# 2.6: First Order Linear Differential Equations

- Page ID
- 381

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

In this section we will concentrate on first order linear differential equations. Recall that this means that only a first derivative appears in the differential equation and that the equation is linear. The general first order linear differential equation has the form

\[ y' + p(x)y = g(x) \]

Before we come up with the general solution we will work out the specific example

\[ y' + \frac{2}{x y} = \ln \, x. \]

The strategy for solving this is to realize that the left hand side looks a little like the product rule for differentiation. The product rule is

\[ (my)' = my' + m'y. \]

This leads us to multiplying both sides of the equation by m which is called an *integrating factor*.

\[ m y' + m \frac{2}{x y} = m \ln\, x. \]

We now search for a \(m\) with

\[ m' = m \frac{2}{x} \]

or

\[ \frac{dm}{m} = \frac{2}{x} \; dx. \]

Integrating both sides, produces

\[ \ln\, m = 2\ln\, x = \ln(x^2) \]

or \( m = x^2 \) by exponentiating both sides.

Going back to the original differential equation and multiplying both sides by \( x^2 \), we get

\[ x^2y' + 2xy = x^2 \ln\, x . \]

Using the product rule in reverse gives

\[ (x^2y)' = x^2 \ln \, x . \]

Now integrate both sides. Note that the integral of the derivative is the original. Integrate by parts to get

\[ \int x^2 \, \ln \, x \, dx. \]

- \(u = \ln x\) and \(dv = x^2 \,dx\)
- \(du = \dfrac{1}{x} dx \) and \( v = \dfrac{1}{3} x^3\)

\[ = \dfrac{x^3 \ln\, x}{3} - \int \dfrac{x^2}{3} \, dx = \dfrac{x^3 \ln \, x}{3} - \dfrac{x^3}{9} +C \]

Hence

\[ x^2y = \frac{1}{3} x^3 \ln x - \frac{1}{9} x^3 + C. \]

Divide by \(x^2\)

\[ y = \dfrac{1}{3} x \ln \, x - \dfrac{1}{9} x + \dfrac{C}{x^2} . \]

Notice that when \(C\) is nonzero, the solutions are undefined at \(x = 0\). Also given an initial value with \(x\) positive, there will be no solution for negative \(x\). Now we will derive the general solution to first order linear differential equations.

Example \(\PageIndex{1}\)

Consider

\[ y' + p(t)y = g(t). \]

We multiply both sides by \(m\) to get

\[ my' + mp(x)y = mg(x). \]

We now search for an \(m\) with

\[ m' = mp(x) \]

or

\[ \dfrac{dm}{m} = p(x) \, dx . \]

Integrating both sides, produces

\[ \text{ln} \, m = \int p(x)\, dx \]

exponentiating both sides

\[ m = e^{\int p(x)\, dx}. \]

Going back to the original differential equation and multiplying both sides by \(m\), we get

\[ my' + mp(x)y = mg(x) \]

\[ (my)' = mg(x) \]

\[ my = \int \mu \, g(x) \, dx. \]

Solving for \(y\) gives

\[ y = e^{-\int p(x)\, dx}\int g(x) e^{\int p(x) \, dx } \, dx .\]

Larry Green (Lake Tahoe Community College)

Integrated by Justin Marshall.