SecondOrderIVPs - Maple Help

ODE Steps for Second Order IVPs

Overview

 • This help page gives a few examples of using the command ODESteps to solve second order initial value problems.
 • See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.

Examples

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
 > $\mathrm{ivp1}≔\left\{\mathrm{diff}\left(y\left(x\right),x,x\right)-\mathrm{diff}\left(y\left(x\right),x\right)-x\mathrm{exp}\left(x\right)=0,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right),x\right),x=0\right)=0,y\left(0\right)=1\right\}$
 ${\mathrm{ivp1}}{≔}\left\{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{0}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{0}\right\}}{=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}$ (1)
 > $\mathrm{ODESteps}\left(\mathrm{ivp1}\right)$
 $\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){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{0}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{0}\right\}}{=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}\\ \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{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{x}{}{{ⅇ}}^{{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{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Characteristic polynomial of homogeneous ODE}\\ {}& {}& {{r}}^{{2}}{-}{r}{=}{0}\\ \text{•}& {}& \text{Factor the characteristic polynomial}\\ {}& {}& {r}{}\left({r}{-}{1}\right){=}{0}\\ \text{•}& {}& \text{Roots of the characteristic polynomial}\\ {}& {}& {r}{=}\left({0}{,}{1}\right)\\ \text{•}& {}& \text{1st solution of the homogeneous ODE}\\ {}& {}& {{y}}_{{1}}{}\left({x}\right){=}{1}\\ \text{•}& {}& \text{2nd solution of the homogeneous ODE}\\ {}& {}& {{y}}_{{2}}{}\left({x}\right){=}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{General solution of the ODE}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{_C1}}{}{{y}}_{{1}}{}\left({x}\right){+}{\mathrm{_C2}}{}{{y}}_{{2}}{}\left({x}\right){+}{{y}}_{{p}}{}\left({x}\right)\\ \text{•}& {}& \text{Substitute in solutions of the homogeneous ODE}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{_C1}}{+}{\mathrm{_C2}}{}{{ⅇ}}^{{x}}{+}{{y}}_{{p}}{}\left({x}\right)\\ \text{▫}& {}& {\text{Find a particular solution}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{{y}}_{{p}}{}\left({x}\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{of the ODE}}\\ {}& \text{◦}& {\text{Use variation of paramaters to find}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{{y}}_{{p}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{here}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{is the forcing function}}\\ {}& {}& \left[{{y}}_{{p}}{}\left({x}\right){=}{-}{{y}}_{{1}}{}\left({x}\right){}\left({\int }\frac{{{y}}_{{2}}{}\left({x}\right){}{f}{}\left({x}\right)}{{W}{}\left({{y}}_{{1}}{}\left({x}\right){,}{{y}}_{{2}}{}\left({x}\right)\right)}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){+}{{y}}_{{2}}{}\left({x}\right){}\left({\int }\frac{{{y}}_{{1}}{}\left({x}\right){}{f}{}\left({x}\right)}{{W}{}\left({{y}}_{{1}}{}\left({x}\right){,}{{y}}_{{2}}{}\left({x}\right)\right)}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){,}{f}{}\left({x}\right){=}{x}{}{{ⅇ}}^{{x}}\right]\\ {}& \text{◦}& \text{Wronskian of solutions of the homogeneous equation}\\ {}& {}& {W}{}\left({{y}}_{{1}}{}\left({x}\right){,}{{y}}_{{2}}{}\left({x}\right)\right){=}\left[\begin{array}{cc}{1}& {{ⅇ}}^{{x}}\\ {0}& {{ⅇ}}^{{x}}\end{array}\right]\\ {}& \text{◦}& \text{Compute Wronskian}\\ {}& {}& {W}{}\left({{y}}_{{1}}{}\left({x}\right){,}{{y}}_{{2}}{}\left({x}\right)\right){=}{{ⅇ}}^{{x}}\\ {}& \text{◦}& {\text{Substitute functions into equation for}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{{y}}_{{p}}{}\left({x}\right)\\ {}& {}& {{y}}_{{p}}{}\left({x}\right){=}{-}\left({\int }{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){+}{{ⅇ}}^{{x}}{}\left({\int }{x}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)\\ {}& \text{◦}& \text{Compute integrals}\\ {}& {}& {{y}}_{{p}}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{2}\right)}{{2}}\\ \text{•}& {}& \text{Substitute particular solution into general solution to