# 2.6: First Order Linear Differential Equations

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 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

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

### Contributors

- Larry Green (Lake Tahoe Community College)

Integrated by Justin Marshall.