Skip to main content

# 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 $$\PageIndex{1}$$

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.