ODE}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{_C1}}{+}{\mathrm{_C2}}{}{{ⅇ}}^{{x}}{+}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{2}\right)}{{2}}\\ \text{•}& {}& {\text{Use initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{}\left({0}\right){=}{1}\\ {}& {}& {1}{=}{\mathrm{_C1}}{+}{\mathrm{_C2}}{+}{1}\\ \text{•}& {}& \text{Compute derivative of the solution}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\mathrm{_C2}}{}{{ⅇ}}^{{x}}{+}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{2}\right)}{{2}}{+}\frac{{{ⅇ}}^{{x}}{}\left({2}{}{x}{-}{2}\right)}{{2}}\\ \text{•}& {}& {\text{Use the initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{0}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{0}\right\}}{=}{0}\\ {}& {}& {0}{=}{\mathrm{_C2}}\\ \text{•}& {}& {\text{Solve for}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{_C1}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{and}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{_C2}}\\ {}& {}& \left\{{\mathrm{_C1}}{=}{0}{,}{\mathrm{_C2}}{=}{0}\right\}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{2}\right)}{{2}}\end{array}$ (2)
 > $\mathrm{ivp2}≔\left\{\mathrm{diff}\left(y\left(x\right),x,x\right)+\frac{5{\mathrm{diff}\left(y\left(x\right),x\right)}^{2}}{y\left(x\right)}=0,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right),x\right),x=1\right)=-3,y\left(1\right)=1\right\}$
 ${\mathrm{ivp2}}{≔}\left\{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{5}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{2}}}{{y}{}\left({x}\right)}{=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{-3}{,}{y}{}\left({1}\right){=}{1}\right\}$ (3)
 > $\mathrm{ODESteps}\left(\mathrm{ivp2}\right)$
 $\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){+}\frac{{5}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{2}}}{{y}{}\left({x}\right)}{=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{-3}{,}{y}{}\left({1}\right){=}{1}\right\}\\ \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{Define new dependent variable}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}\\ {}& {}& {u}{}\left({x}\right){=}\left[{}\right]\\ \text{•}& {}& {\text{Compute}}\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)\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ \text{•}& {}& \text{Use chain rule on the lhs}\\ {}& {}& \left(\left[{}\right]\right){}\left(\left[{}\right]\right){=}\left[{}\right]\\ \text{•}& {}& {\text{Substitute in the definition of}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}\\ {}& {}& {u}{}\left({y}\right){}\left(\left[{}\right]\right){=}\left[{}\right]\\ \text{•}& {}& {\text{Make substitutions}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{u}{}\left({y}\right){\text{,}}\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){=}{u}{}\left({y}\right){}\left(\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({y}\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{to reduce order of ODE}}\\ {}& {}& {u}{}\left({y}\right){}\left(\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({y}\right)\right){+}\frac{{5}{}{{u}{}\left({y}\right)}^{{2}}}{{y}}{=}{0}\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \frac{\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({y}\right)}{{u}{}\left({y}\right)}{=}{-}\frac{{5}}{{y}}\\ \text{•}& {}& {\text{Integrate both sides with respect to}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}\\ {}& {}& {\int }\frac{\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({y}\right)}{{u}{}\left({y}\right)}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}{=}{\int }{-}\frac{{5}}{{y}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}{+}{\mathrm{_C1}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& {\mathrm{ln}}{}\left({u}{}\left({y}\right)\right){=}{-}{5}{}{\mathrm{ln}}{}\left({y}\right){+}{\mathrm{_C1}}\\ \text{•}& {}& {\text{Solve for}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}{}\left({y}\right)\\ {}& {}& {u}{}\left({y}\right){=}\frac{{{ⅇ}}^{{\mathrm{_C1}}}}{{{y}}^{{5}}}\\ \text{•}& {}& {\text{Solve 1st ODE for}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{}\left({y}\right)\\ {}& {}& {u}{}\left({y}\right){=}\frac{{{ⅇ}}^{{\mathrm{_C1}}}}{{{y}}^{{5}}}\\ \text{•}& {}& {\text{Revert to original variables with