 Series Solutions - Maple Help

ODE Steps for Series Solutions Overview

 • This help page gives a few examples of using the command ODESteps to solve ordinary differential equations by means of series expansions.
 • See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence. Examples

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
 > $\mathrm{ode1}≔{x}^{2}\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)+x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+5xy\left(x\right)=0$
 ${\mathrm{ode1}}{≔}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{5}{}{x}{}{y}{}\left({x}\right){=}{0}$ (1)
 > $\mathrm{ODESteps}\left(\mathrm{ode1}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{5}{}{x}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate 2nd derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{{5}{}{y}{}\left({x}\right)}{{x}}{-}\frac{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)}{{x}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)}{{x}}{+}\frac{{5}{}{y}{}\left({x}\right)}{{x}}{=}{0}\\ \text{▫}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& \text{◦}& \text{Define functions}\\ {}& {}& \left[{{P}}_{{2}}{}\left({x}\right){=}\frac{{1}}{{x}}{,}{{P}}_{{3}}{}\left({x}\right){=}\frac{{5}}{{x}}\right]\\ {}& \text{◦}& x\cdot {P}_{2}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=0\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{0}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{0}}{=}{1}\\ {}& \text{◦}& {x}^{2}\cdot {P}_{3}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=0\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{0}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{0}}{=}{0}\\ {}& \text{◦}& x=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& {{x}}_{{0}}{=}{0}\\ \text{•}& {}& \text{Multiply by denominators}\\ {}& {}& \left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{+}{5}{}{y}{}\left({x}\right){+}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Assume series solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}{+}{r}}\\ \text{▫}& {}& \text{Rewrite ODE with series expansions}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}{{x}}^{{k}{+}{r}{-}{1}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{-1}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}}{}\left({k}{+}{1}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\cdot \left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}\left({k}{+}{r}{-}{1}\right){}{{x}}^{{k}{+}{r}{-}{1}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+1\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{-1}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}}{}\left({k}{+}{1}{+}{r}\right){}\left({k}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& {}& \text{Rewrite ODE with series expansions}\\ {}& {}& {{a}}_{{0}}{}{{r}}^{{2}}{}{{x}}^{{-}{1}{+}{r}}{+}\left({\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}\left({{a}}_{{k}{+}{1}}{}{\left({k}{+}{1}{+}{r}\right)}^{{2}}{+}{5}{}{{a}}_{{k}}\right){}{{x}}^{{k}{+}{r}}\right){=}{0}\\ \text{•}& {}& {a}_{0}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{cannot be 0 by assumption, giving the indicial equation}\\ {}& {}& {{r}}^{{2}}{=}{0}\\ \text{•}& {}& \text{Values of r that satisfy the indicial equation}\\ {}& {}& {r}{=}{0}\\ \text{•}& {}& \text{Each term in the series must be 0, giving the recursion relation}\\ {}& {}& {{a}}_{{k}{+}{1}}{}{\left({k}{+}{1}\right)}^{{2}}{+}{5}{}{{a}}_{{k}}{=}{0}\\ \text{•}& {}& \text{Recursion relation that defines series solution to ODE}\\ {}& {}& {{a}}_{{k}{+}{1}}{=}{-}\frac{{5}{}{{a}}_{{k}}}{{\left({k}{+}{1}\right)}^{{2}}}\\ \text{•}& {}& \text{Recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=0\\ {}& {}& {{a}}_{{k}{+}{1}}{=}{-}\frac{{5}{}{{a}}_{{k}}}{{\left({k}{+}{1}\right)}^{{2}}}\\ \text{•}& {}& \text{Solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=0\\ {}& {}& \left[{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}}{,}{{a}}_{{k}{+}{1}}{=}{-}\frac{{5}{}{{a}}_{{k}}}{{\left({k}{+}{1}\right)}^{{2}}}\right]\end{array}$ (2)
