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 x^{2} y" + 7 x(x + 1) y' - 3 y = 0

Here: P(x) = 7(x + 1)/ 2x

Q(x) = -3/2 x^{2}

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! (x^{2}) + 0 + 1/5! (x^{3}) + ....

y(x) = x - 1/3! (x^{2}) + 0 + 1/5! (x^{3}) - ....

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:

x^{2} y " + 2 y' + xy = 0

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

xy" + x^{3} y - 3 y = 0

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

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