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.

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right)\:$

$\mathrm{ode1}≔x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x\mathrm{diff}\left(y\left(x\right)\,x\right)+5xy\left(x\right)=0$

$\mathrm{ode1}≔x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+5xy\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode1}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+5xy\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=-\frac{5y\left(x\right)}{x}-\frac{\frac{\ⅆ}{\ⅆx}y\left(x\right)}{x}\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{\frac{\ⅆ}{\ⅆx}y\left(x\right)}{x}+\frac{5y\left(x\right)}{x}=0\\ \text{▫}& & \text{Check to see if}{x}_0=0\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)\text{is analytic at}x=0\\ & & \genfrac{}{}{0ex}{}{\left(x\cdot P_2\left(x\right)\right)}{\phantom{x=0}}|\genfrac{}{}{0ex}{}{\phantom{\left(x\cdot P_2\left(x\right)\right)}}{x=0}=1\\ & \text{◦}& x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & \genfrac{}{}{0ex}{}{\left(x^2\cdot P_3\left(x\right)\right)}{\phantom{x=0}}|\genfrac{}{}{0ex}{}{\phantom{\left(x^2\cdot P_3\left(x\right)\right)}}{x=0}=0\\ & \text{◦}& x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \text{•}& & \text{Multiply by denominators}\\ & & \left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)x+5y\left(x\right)+\frac{\ⅆ}{\ⅆx}y\left(x\right)=0\\ \text{•}& & \text{Assume series solution for}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}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{to series expansion}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k\left(k+r\right)x^{k+r-1}\\ & \text{◦}& \text{Shift index using}k\text{->}k+1\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\sum _{k=\mathrm{-1}}^{\mathrm{\infty }}{a}_{k+1}\left(k+1+r\right)x^{k+r}\\ & \text{◦}& \text{Convert}x\cdot \left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\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}k\text{->}k+1\\ & & x\cdot \left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)=\sum _{k=\mathrm{-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+5a_k\right)x^{k+r}\right)=0\\ \text{•}& & a_0\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+5a_k=0\\ \text{•}& & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=-\frac{5a_k}{{\left(k+1\right)}^2}\\ \text{•}& & \text{Recursion relation for}r=0\\ & & a_{k+1}=-\frac{5a_k}{{\left(k+1\right)}^2}\\ \text{•}& & \text{Solution for}r=0\\ & & \left[y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\,a_{k+1}=-\frac{5a_k}{{\left(k+1\right)}^2}\right]\end{array}$

$\mathrm{ode2}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x\mathrm{diff}\left(y\left(x\right)\,x\right)+y\left(x\right)=0$

$\mathrm{ode2}≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode2}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\\ \text{▫}& & \text{Rewrite DE with series expansions}\\ & \text{◦}& \text{Convert}x\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}k{x}^k\\ & \text{◦}& \text{Convert}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\text{to series expansion}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\sum _{k=2}^{\mathrm{\infty }}{a}_{k}k\left(k-1\right)x^{k-2}\\ & \text{◦}& \text{Shift index using}k\text{->}k+2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k+2}\left(k+2\right)\left(k+1\right)x^k\\ & & \text{Rewrite DE with series expansions}\\ & & \sum _{k=0}^{\mathrm{\infty }}\left(a_{k+2}\left(k+2\right)\left(k+1\right)+a_k\left(k+1\right)\right)x^k=0\\ \text{•}& & \text{Each term in the series must be 0, giving the recursion relation}\\ & & \left(k+1\right)\left(a_{k+2}\left(k+2\right)+a_k\right)=0\\ \text{•}& & \text{Recursion relation that defines the series solution to the ODE}\\ & & \left[y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\,a_{k+2}=-\frac{a_k}{k+2}\right]\end{array}$

$\mathrm{ode3}≔x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x^2\mathrm{diff}\left(y\left(x\right)\,x\right)+\left(x^3-6\right)y\left(x\right)=0$

