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9.2 *Special cases

For the two special cases I will just give the solution. It requires a substantial amount of algebra to study these two cases.

9.2.1 Two equal roots

If the indicial equation has two equal roots, {γ}_{1} = {γ}_{2}, we have one solution of the form

{y}_{1}(t) = {t}^{{γ}_{1} }{ \mathop{∑ }}_{n=0}^{∞}{c}_{ n}{t}^{n}.
(9.17)

The other solution takes the form

{y}_{2}(t) = {y}_{1}(t)\mathop{ln}\nolimits t + {t}^{{γ}_{1}+1}{ \mathop{∑ }}_{n=0}^{∞}{d}_{ n}{t}^{n}.
(9.18)

Notice that this last solution is always singular at t = 0, whatever the value of {γ}_{1}!

9.2.2 Two roots differing by an integer

If the indicial equation that differ by an integer, {γ}_{1} − {γ}_{2} = n > 0, we have one solution of the form

{y}_{1}(t) = {t}^{{γ}_{1} }{ \mathop{∑ }}_{n=0}^{∞}{c}_{ n}{t}^{n}.
(9.19)

The other solution takes the form

{y}_{2}(t) = a{y}_{1}(t)\mathop{ln}\nolimits t + {t}^{{γ}_{2} }{ \mathop{∑ }}_{n=0}^{∞}{d}_{ n}{t}^{n}.
(9.20)

The constant a is determined by substitution, and in a few relevant cases is even 0, so that the solutions can be of the generalised series form.

9.2.3 Example 1

Find two independent solutions of

{t}^{2}y'' + ty' + ty = 0
(9.21)

near t = 0. The indicial equation is {γ}^{2} = 0, so we get one solution of the series form

{y}_{1}(t) ={ \mathop{∑ }}_{n}{c}_{n}{t}^{n}.
(9.22)

We find

\begin{eqnarray}{ t}^{2}{y''}_{ 1}& =& {\mathop{∑ }}_{n}n(n − 1){c}_{n}{t}^{n} %& \\ t{y'}_{1}& =& {\mathop{∑ }}_{n}n{c}_{n}{t}^{n} %& \\ t{y}_{1}& =& {\mathop{∑ }}_{n}{c}_{n}{t}^{n+1} ={ \mathop{∑ }}_{n'}{c}_{n'−1}{t}^{n'}%&(9.23) \\ \end{eqnarray}

We add terms of equal power in x,

\array{ {t}^{2}{y''}_{1}& =&0&+&0t &+&2{c}_{2}{t}^{2} &+&6{c}_{3}{t}^{3} &+&\mathop{\mathop{…}} \cr t{y'}_{1}& =&0&+&{c}_{1}t &+&2{c}_{2}{t}^{2} &+&3{c}_{3}{t}^{3} &+&\mathop{\mathop{…}} \cr t{y}_{1}& =&0&+&{c}_{0}t &+&{c}_{1}{t}^{2} &+&{c}_{2}{t}^{3} &+&\mathop{\mathop{…}} ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ \cr {t}^{2}y'' + ty' + ty& =&0&+&({c}_{1} + {c}_{0})t&+&(4{c}_{2} + {c}_{1}){t}^{2}&+&(9{c}_{3} + {c}_{2}){t}^{2}&+&\mathop{\mathop{…}}}
(9.24)

Both of these ways give

{t}^{2}y'' + ty' + ty ={ \mathop{∑ }}_{n=1}^{∞}({c}_{ n}{n}^{2} + {c}_{ n−1}){t}^{n},
(9.25)

and lead to the recurrence relation

{c}_{n} = − {1\over { n}^{2}}{c}_{n−1}
(9.26)

which has the solution

{c}_{n} = {(−1)}^{n} {1\over n{!}^{2}}
(9.27)

and thus

{y}_{1}(t) ={ \mathop{∑ }}_{n=0}^{∞}{(−1)}^{n} {1\over n{!}^{2}}{x}^{n}
(9.28)

Let us look at the second solution

{y}_{2}(t) =\mathop{ ln}\nolimits (t){y}_{1}(t) +{\underbrace{ t{\mathop{∑ }}_{n=0}^{∞}{d}_{ n}{t}^{n}} }_{{ y}_{3}(t)}
(9.29)

Her I replace the power series with a symbol, {y}_{3} for convenience. We find

\begin{eqnarray}{ y}_{2}'& =& \mathop{ln}\nolimits (t){y}_{1}' + {{y}_{1}(t)\over t} + {y}_{3}' %& \\ {y}_{2}''& =& \mathop{ln}\nolimits (t){y}_{1}'' + {2{y'}_{1}(t)\over t} −{{y}_{1}(t)\over {t}^{2}} + +{y}_{3}''%&(9.30) \\ \end{eqnarray}

Taking all this together, we have,

\begin{eqnarray}{ t}^{2}{y}_{ 2}'' + t{y}_{2}' + t{y}_{2}& =& \mathop{ln}\nolimits (t)\left ({t}^{2}{y}_{ 1}'' + t{y}_{1}' + t{y}_{1}\right ) − {y}_{1} + 2t{y'}_{1} + {y}_{1} + {t}^{2}{y}_{ 3}'' + t{y}_{3}' + {y}_{3}%& \\ & =& 2t{y}_{1}' + {t}^{2}{y}_{ 3}'' + t{y}_{3}' + t{y}_{3} = 0. %&(9.31) \\ \end{eqnarray}

If we now substitute the series expansions for {y}_{1} and {y}_{3} we get

2{c}_{n} + {d}_{n}{(n + 1)}^{2} + {d}_{ n−1} = 0,
(9.32)

which can be manipulated to the form

Here there is some missing material

9.2.4 Example 2

Find two independent solutions of

{t}^{2}y'' + {t}^{2}y' − ty = 0
(9.33)

near t = 0.

The indicial equation is α(α − 1) = 0, so that we have two roots differing by an integer. The solution for α = 1 is {y}_{1} = t, as can be checked by substitution. The other solution should be found in the form

{y}_{2}(t) = at\mathop{ln}\nolimits t +{ \mathop{∑ }}_{k=0}{d}_{k}{t}^{k}
(9.34)

We find

\begin{eqnarray}{ y}_{2}'& =& a + a\mathop{ln}\nolimits t +{ \mathop{∑ }}_{k=0}k{d}_{k}{t}^{k−1} %& \\ {y}_{2}''& =& a∕t +{ \mathop{∑ }}_{k=0}k(k − 1){d}_{k}{t}^{k−2}%& \\ & & %&(9.35) \\ \end{eqnarray}

We thus find

\begin{eqnarray}{ t}^{2}{y''}_{ 2} + {t}^{2}{y'}_{ 2} − t{y}_{2} = a(t + {t}^{2}) +{ \mathop{∑ }}_{k=q}^{∞}\left [{d}_{ k}k(k − 1) + {d}_{k−1}(k − 2)\right ]{t}^{k}& & %&(9.36) \\ \end{eqnarray}

We find

{d}_{0} = a,\kern 2.77695pt \kern 2.77695pt \kern 2.77695pt 2{d}_{2} + a = 0,\kern 2.77695pt \kern 2.77695pt \kern 2.77695pt {d}_{k} = (k − 2)∕(k(k − 1)){d}_{k−1}\kern 2.77695pt \kern 2.77695pt (k > 2)
(9.37)

On fixing {d}_{0} = 1 we find

{y}_{2}(t) = 1 + t\mathop{ln}\nolimits t +{ \mathop{∑ }}_{k=2}^{∞} {1\over (k − 1)!k!}{(−1)}^{k+1}{t}^{k}
(9.38)

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