# 9.3: Special Cases

- Page ID
- 8336

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

## Case I: Two equal roots

If the indicial equation has two equal roots, \(\gamma_1=\gamma_2\), we have one solution of the form \[y_1(t) = t^{\gamma_1} \sum_{n=0}^\infty c_n t^n.\] The other solution takes the form \[y_2(t) = y_1(t)\ln t +t^{\gamma_1+1} \sum_{n=0}^\infty d_n t^n.\] Notice that this last solution is always singular at \(t=0\), whatever the value of \(\gamma_1\)!

## Case II: Two roots differing by an integer

If the indicial equation that differ by an integer, \(\gamma_1-\gamma_2=n>0\), we have one solution of the form \[y_1(t) = t^{\gamma_1} \sum_{n=0}^\infty c_n t^n.\] The other solution takes the form \[y_2(t) = ay_1(t)\ln t +t^{\gamma_2} \sum_{n=0}^\infty d_n t^n.\] 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.

Example \(\PageIndex{1}\):

Find two independent solutions of \[t^2y''+ty'+ty=0\] near \(t=0\).

**Solution**

The indicial equation is \(\gamma^2=0\), so we get one solution of the series form

\[y_1(t) = \sum_n c_n t^n.\]

We find

\[\begin{align*} t^2y''_1 &= \sum_n n(n-1) c_n t^n \nonumber\\ ty'_1 &= \sum_n n c_n t^n \nonumber\\ ty_1 &= \sum_nc_n t^{n+1} =\sum_{n'}c_{n'-1} t^{n'}\end{align*}\]

We add terms of equal power in \(x\),

\[\begin{array}{rclclclclcl} t^2y''_1 &= 0&+& 0 t&+&2c_2t^2&+&6c_3t^3&+&\ldots\\ ty'_1 &= 0&+& c_1t&+&2c_2t^2&+&3c_3t^3&+&\ldots\\ ty_1 &= 0&+& c_0t&+&c_1t^2&+&c_2t^3&+&\ldots\\ \hline t^2y''+ty'+ty&= 0&+&(c_1+c_0)t&+&(4c_2+c_1)t^2&+&(9c_3+c_2)t^2&+&\ldots \end{array}\]

Both of these ways give \[t^2y''+ty'+ty=\sum_{n=1}^\infty (c_n n^2+c_{n-1})t^n,\]

and lead to the recurrence relation

\[c_n = -\frac{1}{n^2} c_{n-1}\] which has the solution \[c_n = (-1)^n \frac{1}{n!^2}\] and thus \[y_1(t) = \sum_{n=0}^\infty (-1)^n \frac{1}{n!^2} x^n\] Let us look at the second solution

\[y_2(t) =\ln(t) y_1(t)+\underbrace{t\sum_{n=0}^\infty d_n t^n}_{y_3(t)}\]

Here I replace the power series with a symbol, \(y_3\) for convenience. We find

\[\begin{align*} y_2' &= \ln(t) y_1' + \frac{y_1(t)}{t}+y_3'\nonumber\\ y_2'' &= \ln(t) y_1'' + \frac{2y'_1(t)}{t}-\frac{y_1(t)}{t^2}+ +y_3''\end{align*}\]

Taking all this together, we have,

\[\begin{align*} t^2y_2''+ty_2'+ty_2 &= \ln(t)\left(t^2 {y_1}''+t {y_1}'+t{y_1}\right) -y_1+2ty'_1+y_1 + t^2 {y_3}''+t {y_3}'+y_3 \nonumber\\ &= 2t{y_1}'+t^2{y_3}''+t{y_3}'+ty_3=0.\end{align*}\]

If we now substitute the series expansions for \(y_1\) and \(y_3\) we get \[2c_n+d_n(n+1)^2+d_{n-1}=0,\] which can be manipulated to the form

Here there is some missing material

Example \(\PageIndex{2}\):

Find two independent solutions of \[t^2{y'}'+t^2{y}'-ty=0\] near \(t=0\).

**Solution**

The indicial equation is \(\alpha(\alpha-1)=0\), so that we have two roots differing by an integer. The solution for \(\alpha=1\) is \(y_1=t\), as can be checked by substitution. The other solution should be found in the form

\[y_2(t) = at\ln t + \sum_{k=0}d_k t^k\]

We find

\[\begin{align*} y_2' & = & a+a\ln t + \sum_{k=0}kd_k t^{k-1} \nonumber \\ y_2'' & = & a/t + \sum_{k=0}k(k-1)d_k t^{k-2} \nonumber \\\end{align*}\]

We thus find \[\begin{align*} t^2y''_2+t^2y'_2-ty_2= a(t+t^2)+ \sum_{k=q}^\infty \left[d_k k(k-1)+d_{k-1}(k-2)\right] t^k\end{align*}\]

We find

\[d_0 = a,\;\;\;2 d_2+a=0,\;\;\;d_k = (k-2)/(k(k-1))d_{k-1}\;\;(k>2)\]

On fixing \(d_0=1\) we find \[y_2(t) = 1 + t \ln t + \sum_{k=2}^\infty \frac{1}{(k-1)!k!}(-1)^{k+1} t^k\]