 > $\mathrm{ode2}≔\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)+x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+y\left(x\right)=0$
 ${\mathrm{ode2}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{y}{}\left({x}\right){=}{0}$ (3)
 > $\mathrm{ODESteps}\left(\mathrm{ode2}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Assume series solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}}\\ \text{▫}& {}& \text{Rewrite DE with series expansions}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\cdot \left(\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{k}{}{{x}}^{{k}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{2}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{k}{}\left({k}{-}{1}\right){}{{x}}^{{k}{-}{2}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{2}}{}\left({k}{+}{2}\right){}\left({k}{+}{1}\right){}{{x}}^{{k}}\\ {}& {}& \text{Rewrite DE with series expansions}\\ {}& {}& {\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}\left({{a}}_{{k}{+}{2}}{}\left({k}{+}{2}\right){}\left({k}{+}{1}\right){+}{{a}}_{{k}}{}\left({k}{+}{1}\right)\right){}{{x}}^{{k}}{=}{0}\\ \text{•}& {}& \text{Each term in the series must be 0, giving the recursion relation}\\ {}& {}& \left({k}{+}{1}\right){}\left({{a}}_{{k}{+}{2}}{}\left({k}{+}{2}\right){+}{{a}}_{{k}}\right){=}{0}\\ \text{•}& {}& \text{Recursion relation that defines the series solution to the ODE}\\ {}& {}& \left[{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}}{,}{{a}}_{{k}{+}{2}}{=}{-}\frac{{{a}}_{{k}}}{{k}{+}{2}}\right]\end{array}$ (4)
 > $\mathrm{ode3}≔{x}^{2}\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)+{x}^{2}\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+\left({x}^{3}-6\right)y\left(x\right)=0$
 ${\mathrm{ode3}}{≔}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{{x}}^{{2}}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({{x}}^{{3}}{-}{6}\right){}{y}{}\left({x}\right){=}{0}$ (5)
 > $\mathrm{ODESteps}\left(\mathrm{ode3}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{{x}}^{{2}}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({{x}}^{{3}}{-}{6}\right){}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate 2nd derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{\left({{x}}^{{3}}{-}{6}\right){}{y}{}\left({x}\right)}{{{x}}^{{2}}}{-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{\left({{x}}^{{3}}{-}{6}\right){}{y}{}\left({x}\right)}{{{x}}^{{2}}}{=}{0}\\ \text{▫}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& \text{◦}& \text{Define functions}\\ {}& {}& \left[{{P}}_{{2}}{}\left({x}\right){=}{1}{,}{{P}}_{{3}}{}\left({x}\right){=}\frac{{{x}}^{{3}}{-}{6}}{{{x}}^{{2}}}\right]\\ {}& \text{◦}& x\cdot {P}_{2}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=0\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{0}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{0}}{=}{0}\\ {}& \text{◦}& {x}^{2}\cdot {P}_{3}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=0\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{0}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{0}}{=}{-6}\\ {}& \text{◦}& x=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& {{x}}_{{0}}{=}{0}\\ \text{•}& {}& \text{Multiply by denominators}\\ {}& {}& {{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{{x}}^{{2}}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({{x}}^{{3}}{-}{6}\right){}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Assume series solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}{+}{r}}\\ \text{▫}& {}& \text{Rewrite ODE with series expansions}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}^{m}\cdot y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}m=0..3\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}{+}{r}{+}{m}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k-m\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{m}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{-}{m}}{}{{x}}^{{k}{+}{r}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}^{2}\cdot \left(\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}{{x}}^{{k}{+}{r}{+}{1}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k-1\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{1}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{-}{1}}{}\left({k}{-}{1}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}^{2}\cdot \left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}\left({k}{-}{1}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& {}& \text{Rewrite ODE with series expansions}\\ {}& {}& {{a}}_{{0}}{}\left({2}{+}{r}\right){}\left({-}{3}{+}{r}\right){}{{x}}^{{r}}{+}\left({{a}}_{{1}}{}\left({3}{+}{r}\right){}\left({-}{2}{+}{r}\right){+}{{a}}_{{0}}{}{r}\right){}{{x}}^{{1}{+}{r}}{+}\left({{a}}_{{2}}{}\left({4}{+}{r}\right){}\left({-}{1}{+}{r}\right){+}{{a}}_{{1}}{}\left({1}{+}{r}\right)\right){}{{x}}^{{2}{+}{r}}{+}\left({\sum }_{{k}{=}{3}}^{{\mathrm{\infty }}}{}\left({{a}}_{{k}}{}\left({k}{+}{r}{+}{2}\right){}\left({k}{+}{r}{-}{3}\right){+}{{a}}_{{k}{-}{1}}{}\left({k}{-}{1}{+}{r}\right){+}{{a}}_{{k}{-}{3}}\right){}{{x}}^{{k}{+}{r}}\right){=}{0}\\ \text{•}& {}& {a}_{0}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{cannot be 0 by assumption, giving the indicial equation}\\ {}& {}& \left({2}{+}{r}\right){}\left({-}{3}{+}{r}\right){=}{0}\\ \text{•}& {}& \text{Values of r that satisfy the indicial equation}\\ {}& {}& {r}{\in }\left\{{-2}{,}{3}\right\}\\ \text{•}& {}& \text{The coefficients of each power of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{must be 0}\\ {}& {}& \left[{{a}}_{{1}}{}\left({3}{+}{r}\right){}\left({-}{2}{+}{r}\right){+}{{a}}_{{0}}{}{r}{=}{0}{,}{{a}}_{{2}}{}\left({4}{+}{r}\right){}\left({-}{1}{+}{r}\right){+}{{a}}_{{1}}{}\left({1}{+}{r}\right){=}{0}\right]\\ \text{•}& {}& \text{Solve for the dependent coefficient(s)}\\ {}& {}& \left\{{{a}}_{{1}}{=}{-}\frac{{{a}}_{{0}}{}{r}}{{{r}}^{{2}}{+}{r}{-}{6}}{,}{{a}}_{{2}}{=}\frac{{{a}}_{{0}}{}{r}{}\left({1}{+}{r}\right)}{{{r}}^{{4}}{+}{4}{}{{r}}^{{3}}{-}{7}{}{{r}}^{{2}}{-}{22}{}{r}{+}{24}}\right\}\\ \text{•}& {}& \text{Each term in the series must be 0, giving the recursion relation}\\ {}& {}& {{a}}_{{k}}{}\left({k}{+}{r}{+}{2}\right){}\left({k}{+}{r}{-}{3}\right){+}{{a}}_{{k}{-}{1}}{}{k}{+}{{a}}_{{k}{-}{1}}{}{r}{+}{{a}}_{{k}{-}{3}}{-}{{a}}_{{k}{-}{1}}{=}{0}\\ \text{•}& {}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+3\\ {}& {}& {{a}}_{{k}{+}{3}}{}\left({k}{+}{5}{+}{r}\right){}\left({k}{+}{r}\right){+}{{a}}_{{k}{+}{2}}{}\left({k}{+}{3}\right){+}{{a}}_{{k}{+}{2}}{}{r}{+}{{a}}_{{k}}{-}{{a}}_{{k}{+}{2}}{=}{0}\\ \text{•}& {}& \text{Recursion relation that defines series solution to ODE}\\ {}& {}& {{a}}_{{k}{+}{3}}{=}{-}\frac{{k}{}{{a}}_{{k}{+}{2}}{+}{{a}}_{{k}{+}{2}}{}{r}{+}{{a}}_{{k}}{+}{2}{}{{a}}_{{k}{+}{2}}}{\left({k}{+}{5}{+}{r}\right){}\left({k}{+}{r}\right)}\\ \text{•}& {}& \text{Recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=-2\\ {}& {}& {{a}}_{{k}{+}{3}}{=}{-}\frac{{k}{}{{a}}_{{k}{+}{2}}{+}{{a}}_{{k}}}{\left({k}{+}{3}\right){}\left({k}{-}{2}\right)}\\ \text{•}& {}& \text{Series not valid for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=-2\text{, division by}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{in the recursion relation at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=2\\ {}& {}& {{a}}_{{k}{+}{3}}{=}{-}\frac{{k}{}{{a}}_{{k}{+}{2}}{+}{{a}}_{{k}}}{\left({k}{+}{3}\right){}\left({k}{-}{2}\right)}\\ \text{•}& {}& \text{Recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=3\\ {}& {}& {{a}}_{{k}{+}{3}}{=}{-}\frac{{k}{}{{a}}_{{k}{+}{2}}{+}{{a}}_{{k}}{+}{5}{}{{a}}_{{k}{+}{2}}}{\left({k}{+}{8}\right){}\left({k}{+}{3}\right)}\\ \text{•}& {}& \text{Solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=3\\ {}& {}& \left[{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}{+}{3}}{,}{{a}}_{{k}{+}{3}}{=}{-}\frac{{k}{}{{a}}_{{k}{+}{2}}{+}{{a}}_{{k}}{+}{5}{}{{a}}_{{k}{+}{2}}}{\left({k}{+}{8}\right){}\left({k}{+}{3}\right)}{,}{{a}}_{{1}}{=}{-}\frac{{{a}}_{{0}}}{{2}}{,}{{a}}_{{2}}{=}\frac{{{a}}_{{0}}}{{7}}\right]\end{array}$ (6)
