It's generally agreed that the optimal technique for finding the solution of a general linear differential equation - with variable coefficients - is based on the use of power series. For starters, recall when we looked at Fourier series (Nov. 24 post) we began by looking at the Taylor series, e.g.
f(x) = f(x o ) + f ’(x o) × (x – x o) + f ” (x o) (x – x o)2 / 2! +
f ”’ (x o) (x – x o)3 / 2! + f n (x o) (x– x o)n / n! + ….
And note here that a function f(x) may be represented by such a series, provided all derivatives exist at x = x o . From here we can further assert a function f(x) is analytic at x = x o if f(x) can be expanded in a power series valid about some point.
To fix ideas consider the general form for the linear differential equation:
a o (x ) + D n x y + a 1 (x ) + D n-1 x y + …+ a n (x ) y = f(x)
In which a i (x ), (i = 0,.....n) are polynomials. Then:
The point x = x o is called an ordinary point of the equation: a o (x o ) ≠ 0
Any point x = x 1 for which a o (x 1 ) = 0 is called a singular point of the differential equation.
The point x = x 1 is called a regular singular point of the general linear DE if the equation can be written in the form:
[(x - x 1 ) n D n + (x - x 1 ) b1 (x) D n-1 +
(x - x 1 ) n- 2 b1 (x) D n-2 +. .. .....(x - x 1 ) n- 1 b n- 1 (x) D + b n (x) ] y = 0
Where b i (x), (i = 1,.......n) are analytic at x = x 1
A key to knowing whether a power series type solution is applicable is to first get the differential equation in the form:
y " + P (x) y' + Q (x) y = j((x)
Where:
P(x) = b 1 (x)/ b 2 (x)
Q(x) = b 0 (x)/ b 2 (x)
j((x) = g(x) / b 2 (x)
Then one can determine if x = 0 is an ordinary point, fulfilling that the power series method is applicable. Consider for example, the DE:
y" - xy' + 2y = 0
Here, P(x) = -x and Q(x) = 2, therefor because they are analytic every value of x, in particular x = 0, is an ordinary point. Again, the power series method is only applicable when x = 0 is an ordinary point.
Example 2: Determine whether x = 0 is an ordinary point of the differential equation:
2 x2 y" + 7 x(x + 1) y' - 3 y = 0
Here: P(x) = 7(x + 1)/ 2x
Q(x) = -3/2 x2
Here, neither is analytic at x = 0 (both denominators = 0) so x = 0 is not an ordinary point but a singular point. At this juncture we note that the use of power series to express a function which is a solution of a given differential equation, is based on the fact there exists a series which contains the necessary arbitrary constants and which converges inside a circle with x = x o and extends out to the singular point nearest to x = x o .
Example 3: Find a power series solution to:
(x - 1) y"' + y" + (x - 1) y' + y = 0
If: y = y" = 0, y ' = 1 at x = 0
Solution:
We proceed by first solving for y"' and differentiating:
y"' = - (x - 1)- 1 y" - y' - (x - 1)- 1 y
y iv = - (x - 1)- 1 y"' + [(x - 1)- 2 - 1] y" - (x - 1)- 1 y' +
(x - 1) - 2 y
y v = - (x - 1)- 1 y iv + 2[(x - 1)- 2 - 1] y"'
- 2[(x - 1)- 3 + (x - 1)- 1 ] y" + 2(x- 1)- 2 y' - 2( x -1 )- 2 y
At x = 0 we have:
y"'(0) = -1, y iv (0) = 0, yv (0) = 1
The resulting series is then:
y(x) = 0 + x + 0 - 1/3! (x2) + 0 + 1/5! (x3) + ....
y(x) = x - 1/3! (x2) + 0 + 1/5! (x3) - ....
y(x) = sin x
Problems:
1. Give the singular points for each of the following:
(a) (x - 3) y" + (x + 1)y = 0
(b) (x2 + 1) y"' + y" - x2 y = 0
2. Determine whether x = 0 is an ordinary point of the differential equation:
x2 y " + 2 y' + xy = 0
3. Find the power series solution for the differential equation:
xy" + x3 y - 3xy = 0
That satisfies: y = 0 and y' = 2 at x = 1
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