First Order Linear Differential Equations

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' + 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 2/x y = m ln x

We now search for a m with

m' = m 2/x

dm/m = 2/x dx

Integrating both sides, produces

ln m = 2ln x = ln(x²)

m = x² exponentiating both sides

Going back to the original differential equation and multiplying both sides by x², we get

x²y' + 2x y = x² ln x

Using the product rule in reverse gives

(x²y)' = x²ln x

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

u = ln x dv = x² dx
du = 1/x dx v = 1/3 x³

Hence

x²y = 1/3 x³ ln x - 1/9 x³ + C

y = 1/3 x ln x - 1/9 x + C/x² Divide by x²

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.

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 a m with

m' = mp(x)

dm/m = p(x) dx

Integrating both sides, produces

ln m =