$\mathrm{ode3}≔x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x^2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+\left(x^3-6\right)y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode3}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & x^2\left(\frac{{\ⅆ}^2}{\ⅆx^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\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=-\frac{\left(x^3-6\right)y\left(x\right)}{x^2}-\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{\ⅆ}{\ⅆx}y\left(x\right)+\frac{\left(x^3-6\right)y\left(x\right)}{x^2}=0\\ \text{▫}& & \text{Check to see if}{x}_0=0\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)\text{is analytic at}x=0\\ & & \genfrac{}{}{0ex}{}{\left(x\cdot P_2\left(x\right)\right)}{\phantom{x=0}}|\genfrac{}{}{0ex}{}{\phantom{\left(x\cdot P_2\left(x\right)\right)}}{x=0}=0\\ & \text{◦}& x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & \genfrac{}{}{0ex}{}{\left(x^2\cdot P_3\left(x\right)\right)}{\phantom{x=0}}|\genfrac{}{}{0ex}{}{\phantom{\left(x^2\cdot P_3\left(x\right)\right)}}{x=0}=\mathrm{-6}\\ & \text{◦}& x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \text{•}& & \text{Multiply by denominators}\\ & & x^2\left(\frac{{\ⅆ}^2}{\ⅆx^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\\ \text{•}& & \text{Assume series solution for}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}{x}^m\cdot y\left(x\right)\text{to series expansion for}m=0..3\\ & & x^m\cdot y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^{k+r+m}\\ & \text{◦}& \text{Shift index using}k\text{->}k-m\\ & & x^m\cdot y\left(x\right)=\sum _{k=m}^{\mathrm{\infty }}{a}_{k-m}{x}^{k+r}\\ & \text{◦}& \text{Convert}{x}^2\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\text{to series expansion}\\ & & x^2\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k\left(k+r\right)x^{k+r+1}\\ & \text{◦}& \text{Shift index using}k\text{->}k-1\\ & & x^2\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=\sum _{k=1}^{\mathrm{\infty }}{a}_{k-1}\left(k-1+r\right)x^{k+r}\\ & \text{◦}& \text{Convert}{x}^2\cdot \left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)\text{to series expansion}\\ & & x^2\cdot \left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\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\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\{\mathrm{-2}\,3\right\}\\ \text{•}& & \text{The coefficients of each power of}x\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+4r^3-7r^2-22r+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}k\text{->}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+2a_{k+2}}{\left(k+5+r\right)\left(k+r\right)}\\ \text{•}& & \text{Recursion relation for}r=\mathrm{-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}r=\mathrm{-2}\text{, division by}0\text{in the recursion relation at}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}r=3\\ & & a_{k+3}=-\frac{k{a}_{k+2}+a_k+5a_{k+2}}{\left(k+8\right)\left(k+3\right)}\\ \text{•}& & \text{Solution for}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+5a_{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}$

$\mathrm{ode4}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+\mathrm{diff}\left(y\left(x\right)\,x\right)+x^{2}y\left(x\right)=0$

$\mathrm{ode4}≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{\ⅆ}{\ⅆx}y\left(x\right)+x^{2}y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode4}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{\ⅆ}{\ⅆx}y\left(x\right)+x^{2}y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Assume series solution for}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}{x}^2\cdot y\left(x\right)\text{to series expansion}\\ & & x^2\cdot y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^{k+2}\\ & \text{◦}& \text{Shift index using}k\text{->}k-2\\ & & x^2\cdot y\left(x\right)=\sum _{k=2}^{\mathrm{\infty }}{a}_{k-2}{x}^k\\ & \text{◦}& \text{Convert}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{to series expansion}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}{a}_{k}k{x}^{k-1}\\ & \text{◦}& \text{Shift index using}k\text{->}k+1\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k+1}\left(k+1\right)x^k\\ & \text{◦}& \text{Convert}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\text{to series expansion}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\sum _{k=2}^{\mathrm{\infty }}{a}_{k}k\left(k-1\right)x^{k-2}\\ & \text{◦}& \text{Shift index using}k\text{->}k+2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}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}\\ & & 2a_2+a_1+\left(6a_3+2a_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}x\text{must be 0}\\ & & \left[2a_2+a_1=0\,6a_3+2a_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+3k+2\right)a_{k+2}+a_{k+1}k+a_{k-2}+a_{k+1}=0\\ \text{•}& & \text{Shift index using}k\text{->}k+2\\ & & \left({\left(k+2\right)}^2+3k+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+3a_{k+3}}{k^2+7k+12}\,a_2=-\frac{a_1}{2}\,a_3=\frac{a_1}{6}\right]\end{array}$

$\mathrm{ode5}≔\mathrm{diff}\left(\left(-x^2+1\right)\mathrm{diff}\left(y\left(x\right)\,x\right)\,x\right)+12y\left(x\right)=0$

