Series Solutions to Second Order Linear Differential Equations

Series Solutions to Second Order Linear Differential Equations

We have fully investigated solving second order linear differential equations with constant coefficients. Now we will explore how to find solutions to second order linear differential equations whose coefficients are not necessarily constant. Let

P(x)y'' + Q(x)y' + R(x)y = g(x)

Be a second order differential equation with P, Q, R, and g all continuous. Then x₀ is a singular point if P(x₀) = 0, but Q and R do not both vanish at x₀. Otherwise we say that x₀ is an ordinary point. For now, we will investigate only ordinary points.

Example

Find a solution to

y'' + xy' + y = 0 y(0) = 0 y'(0) = 1

Solution

Since the differential equation has non-constant coefficients, we cannot assume that a solution is in the form y = e^rt. Instead, we use the fact that the second order linear differential equation must have a unique solution. We can express this unique solution as a power series

If we can determine the a_n for all n, then we know the solution. Fortunately, we can easily take derivatives

Now we plug these into the original differential equation

We can multiply the x into the second term to get

We would like to combine like terms, but there are two problems. The first is the powers of x do not match and the second is that the summations begin in differently. We will first deal with the powers of x. We shift the index of the first summation by letting

u = n - 2 n = u + 2

We arrive at

Since u is a dummy variable, we can call it n instead to get

Next we deal with the second issue. The second summation begins at 1 while the first and third begin at 0. We deal with this by pulling out the 0^th term. We plug in 0 into the first and third series to get

(0 + 2)(0 + 1)a₀₊₂x₀ = 2a₂

and

a₀x⁰ = a₀

We can write the series as

The initial conditions give us that

a₀ = 0 and a₁ = 1

Now we equate coefficients. The terms in the series begin with the first power of x, hence the constant term gives us

2a₂ + a₀ = 0

Since a₀ = 0, so is a₂. Now the coefficient in front of xⁿ is zero for all n. We have

(n + 2)(n + 1)a_n+2 + (n + 1)a_n = 0

Solving for a_n+2 gives

-a_n a_n+2 =
n+2

We immediately see that

a_n = 0

for n even. Now compute the odd a_n

                                      -1                      1                          -1
        a₁ = 1        a₃ =                 a₅ =                    a₇ =
                                        3                     3^.5                        3^.5^.7

In general

(-1)ⁿ 2ⁿ(n!)(-1)ⁿ a_2n+1 = =
3^.5^.7^....^.(2n+1) (2n + 1)!

The final solution is

This cannot be written in terms of elementary functions, however a computer can graph or calculate a value with as many decimal places as needed.

Example

Find the the first three nonzero terms of two linearly independent solutions to

xy'' + 2y = 0

Solution

Notice that 0 is a singular point of this differential equation. We will not be able to find a solution in the form Sa_nyⁿ, since the solution will not be differentiable at zero. Alternatively, we find a solution in the form