Javascript required
Skip to content Skip to sidebar Skip to footer

Find the Solutions to a Differential Equation Calculator

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  =  ert .  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 an 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 0th 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 x n is zero for all n.  We have

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

Solving for an+2 gives

                       -an
     an+2  =
n+2

We immediately see that

        an  =  0

for n even.  Now compute the odd an

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

In general

                                 (-1)n                            2n(n!)(-1)n
        a2n+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 Sanyn , since the solution will not be differentiable at zero.  Alternatively, we find a solution in the form

This is the power series centered about x  =  1, which is not a singular point.  Now take derivatives

Plugging into the differential equation gives

Writing

x  =  (x - 1) + 1

and multiplying through gives

Let u  =  n - 2 in the first summation, u  =  n - 2 in the second and then changing the index variable back to n gives

Now plugging in n  =  0 into the second and third series we get

Now we can equate coefficients to find

        2a2 + 2a0  =  0

        (n + 1)nan+1 + (n + 2)(n + 1)an+2 + 2an  =  0

The first equation says that

        a2  =  -a0

The recursion relationship says

                        -(n + 1)nan+1 - 2an
     an+2  =
(n + 2)(n + 1)

We want to find two linearly independent solutions.  To do this, we can choose the first two terms of the series.  The easiest choices are

     a0  =  0    a1  =  1      and a0  =  1    a1  =  0

Plugging the first pair, we get

        a0  =  0    a1  =  1    a2  =  0    a3  =  [-2(0) - 2(1)]/6  =  -1/3

        a4  =  [-2(3)(-1/3) - 2(0)]/12  =  1/6

Plugging in the second pair, we get

        a0  =  1    a1  =  0    a2  =  -1    a3  =  [-2(-1) - 2(0)]/6  =  1/3

We can write

        y1  =  (x - 1) - 1/3 (x - 1)3 + 1/6 (x - 1)4 + ...

        y2  =  1 - (x - 1)2 + 1/3 (x - 1)3 + ...

Click here for an additional example


Back to the Power Series Methods and Laplace Transforms Home Page

Back to the Differential Equations Home Page

Back to the Math Department Home Page

e-mail Questions and Suggestions

Find the Solutions to a Differential Equation Calculator

Source: https://ltcconline.net/greenl/courses/204/PowerLaplace/seriesSolutions1.htm