$\mathrm{ode5}≔-2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+\left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+12y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode5}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & -2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+\left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+12y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{12y\left(x\right)}{x^2-1}-\frac{2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x^2-1}\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x^2-1}-\frac{12y\left(x\right)}{x^2-1}=0\\ \text{▫}& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \text{◦}& \text{Define functions}\\ & & \left[P_2\left(x\right)=\frac{2x}{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)\text{is analytic at}x=\mathrm{-1}\\ & & \genfrac{}{}{0ex}{}{\left(\left(x+1\right)\cdot P_2\left(x\right)\right)}{\phantom{x=\mathrm{-1}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\left(x+1\right)\cdot P_2\left(x\right)\right)}}{x=\mathrm{-1}}=1\\ & \text{◦}& {\left(x+1\right)}^2\cdot P_3\left(x\right)\text{is analytic at}x=\mathrm{-1}\\ & & \genfrac{}{}{0ex}{}{\left({\left(x+1\right)}^2\cdot P_3\left(x\right)\right)}{\phantom{x=\mathrm{-1}}}|\genfrac{}{}{0ex}{}{\phantom{\left({\left(x+1\right)}^2\cdot P_3\left(x\right)\right)}}{x=\mathrm{-1}}=0\\ & \text{◦}& x=\mathrm{-1}\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=\mathrm{-1}\\ \text{•}& & \text{Multiply by denominators}\\ & & \left(x^2-1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-12y\left(x\right)=0\\ \text{•}& & \text{Change variables using}x=u-1\text{so that the regular singular point is at}u=0\\ & & \left(u^2-2u\right)\left(\frac{{\ⅆ}^2}{\ⅆu^2}y\left(u\right)\right)+\left(2u-2\right)\left(\frac{\ⅆ}{\ⅆu}y\left(u\right)\right)-12y\left(u\right)=0\\ \text{•}& & \text{Assume series solution for}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}{u}^m\cdot \left(\frac{\ⅆ}{\ⅆu}y\left(u\right)\right)\text{to series expansion for}m=0..1\\ & & u^m\cdot \left(\frac{\ⅆ}{\ⅆu}y\left(u\right)\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k\left(k+r\right)u^{k+r-1+m}\\ & \text{◦}& \text{Shift index using}k\text{->}k+1-m\\ & & u^m\cdot \left(\frac{\ⅆ}{\ⅆu}y\left(u\right)\right)=\sum _{k=-1+m}^{\mathrm{\infty }}{a}_{k+1-m}\left(k+1-m+r\right)u^{k+r}\\ & \text{◦}& \text{Convert}{u}^m\cdot \left(\frac{{\ⅆ}^2}{\ⅆu^2}y\left(u\right)\right)\text{to series expansion for}m=1..2\\ & & u^m\cdot \left(\frac{{\ⅆ}^2}{\ⅆu^2}y\left(u\right)\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}k\text{->}k+2-m\\ & & u^m\cdot \left(\frac{{\ⅆ}^2}{\ⅆu^2}y\left(u\right)\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}\\ & & -2a_0{r}^2{u}^{-1+r}+\left(\sum _{k=0}^{\mathrm{\infty }}\left(-2a_{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\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -2r^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}\\ & & -2a_{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}r=0\text{; series terminates at}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}k=0\\ & & a_1=-6a_0\\ \text{•}& & \text{Apply recursion relation for}k=1\\ & & a_2=-\frac{5a_1}{4}\\ \text{•}& & \text{Express in terms of}{a}_0\\ & & a_2=\frac{15a_0}{2}\\ \text{•}& & \text{Apply recursion relation for}k=2\\ & & a_3=-\frac{a_2}{3}\\ \text{•}& & \text{Express in terms of}{a}_0\\ & & a_3=-\frac{5a_0}{2}\\ \text{•}& & \text{Terminating series solution of the ODE for}r=0\text{. Use reduction of order to find the second linearly independent solution}\\ & & y\left(u\right)=a_0\cdot \left(1-6u+\frac{15}{2}{u}^2-\frac{5}{2}{u}^3\right)\\ \text{•}& & \text{Revert the change of variables}u=x+1\\ & & \left[y\left(x\right)=a_0\left(\frac{3}{2}x-\frac{5}{2}{x}^3\right)\right]\\ \text{•}& & \text{Use this particular solution to reduce the order of the ODE and find a complete solution}\\ & & y\left(x\right)=\left(\frac{3}{2}x-\frac{5}{2}{x}^3\right)\left(\mathrm{c\_\_1}\left(\frac{5x}{36\left(x^2-\frac{3}{5}\right)}-\frac{\ln \left(x+1\right)}{8}+\frac{\ln \left(-1+x\right)}{8}+\frac{1}{9x}\right)+\mathrm{c\_\_2}\right)\end{array}$