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Can someone please explain some steps of a worked example.

Q. Find the power series solutions about z = 0 of 4zy'' + 2y' + y = 0.

(note: y = y(z))

Writing the equation in standard form:

[tex]

y'' + \frac{1}{{2z}}y' + \frac{1}{{4z}}y = 0

[/tex]

Let [tex]y = z^\sigma \sum\limits_{n = 0}^\infty {a_n z^n }[/tex]

The indicial equation has roots zero and (1/2).

Using the substitution for y above the DE becomes:

[tex]

\sum\limits_{n = 0}^\infty {\left[ {\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right) + \frac{1}{2}\left( {n + \sigma } \right) + \frac{1}{4}z} \right]} a_n z^n = 0

[/tex]...(1)

Demanding that the coefficients of z^n vanish separately in the above equation we obtain the recurrence relation:

[tex]

\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0

[/tex]

If we choose the larger root, 1/2, then the recurrence relation becomes [tex]a_n = \frac{{ - a_{n - 1} }}{{2n\left( {n + 1} \right)}}[/tex]

Setting a_0 = 1 we find...etc

Any help would be good thanks.

Q. Find the power series solutions about z = 0 of 4zy'' + 2y' + y = 0.

(note: y = y(z))

Writing the equation in standard form:

[tex]

y'' + \frac{1}{{2z}}y' + \frac{1}{{4z}}y = 0

[/tex]

Let [tex]y = z^\sigma \sum\limits_{n = 0}^\infty {a_n z^n }[/tex]

The indicial equation has roots zero and (1/2).

Using the substitution for y above the DE becomes:

[tex]

\sum\limits_{n = 0}^\infty {\left[ {\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right) + \frac{1}{2}\left( {n + \sigma } \right) + \frac{1}{4}z} \right]} a_n z^n = 0

[/tex]...(1)

Demanding that the coefficients of z^n vanish separately in the above equation we obtain the recurrence relation:

[tex]

\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0

[/tex]

*What happened to the factor of z which was multiplied by (1/4) in (1)?*If we choose the larger root, 1/2, then the recurrence relation becomes [tex]a_n = \frac{{ - a_{n - 1} }}{{2n\left( {n + 1} \right)}}[/tex]

Setting a_0 = 1 we find...etc

*I am told that the larger of the two roots of the indicial equation always leads to a solution. However, I don't understand why a_0 is set equal to one? Is that an arbitrary selection or is it convenient?*Any help would be good thanks.

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