Solving Differential Equations By Power Series

 The optimal technique for finding the solution of a general linear differential equation - with variable coefficients - is based on the use of power series.   A key convenience from the calculus in this endeavor is 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!  +  ….

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 differential equation if the equation can be written in the form:

[(x - x ₁ ) ⁿ  D ⁿ  +  (x - x ₁ )   b ₁  (x) D ^n-1  + 

(x - x ₁  )^{n- 2} b ₁ (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.

Example:

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 the differential equation:

(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,  y^v (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)  -   ....

which is easily recognizable as the function:

y(x) =  sin x  

 

Suggested Problems:

1. Determine whether x = 0 is an ordinary point of the differential equation:

x² y " + 2 y'  +  xy = 0 

2. Find the power series solution for the differential equation:

xy" +  x³ y   - 3 y = 0   

That satisfies: y = 0 and y' = 2  at x = 1