 > $\mathrm{ode4}≔\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)+\frac{ⅆ}{ⅆx}y\left(x\right)+{x}^{2}y\left(x\right)=0$
 ${\mathrm{ode4}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{{x}}^{{2}}{}{y}{}\left({x}\right){=}{0}$ (7)
 > $\mathrm{ODESteps}\left(\mathrm{ode4}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{{x}}^{{2}}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Assume series solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}}\\ \text{▫}& {}& \text{Rewrite ODE with series expansions}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}^{2}\cdot y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}{+}{2}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k-2\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{2}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{-}{2}}{}{{x}}^{{k}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{1}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{k}{}{{x}}^{{k}{-}{1}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}}{}\left({k}{+}{1}\right){}{{x}}^{{k}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{2}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{k}{}\left({k}{-}{1}\right){}{{x}}^{{k}{-}{2}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{2}}{}\left({k}{+}{2}\right){}\left({k}{+}{1}\right){}{{x}}^{{k}}\\ {}& {}& \text{Rewrite ODE with series expansions}\\ {}& {}& {2}{}{{a}}_{{2}}{+}{{a}}_{{1}}{+}\left({6}{}{{a}}_{{3}}{+}{2}{}{{a}}_{{2}}\right){}{x}{+}\left({\sum }_{{k}{=}{2}}^{{\mathrm{\infty }}}{}\left({{a}}_{{k}{+}{2}}{}\left({k}{+}{2}\right){}\left({k}{+}{1}\right){+}{{a}}_{{k}{+}{1}}{}\left({k}{+}{1}\right){+}{{a}}_{{k}{-}{2}}\right){}{{x}}^{{k}}\right){=}{0}\\ \text{•}& {}& \text{The coefficients of each power of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{must be 0}\\ {}& {}& \left[{2}{}{{a}}_{{2}}{+}{{a}}_{{1}}{=}{0}{,}{6}{}{{a}}_{{3}}{+}{2}{}{{a}}_{{2}}{=}{0}\right]\\ \text{•}& {}& \text{Solve for the dependent coefficient(s)}\\ {}& {}& \left\{{{a}}_{{2}}{=}{-}\frac{{{a}}_{{1}}}{{2}}{,}{{a}}_{{3}}{=}\frac{{{a}}_{{1}}}{{6}}\right\}\\ \text{•}& {}& \text{Each term in the series must be 0, giving the recursion relation}\\ {}& {}& \left({{k}}^{{2}}{+}{3}{}{k}{+}{2}\right){}{{a}}_{{k}{+}{2}}{+}{{a}}_{{k}{+}{1}}{}{k}{+}{{a}}_{{k}{-}{2}}{+}{{a}}_{{k}{+}{1}}{=}{0}\\ \text{•}& {}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+2\\ {}& {}& \left({\left({k}{+}{2}\right)}^{{2}}{+}{3}{}{k}{+}{8}\right){}{{a}}_{{k}{+}{4}}{+}{{a}}_{{k}{+}{3}}{}\left({k}{+}{2}\right){+}{{a}}_{{k}}{+}{{a}}_{{k}{+}{3}}{=}{0}\\ \text{•}& {}& \text{Recursion relation that defines the series solution to the ODE}\\ {}& {}& \left[{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}}{,}{{a}}_{{k}{+}{4}}{=}{-}\frac{{k}{}{{a}}_{{k}{+}{3}}{+}{{a}}_{{k}}{+}{3}{}{{a}}_{{k}{+}{3}}}{{{k}}^{{2}}{+}{7}{}{k}{+}{12}}{,}{{a}}_{{2}}{=}{-}\frac{{{a}}_{{1}}}{{2}}{,}{{a}}_{{3}}{=}\frac{{{a}}_{{1}}}{{6}}\right]\end{array}$ (8)
 > $\mathrm{ode5}≔\frac{ⅆ}{ⅆx}\left(\left(-{x}^{2}+1\right)\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)\right)+12y\left(x\right)=0$
 ${\mathrm{ode5}}{≔}{-}{2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({-}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{12}{}{y}{}\left({x}\right){=}{0}$ (9)
