Cauchy-Euler Equations - Maple Help

ODE Steps for Cauchy-Euler Equations

Overview

 • This help page gives a few examples of using the command ODESteps to solve Cauchy-Euler equations.
 • See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.

Examples

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
 > $\mathrm{ode1}≔{x}^{2}\mathrm{diff}\left(y\left(x\right),x,x\right)-4x\mathrm{diff}\left(y\left(x\right),x\right)+2y\left(x\right)=0$
 ${\mathrm{ode1}}{≔}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{4}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{2}{}{y}{}\left({x}\right){=}{0}$ (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){-}{4}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{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{Isolate 2nd derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{{2}{}{y}{}\left({x}\right)}{{{x}}^{{2}}}{+}\frac{{4}{}\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{{4}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{{x}}{+}\frac{{2}{}{y}{}\left({x}\right)}{{{x}}^{{2}}}{=}{0}\\ \text{•}& {}& \text{Multiply by denominators of the ODE}\\ {}& {}& {{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{4}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{2}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Make a change of variables}\\ {}& {}& {t}{=}{\mathrm{ln}}{}\left({x}\right)\\ \text{▫}& {}& \text{Substitute the change of variables back into the ODE}\\ {}& \text{◦}& \text{Calculate the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{1st}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{derivative of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{y}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{x}\text{, using the chain rule}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{t}{}\left({x}\right)\right)\\ {}& \text{◦}& \text{Compute derivative}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)}{{x}}\\ {}& \text{◦}& \text{Calculate the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{2nd}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{derivative of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{y}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{x}\text{, using the chain rule}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right){}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{t}{}\left({x}\right)\right)}^{{2}}{+}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{t}{}\left({x}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right)\\ {}& \text{◦}& \text{Compute derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)}{{{x}}^{{2}}}{-}\frac{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)}{{{x}}^{{2}}}\\ {}& {}& \text{Substitute the change of variables back into the ODE}\\ {}& {}& {{x}}^{{2}}{}\left(\frac{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)}{{{x}}^{{2}}}{-}\frac{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)}{{{x}}^{{2}}}\right){-}{4}{}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){+}{2}{}{y}{}\left({t}\right){=}{0}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){-}{5}{}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){+}{2}{}{y}{}\left({t}\right){=}{0}\\ \text{•}& {}& \text{Characteristic polynomial of ODE}\\ {}& {}& {{r}}^{{2}}{-}{5}{}{r}{+}{2}{=}{0}\\ \text{•}& {}& \text{Use quadratic formula to solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}r\\ {}& {}& {r}{=}\frac{{5}{±}\left(\sqrt{{17}}\right)}{{2}}\\ \text{•}& {}& \text{Roots of the characteristic polynomial}\\ {}& {}& {r}{=}\left(\frac{{5}}{{2}}{-}\frac{\sqrt{{17}}}{{2}}{,}\frac{{5}}{{2}}{+}\frac{\sqrt{{17}}}{{2}}\right)\\ \text{•}& {}& \text{1st solution of the ODE}\\ {}& {}& {{y}}_{{1}}{}\left({t}\right){=}{{ⅇ}}^{\left(\frac{{5}}{{2}}{-}\frac{\sqrt{{17}}}{{2}}\right){}{t}}\\ \text{•}& {}& \text{2nd solution of the ODE}\\ {}& {}& {{y}}_{{2}}{}\left({t}\right){=}{{ⅇ}}^{\left(\frac{{5}}{{2}}{+}\frac{\sqrt{{17}}}{{2}}\right){}{t}}\\ \text{•}& {}& \text{General solution of the ODE}\\ {}& {}& {y}{}\left({t}\right){=}{\mathrm{C1}}{}{{y}}_{{1}}{}\left({t}\right){+}{\mathrm{C2}}{}{{y}}_{{2}}{}\left({t}\right)\\ \text{•}& {}& \text{Substitute in solutions}\\ {}& {}& {y}{}\left({t}\right){=}{\mathrm{C1}}{}{{ⅇ}}^{\left(\frac{{5}}{{2}}{-}\frac{\sqrt{{17}}}{{2}}\right){}{t}}{+}{\mathrm{C2}}{}{{ⅇ}}^{\left(\frac{{5}}{{2}}{+}\frac{\sqrt{{17}}}{{2}}\right){}{t}}\\ \text{•}& {}& \text{Change variables back using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}t=\mathrm{ln}{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{C1}}{}{{ⅇ}}^{\left(\frac{{5}}{{2}}{-}\frac{\sqrt{{17}}}{{2}}\right){}{\mathrm{ln}}{}\left({x}\right)}{+}{\mathrm{C2}}{}{{ⅇ}}^{\left(\frac{{5}}{{2}}{+}\frac{\sqrt{{17}}}{{2}}\right){}{\mathrm{ln}}{}\left({x}\right)}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {y}{}\left({x}\right){=}{{x}}^{{5}}{{2}}}{}\left({{x}}^{{-}\frac{\sqrt{{17}}}{{2}}}{}{\mathrm{C1}}{+}{{x}}^{\frac{\sqrt{{17}}}{{2}}}{}{\mathrm{C2}}\right)\end{array}$ (2)
 > $\mathrm{ode2}≔{x}^{3}\mathrm{diff}\left(y\left(x\right),x,x,x\right)+3{x}^{2}\mathrm{diff}\left(y\left(x\right),x,x\right)-6x\mathrm{diff}\left(y\left(x\right),x\right)-6y\left(x\right)=0$
 ${\mathrm{ode2}}{≔}{{x}}^{{3}}{}\left(\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{x}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{3}{}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{6}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{6}{}{y}{}\left({x}\right){=}{0}$ (3)
 > $\mathrm{ODESteps}\left(\mathrm{ode2}\right)$