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

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

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