 > $\mathrm{ODESteps}\left(\mathrm{ode5}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {-}{2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({-}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{12}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate 2nd derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{12}{}{y}{}\left({x}\right)}{{{x}}^{{2}}{-}{1}}{-}\frac{{2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{{x}}^{{2}}{-}{1}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{{x}}^{{2}}{-}{1}}{-}\frac{{12}{}{y}{}\left({x}\right)}{{{x}}^{{2}}{-}{1}}{=}{0}\\ \text{▫}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& \text{◦}& \text{Define functions}\\ {}& {}& \left[{{P}}_{{2}}{}\left({x}\right){=}\frac{{2}{}{x}}{{{x}}^{{2}}{-}{1}}{,}{{P}}_{{3}}{}\left({x}\right){=}{-}\frac{{12}}{{{x}}^{{2}}{-}{1}}\right]\\ {}& \text{◦}& \left(x+1\right)\cdot {P}_{2}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=-1\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{-1}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{-1}}{=}{1}\\ {}& \text{◦}& {\left(x+1\right)}^{2}\cdot {P}_{3}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=-1\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{-1}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{-1}}{=}{0}\\ {}& \text{◦}& x=-1\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& {{x}}_{{0}}{=}{-1}\\ \text{•}& {}& \text{Multiply by denominators}\\ {}& {}& \left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}\left({{x}}^{{2}}{-}{1}\right){+}{2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{12}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Change variables using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=u-1\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{so that the regular singular point is at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u=0\\ {}& {}& \left({{u}}^{{2}}{-}{2}{}{u}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{u}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({u}\right)\right){+}\left({2}{}{u}{-}{2}\right){}\left(\frac{{ⅆ}}{{ⅆ}{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({u}\right)\right){-}{12}{}{y}{}\left({u}\right){=}{0}\\ \text{•}& {}& \text{Assume series solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(u\right)\\ {}& {}& {y}{}\left({u}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{u}}^{{k}{+}{r}}\\ \text{▫}& {}& \text{Rewrite ODE with series expansions}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}^{m}\cdot \left(\frac{ⅆ}{ⅆu}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(u\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}m=0..1\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}{{u}}^{{k}{+}{r}{-}{1}{+}{m}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+1-m\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{-}{1}{+}{m}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}{-}{m}}{}\left({k}{+}{1}{-}{m}{+}{r}\right){}{{u}}^{{k}{+}{r}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}^{m}\cdot \left(\frac{{ⅆ}^{2}}{ⅆ{u}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(u\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}m=1..2\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}\left({k}{+}{r}{-}{1}\right){}{{u}}^{{k}{+}{r}{-}{2}{+}{m}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+2-m\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{-}{2}{+}{m}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{2}{-}{m}}{}\left({k}{+}{2}{-}{m}{+}{r}\right){}\left({k}{+}{1}{-}{m}{+}{r}\right){}{{u}}^{{k}{+}{r}}\\ {}& {}& \text{Rewrite ODE with series expansions}\\ {}& {}& {-}{2}{}{{a}}_{{0}}{}{{r}}^{{2}}{}{{u}}^{{-}{1}{+}{r}}{+}\left({\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}\left({-}{2}{}{{a}}_{{k}{+}{1}}{}{\left({k}{+}{1}{+}{r}\right)}^{{2}}{+}{{a}}_{{k}}{}\left({k}{+}{r}{+}{4}\right){}\left({k}{+}{r}{-}{3}\right)\right){}{{u}}^{{k}{+}{r}}\right){=}{0}\\ \text{•}& {}& {a}_{0}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{cannot be 0 by assumption, giving the indicial equation}\\ {}& {}& {-}{2}{}{{r}}^{{2}}{=}{0}\\ \text{•}& {}& \text{Values of r that satisfy the indicial equation}\\ {}& {}& {r}{=}{0}\\ \text{•}& {}& \text{Each term in the series must be 0, giving the recursion relation}\\ {}& {}& {-}{2}{}{{a}}_{{k}{+}{1}}{}{\left({k}{+}{1}\right)}^{{2}}{+}{{a}}_{{k}}{}\left({k}{+}{4}\right){}\left({k}{-}{3}\right){=}{0}\\ \text{•}& {}& \text{Recursion relation that defines