substitution}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}{}\left({y}\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){\text{,}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{=}{y}{}\left({x}\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{\mathrm{_C1}}}}{{{y}{}\left({x}\right)}^{{5}}}\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{{y}{}\left({x}\right)}^{{5}}{=}{{ⅇ}}^{{\mathrm{_C1}}}\\ \text{•}& {}& {\text{Integrate both sides with respect to}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{{y}{}\left({x}\right)}^{{5}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }{{ⅇ}}^{{\mathrm{_C1}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{_C2}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& \frac{{{y}{}\left({x}\right)}^{{6}}}{{6}}{=}{{ⅇ}}^{{\mathrm{_C1}}}{}{x}{+}{\mathrm{_C2}}\\ \text{•}& {}& {\text{Solve for}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{}\left({x}\right)\\ {}& {}& \left\{{y}{}\left({x}\right){=}{\left({6}{}{{ⅇ}}^{{\mathrm{_C1}}}{}{x}{+}{6}{}{\mathrm{_C2}}\right)}^{{1}}{{6}}}{,}{y}{}\left({x}\right){=}{-}{\left({6}{}{{ⅇ}}^{{\mathrm{_C1}}}{}{x}{+}{6}{}{\mathrm{_C2}}\right)}^{{1}}{{6}}}\right\}\\ \text{•}& {}& {\text{Use initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{}\left({1}\right){=}{1}\\ {}& {}& {1}{=}{\left({6}{}{{ⅇ}}^{{\mathrm{_C1}}}{+}{6}{}{\mathrm{_C2}}\right)}^{{1}}{{6}}}\\ \text{•}& {}& \text{Compute derivative of the solution}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{\mathrm{_C1}}}}{{\left({6}{}{{ⅇ}}^{{\mathrm{_C1}}}{}{x}{+}{6}{}{\mathrm{_C2}}\right)}^{{5}}{{6}}}}\\ \text{•}& {}& {\text{Use the initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{-3}\\ {}& {}& {-3}{=}\frac{{{ⅇ}}^{{\mathrm{_C1}}}}{{\left({6}{}{{ⅇ}}^{{\mathrm{_C1}}}{+}{6}{}{\mathrm{_C2}}\right)}^{{5}}{{6}}}}\\ \text{•}& {}& \text{Solution is complex}\\ \text{•}& {}& {\text{Use initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{}\left({1}\right){=}{1}\\ {}& {}& {1}{=}{-}{\left({6}{}{{ⅇ}}^{{\mathrm{_C1}}}{+}{6}{}{\mathrm{_C2}}\right)}^{{1}}{{6}}}\\ \text{•}& {}& \text{Compute derivative of the solution}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{{{ⅇ}}^{{\mathrm{_C1}}}}{{\left({6}{}{{ⅇ}}^{{\mathrm{_C1}}}{}{x}{+}{6}{}{\mathrm{_C2}}\right)}^{{5}}{{6}}}}\\ \text{•}& {}& {\text{Use the initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{-3}\\ {}& {}& {-3}{=}{-}\frac{{{ⅇ}}^{{\mathrm{_C1}}}}{{\left({6}{}{{ⅇ}}^{{\mathrm{_C1}}}{+}{6}{}{\mathrm{_C2}}\right)}^{{5}}{{6}}}}\\ \text{•}& {}& \text{Solution is complex}\end{array}$ (4)
 > $\mathrm{ivp3}≔\left\{\mathrm{diff}\left(y\left(x\right),x,x\right)-\mathrm{diff}\left(y\left(x\right),x\right)-6y\left(x\right)=0,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right),x\right),x=1\right)=a,y\left(1\right)=0\right\}$
 ${\mathrm{ivp3}}{≔}\left\{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{6}{}{y}{}\left({x}\right){=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{a}{,}{y}{}\left({1}\right){=}{0}\right\}$ (5)
 > $\mathrm{ODESteps}\left(\mathrm{ivp3}\right)$
 $\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){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{6}{}{y}{}\left({x}\right){=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{a}{,}{y}{}\left({1}\right){=}{0}\right\}\\ \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{Characteristic polynomial of ODE}\\ {}& {}& {{r}}^{{2}}{-}{r}{-}{6}{=}{0}\\ \text{•}& {}& \text{Factor the characteristic polynomial}\\ {}& {}& \left({r}{+}{2}\right){}\left({r}{-}{3}\right){=}{0}\\ \text{•}& {}& \text{Roots of the characteristic polynomial}\\ {}& {}& {r}{=}\left({-2}{,}{3}\right)\\ \text{•}& {}& \text{1st solution of the ODE}\\ {}& {}& {{y}}_{{1}}{}\left({x}\right){=}{{ⅇ}}^{{-}{2}{}{x}}\\ \text{•}& {}& \text{2nd solution of the ODE}\\ {}& {}& {{y}}_{{2}}{}\left({x}\right){=}{{ⅇ}}^{{3}{}{x}}\\ \text{•}& {}& \text{General solution of the ODE}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{_C1}}{}{{y}}_{{1}}{}\left({x}\right){+}{\mathrm{_C2}}{}{{y}}_{{2}}{}\left({x}\right)\\ \text{•}& {}& \text{Substitute in solutions}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{_C1}}{}{{ⅇ}}^{{-}{2}{}{x}}{+}{\mathrm{_C2}}{}{{ⅇ}}^{{3}{}{x}}\\ \text{•}& {}& {\text{Use initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{}\left({1}\right){=}{0}\\ {}& {}& {0}{=}{\mathrm{_C1}}{}{{ⅇ}}^{{-2}}{+}{\mathrm{_C2}}{}{{ⅇ}}^{{3}}\\ \text{•}& {}& \text{Compute derivative of the solution}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}{2}{}{\mathrm{_C1}}{}{{ⅇ}}^{{-}{2}{}{x}}{+}{3}{}{\mathrm{_C2}}{}{{ⅇ}}^{{3}{}{x}}\\ \text{•}& {}& {\text{Use the initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{a}\\ {}& {}& {a}{=}{-}{2}{}{\mathrm{_C1}}{}{{ⅇ}}^{{-2}}{+}{3}{}{\mathrm{_C2}}{}{{ⅇ}}^{{3}}\\ \text{•}& {}& {\text{Solve for}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{_C1}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{and}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{_C2}}\\ {}& {}& \left\{{\mathrm{_C1}}{=}{-}\frac{{a}}{{5}{}{{ⅇ}}^{{-2}}}{,}{\mathrm{_C2}}{=}\frac{{a}}{{5}{}{{ⅇ}}^{{3}}}\right\}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{a}{}{{ⅇ}}^{{-}{2}{}{x}}}{{5}{}{{ⅇ}}^{{-2}}}{+}\frac{{a}{}{{ⅇ}}^{{3}{}{x}}}{{5}{}{{ⅇ}}^{{3}}}\end{array}$ (6)
 > $\mathrm{ivp4}≔\left\{{x}^{2}\mathrm{diff}\left(y\left(x\right),x,x\right)-4x\mathrm{diff}\left(y\left(x\right),x\right)+2y\left(x\right)=0,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right),x\right),x=1\right)=10,y\left(1\right)=-1\right\}$
 ${\mathrm{ivp4}}{≔}\left\{{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{4}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{2}{}{y}{}\left({x}\right){=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{10}{,}{y}{}\left({1}\right){=}{-1}\right\}$ (7)
 > $\mathrm{ODESteps}\left(\mathrm{ivp4}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{{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}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{10}{,}{y}{}\left({1}\right){=}{-1}\right\}\\ \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 1st derivative of}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{with respect to}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}{\text{, using the chain rule}}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ {}& \text{◦}& \text{Compute derivative}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}\frac{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)}{{x}}\\ {}& \text{◦}& {\text{Calculate the 2nd derivative of}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{with respect to}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}{\text{, using the chain rule}}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ {}& \text{◦}& \text{Compute derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}{-}\frac{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)}{{{x}}^{{2}}}{+}\frac{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\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}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)}{{{x}}^{{2}}}{+}\frac{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\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}\\ {}& {}& {-}{5}{}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){+}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\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(\left[{}\right]\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){}\left[{}\right]}{+}{\mathrm{_C2}}{}{{ⅇ}}^{\left(\frac{{5}}{{2}}{+}\frac{\sqrt{{17}}}{{2}}\right){}\left[{}\right]}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{_C1}}{}{{ⅇ}}^{{-}\frac{\left({-}{5}{+}\sqrt{{17}}\right){}\left[{}\right]}{{2}}}{+}{\mathrm{_C2}}{}{{ⅇ}}^{\frac{\left({5}{+}\sqrt{{17}}\right){}\left[{}\right]}{{2}}}\\ \text{•}& {}& {\text{Use initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}{}\left({1}\right){=}{-1}\\ {}& {}& {-1}{=}{\mathrm{_C1}}{+}{\mathrm{_C2}}\\ \text{•}& {}& \text{Compute derivative of the solution}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{{\mathrm{_C1}}{}\left({-}{5}{+}\sqrt{{17}}\right){}{{ⅇ}}^{{-}\frac{\left({-}{5}{+}\sqrt{{17}}\right){}{\mathrm{ln}}{}\left({x}\right)}{{2}}}}{{2}{}{x}}{+}\frac{{\mathrm{_C2}}{}\left({5}{+}\sqrt{{17}}\right){}{{ⅇ}}^{\frac{\left({5}{+}\sqrt{{17}}\right){}{\mathrm{ln}}{}\left({x}\right)}{{2}}}}{{2}{}{x}}\\ \text{•}& {}& {\text{Use the initial condition}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}\right)}{}\end{array}$