 SeriesSolutions - 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

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
 > $\mathrm{ode1}≔{x}^{2}\mathrm{diff}\left(y\left(x\right),x,x\right)+x\mathrm{diff}\left(y\left(x\right),x\right)+5xy\left(x\right)=0$
 ${\mathrm{ode1}}{≔}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{5}{}{x}{}{y}{}\left({x}\right){=}{0}$ (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{◦}& {{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}\\ {}& {}& \left[{}\right]{=}{1}\\ {}& \text{◦}& {{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}\\ {}& {}& \left[{}\right]{=}{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}}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{to series expansion}}\\ {}& {}& \left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{1}}{}\left({k}{+}{r}{+}{1}\right){}\left({k}{+}{r}\right){}{{x}}^{{k}{+}{r}}\\ {}& \text{◦}& {\text{Convert}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{5}{}{y}{}\left({x}\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{to series expansion}}\\ {}& {}& {5}{}{y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{5}{}{{a}}_{{k}}{}{{x}}^{{k}{+}{r}}\\ {}& \text{◦}& {\text{Convert}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{ⅆ}}{{ⅆ}{x}}\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}{+}{1}}{}\left({k}{+}{r}{+}{1}\right){}{{x}}^{{k}{+}{r}}\\ {}& {}& \text{Rewrite ODE with series expansions}\\ {}& {}& {\mathrm{Sum}}{}\left(\left({{a}}_{{k}{+}{1}}{}\left({k}{+}{r}{+}{1}\right){+}{{a}}_{{k}{+}{1}}{}\left({k}{+}{r}{+}{1}\right){}\left({k}{+}{r}\right){+}{5}{}{{a}}_{{k}}\right){}{{x}}^{{k}{+}{r}}{,}{0}{..}{\mathrm{\infty }}\right){=}{0}\\ \text{•}& {}& {{a}}_{{0}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{cannot be 0 by assumption giving the indicial equation}}\\ {}& {}& {r}{=}{0}\\ \text{•}& {}& \text{Values of r that satisfy the indicial equation}\\ {}& {}& {r}{=}{0}\\ \text{•}& {}& \text{Each term must evaluate to 0 giving the recursion relation}\\ {}& {}& {\left({k}{+}{r}{+}{1}\right)}^{{2}}{}{{a}}_{{k}{+}{1}}{+}{5}{}{{a}}_{{k}}\\ \text{•}& {}& \text{Recursion relation that defines series solution to ODE}\\ {}& {}& {{a}}_{{k}{+}{1}}{=}{-}\frac{{5}{}{{a}}_{{k}}}{{\left({k}{+}{r}{+}{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)
 > $\mathrm{ode2}≔\mathrm{diff}\left(y\left(x\right),x,x\right)+x\mathrm{diff}\left(y\left(x\right),x\right)+y\left(x\right)=0$
 ${\mathrm{ode2}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{y}{}\left({x}\right){=}{0}$ (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 ODE with series expansions}\\ {}& \text{◦}& {\text{Convert}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\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}}\\ {}& {}& {x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\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}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}{+}{2}}{}\left({k}{+}{2}\right){}\left({k}{+}{1}\right){}{{x}}^{{k}}\\ {}& \text{◦}& {\text{Convert}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{}\left({x}\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{to series expansion}}\\ {}& {}& {y}{}\left({x}\right){=}{\sum }_{{k}{=}{0}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{{x}}^{{k}}\\ {}& {}& \text{Rewrite ODE with series expansions}\\ {}& {}& {\mathrm{Sum}}{}\left(\left({{a}}_{{k}}{+}{{a}}_{{k}}{}{k}{+}{{a}}_{{k}{+}{2}}{}\left({k}{+}{2}\right){}\left({k}{+}{1}\right)\right){}{{x}}^{{k}}{,}{0}{..}{\mathrm{\infty }}\right){=}{0}\\ \text{•}& {}& \text{Each term must evaluate to 0 giving the recursion relation}\\ {}& {}& \left({k}{+}{1}\right){}\left(\left({k}{+}{2}\right){}{{a}}_{{k}{+}{2}}{+}{{a}}_{{k}}\right){=}{0}\\ \text{•}& {}& \text{Recursion relation that defines series solution to 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)
 > $\mathrm{ode3}≔{x}^{2}\mathrm{diff}\left(y\left(x\right),x,x\right)+{x}^{2}\mathrm{diff}\left(y\left(x\right),x\right)+\left({x}^{3}-6\right)y\left(x\right)=0$
 ${\mathrm{ode3}}{≔}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{{x}}^{{2}}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({{x}}^{{3}}{-}{6}\right){}{y}{}\left({x}\right){=}{0}$ (5)
 > $\mathrm{ODESteps}\left(\mathrm{ode3}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{}}{}\right)\end{array}$