series solution to ODE}\\ {}& {}& {{a}}_{{k}{+}{1}}{=}\frac{{{a}}_{{k}}{}\left({k}{+}{4}\right){}\left({k}{-}{3}\right)}{{2}{}{\left({k}{+}{1}\right)}^{{2}}}\\ \text{•}& {}& \text{Recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=0\text{; series terminates at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=3\\ {}& {}& {{a}}_{{k}{+}{1}}{=}\frac{{{a}}_{{k}}{}\left({k}{+}{4}\right){}\left({k}{-}{3}\right)}{{2}{}{\left({k}{+}{1}\right)}^{{2}}}\\ \text{•}& {}& \text{Apply recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=0\\ {}& {}& {{a}}_{{1}}{=}{-}{6}{}{{a}}_{{0}}\\ \text{•}& {}& \text{Apply recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=1\\ {}& {}& {{a}}_{{2}}{=}{-}\frac{{5}{}{{a}}_{{1}}}{{4}}\\ \text{•}& {}& \text{Express in terms of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{a}_{0}\\ {}& {}& {{a}}_{{2}}{=}\frac{{15}{}{{a}}_{{0}}}{{2}}\\ \text{•}& {}& \text{Apply recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=2\\ {}& {}& {{a}}_{{3}}{=}{-}\frac{{{a}}_{{2}}}{{3}}\\ \text{•}& {}& \text{Express in terms of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{a}_{0}\\ {}& {}& {{a}}_{{3}}{=}{-}\frac{{5}{}{{a}}_{{0}}}{{2}}\\ \text{•}& {}& \text{Terminating series solution of the ODE for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=0\text{. Use reduction of order to find the second linearly independent solution}\\ {}& {}& {y}{}\left({u}\right){=}\left[{}\right]\\ \text{•}& {}& \text{Revert the change of variables}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u=x+1\\ {}& {}& \left[{y}{}\left({x}\right){=}{{a}}_{{0}}{}\left(\frac{{3}}{{2}}{}{x}{-}\frac{{5}}{{2}}{}{{x}}^{{3}}\right)\right]\end{array}$ (10)
 > $\mathrm{ode6}≔x\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)+\left(1-x\right)\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+4y\left(x\right)=0$
 ${\mathrm{ode6}}{≔}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{+}\left({1}{-}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{4}{}{y}{}\left({x}\right){=}{0}$ (11)
 > $\mathrm{ODESteps}\left(\mathrm{ode6}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{+}\left({1}{-}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{4}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate 2nd derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{{4}{}{y}{}\left({x}\right)}{{x}}{+}\frac{\left({-}{1}{+}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{x}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{\left({-}{1}{+}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{x}}{+}\frac{{4}{}{y}{}\left({x}\right)}{{x}}{=}{0}\\ \text{▫}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& \text{◦}& \text{Define functions}\\ {}& {}& \left[{{P}}_{{2}}{}\left({x}\right){=}{-}\frac{{-}{1}{+}{x}}{{x}}{,}{{P}}_{{3}}{}\left({x}\right){=}\frac{{4}}{{x}}\right]\\ {}& \text{◦}& x\cdot {P}_{2}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=0\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{0}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{0}}{=}{1}\\ {}& \text{◦}& {x}^{2}\cdot {P}_{3}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=0\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{0}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{0}}{=}{0}\\ {}& \text{◦}& x=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& {{x}}_{{0}}{=}{0}\\ \text{•}& {}& \text{Multiply by denominators}\\ {}& {}& \left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{+}\left({1}{-}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{4}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Assume series solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}{+}{r}}\\ \text{▫}& {}& \text{Rewrite ODE with series expansions}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}^{m}\cdot \left(\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}m=0..1\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}{{x}}^{{k}{+}{r}{-}{1}{+}{m}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+1-m\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{-}{1}{+}{m}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}{-}{m}}{}\left({k}{+}{1}{-}{m}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\cdot \left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}\left({k}{+}{r}{-}{1}\right){}{{x}}^{{k}{+}{r}{-}{1}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+1\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{-1}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}}{}\left({k}{+}{1}{+}{r}\right){}\left({k}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& {}& \text{Rewrite ODE with series expansions}\\ {}& {}& {{a}}_{{0}}{}{{r}}^{{2}}{}{{x}}^{{-}{1}{+}{r}}{+}\left({\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}\left({{a}}_{{k}{+}{1}}{}{\left({k}{+}{1}{+}{r}\right)}^{{2}}{-}{{a}}_{{k}}{}\left({k}{+}{r}{-}{4}\right)\right){}{{x}}^{{k}{+}{r}}\right){=}{0}\\ \text{•}& {}& {a}_{0}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{cannot be 0 by assumption, giving the indicial equation}\\ {}& {}& {{r}}^{{2}}{=}{0}\\ \text{•}& {}& \text{Values of r that satisfy the indicial equation}\\ {}& {}& {r}{=}{0}\\ \text{•}& {}& \text{Each term in the series must be 0, giving the recursion relation}\\ {}& {}& {{a}}_{{k}{+}{1}}{}{\left({k}{+}{1}\right)}^{{2}}{-}{{a}}_{{k}}{}\left({k}{-}{4}\right){=}{0}\\ \text{•}& {}& \text{Recursion relation that defines series solution to ODE}\\ {}& {}& {{a}}_{{k}{+}{1}}{=}\frac{{{a}}_{{k}}{}\left({k}{-}{4}\right)}{{\left({k}{+}{1}\right)}^{{2}}}\\ \text{•}& {}& \text{Recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=0\text{; series terminates at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=4\\ {}& {}& {{a}}_{{k}{+}{1}}{=}\frac{{{a}}_{{k}}{}\left({k}{-}{4}\right)}{{\left({k}{+}{1}\right)}^{{2}}}\\ \text{•}& {}& \text{Apply recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=0\\ {}& {}& {{a}}_{{1}}{=}{-}{4}{}{{a}}_{{0}}\\ \text{•}& {}& \text{Apply recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=1\\ {}& {}& {{a}}_{{2}}{=}{-}\frac{{3}{}{{a}}_{{1}}}{{4}}\\ \text{•}& {}& \text{Express in terms of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{a}_{0}\\ {}& {}& {{a}}_{{2}}{=}{3}{}{{a}}_{{0}}\\ \text{•}& {}& \text{Apply recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=2\\ {}& {}& {{a}}_{{3}}{=}{-}\frac{{2}{}{{a}}_{{2}}}{{9}}\\ \text{•}& {}& \text{Express in terms of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{a}_{0}\\ {}& {}& {{a}}_{{3}}{=}{-}\frac{{2}{}{{a}}_{{0}}}{{3}}\\ \text{•}& {}& \text{Apply recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=3\\ {}& {}& {{a}}_{{4}}{=}{-}\frac{{{a}}_{{3}}}{{16}}\\ \text{•}& {}& \text{Express in terms of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{a}_{0}\\ {}& {}& {{a}}_{{4}}{=}\frac{{{a}}_{{0}}}{{24}}\\ \text{•}& {}& \text{Terminating series solution of the ODE for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=0\text{. Use reduction of order to find the second linearly independent solution}\\ {}& {}& {y}{}\left({x}\right){=}\left[{}\right]\end{array}$ (12)
 > $\mathrm{ode7}≔x\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)+\left(1-x\right)\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+6y\left(x\right)=0$
 ${\mathrm{ode7}}{≔}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{+}\left({1}{-}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{6}{}{y}{}\left({x}\right){=}{0}$ (13)
