be used for second-order linear equations and systems of equations.

10Scope of the Method

The integrating-factor method applies when an equation can be written in the linear form

If $$p$$p and $$q$$q are continuous on an interval $$I$$I, then an initial value prescribed at a point in $$I$$I determines a unique solution on all of $$I$$I.

An equation of the form

is not linear in $$y$$y as written. A Bernoulli equation of this form can sometimes be transformed into a first-order linear equation by substitution.

In the next lecture, the integrating-factor concept is generalized to a factor that transforms a differential form into a total differential.

11Exercises