 > $\mathrm{ODESteps}\left(\mathrm{ode7}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{+}\left({1}{-}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{6}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate 2nd derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{{6}{}{y}{}\left({x}\right)}{{x}}{+}\frac{\left({-}{1}{+}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{x}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{\left({-}{1}{+}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{x}}{+}\frac{{6}{}{y}{}\left({x}\right)}{{x}}{=}{0}\\ \text{▫}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& \text{◦}& \text{Define functions}\\ {}& {}& \left[{{P}}_{{2}}{}\left({x}\right){=}{-}\frac{{-}{1}{+}{x}}{{x}}{,}{{P}}_{{3}}{}\left({x}\right){=}\frac{{6}}{{x}}\right]\\ {}& \text{◦}& x\cdot {P}_{2}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=0\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{0}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{0}}{=}{1}\\ {}& \text{◦}& {x}^{2}\cdot {P}_{3}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is analytic at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=0\\ {}& {}& \genfrac{}{}{0}{}{\left(\left[{}\right]\right)}{\phantom{{x}{=}{0}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\left[{}\right]\right)}}{{x}{=}{0}}{=}{0}\\ {}& \text{◦}& x=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& \text{Check to see if}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}_{0}=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a regular singular point}\\ {}& {}& {{x}}_{{0}}{=}{0}\\ \text{•}& {}& \text{Multiply by denominators}\\ {}& {}& \left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{+}\left({1}{-}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{6}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Assume series solution for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}{+}{r}}\\ \text{▫}& {}& \text{Rewrite ODE with series expansions}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}^{m}\cdot \left(\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}m=0..1\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}{{x}}^{{k}{+}{r}{-}{1}{+}{m}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+1-m\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{-}{1}{+}{m}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}{-}{m}}{}\left({k}{+}{1}{-}{m}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\cdot \left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to series expansion}\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}\left({k}{+}{r}\right){}\left({k}{+}{r}{-}{1}\right){}{{x}}^{{k}{+}{r}{-}{1}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+1\\ {}& {}& \left[{}\right]{=}{\sum }_{{k}{=}{-1}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}}{}\left({k}{+}{1}{+}{r}\right){}\left({k}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& {}& \text{Rewrite ODE with series expansions}\\ {}& {}& {{a}}_{{0}}{}{{r}}^{{2}}{}{{x}}^{{-}{1}{+}{r}}{+}\left({\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}\left({{a}}_{{k}{+}{1}}{}{\left({k}{+}{1}{+}{r}\right)}^{{2}}{-}{{a}}_{{k}}{}\left({k}{+}{r}{-}{6}\right)\right){}{{x}}^{{k}{+}{r}}\right){=}{0}\\ \text{•}& {}& {a}_{0}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{cannot be 0 by assumption, giving the indicial equation}\\ {}& {}& {{r}}^{{2}}{=}{0}\\ \text{•}& {}& \text{Values of r that satisfy the indicial equation}\\ {}& {}& {r}{=}{0}\\ \text{•}& {}& \text{Each term in the series must be 0, giving the recursion relation}\\ {}& {}& {{a}}_{{k}{+}{1}}{}{\left({k}{+}{1}\right)}^{{2}}{-}{{a}}_{{k}}{}\left({k}{-}{6}\right){=}{0}\\ \text{•}& {}& \text{Recursion relation that defines series solution to ODE}\\ {}& {}& {{a}}_{{k}{+}{1}}{=}\frac{{{a}}_{{k}}{}\left({k}{-}{6}\right)}{{\left({k}{+}{1}\right)}^{{2}}}\\ \text{•}& {}& \text{Recursion relation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=0\text{; series terminates at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k=6\\ {}& {}& {{a}}_{{k}{+}{1}}{=}\frac{{{a}}_{{k}}{}\left({k}{-}{6}\right)}{{\left({k}{+}{1}\right)}^{{2}}}\\ \text{•}& {}& \text{Recursion relation that defines the terminating series solution of the ODE for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r=0\\ {}& {}& \left[{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{5}}{}{{a}}_{{k}}{}{{x}}^{{k}}{,}{{a}}_{{k}{+}{1}}{=}\frac{{{a}}_{{k}}{}\left({k}{-}{6}\right)}{{\left({k}{+}{1}\right)}^{{2}}}\right]\end{array}$ (14)