Second Order IVPs - 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

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
 > $\mathrm{ivp1}≔\left\{\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)-\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)-x{ⅇ}^{x}=0,\genfrac{}{}{0}{}{\frac{ⅆ}{ⅆx}y\left(x\right)}{\phantom{x=0}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{\frac{ⅆ}{ⅆx}y\left(x\right)}}{x=0}=0,y\left(0\right)=1\right\}$
 ${\mathrm{ivp1}}{≔}\left\{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{0}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{0}\right\}}{=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}$ (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 parameters 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){=}{{ⅇ}}^{{x}}{}\left({1}{-}{x}{+}\frac{{1}}{{2}}{}{{x}}^{{2}}\right)\\ \text{•}& {}& \text{Substitute particular solution into general solution to ODE}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{C1}}{+}{\mathrm{C2}}{}{{ⅇ}}^{{x}}{+}{{ⅇ}}^{{x}}{}\left({1}{-}{x}{+}\frac{{1}}{{2}}{}{{x}}^{{2}}\right)\\ \text{▫}& {}& \text{Check validity of solution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)=\mathrm{_C1}+\mathrm{_C2}{}{ⅇ}^{x}+{ⅇ}^{x}{}\left(1-x+\frac{1}{2}{}{x}^{2}\right)\\ {}& \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}}{+}{{ⅇ}}^{{x}}{}\left({1}{-}{x}{+}\frac{{1}}{{2}}{}{{x}}^{{2}}\right){+}\left({x}{-}{1}\right){}{{ⅇ}}^{{x}}\\ {}& \text{◦}& \text{Use the initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\genfrac{}{}{0}{}{\left(\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0}{}{\phantom{\left(\frac{ⅆ}{ⅆx}\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{Substitute constant values into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{2}\right)}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{-}{2}{}{x}{+}{2}\right)}{{2}}\end{array}$ (2)
 > $\mathrm{ivp2}≔\left\{\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)+\frac{5{\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)}^{2}}{y\left(x\right)}=0,\genfrac{}{}{0}{}{\frac{ⅆ}{ⅆx}y\left(x\right)}{\phantom{x=1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{\frac{ⅆ}{ⅆx}y\left(x\right)}}{x=1}=-3,y\left(1\right)=1\right\}$
 ${\mathrm{ivp2}}{≔}\left\{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{5}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{2}}}{{y}{}\left({x}\right)}{=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{-3}{,}{y}{}\left({1}\right){=}{1}\right\}$ (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}}u\\ {}& {}& {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}{ⅆx}\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}{ⅆy}\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}}u{}\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}{ⅆx}\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{Check validity of solution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)={\left(6{}{ⅇ}^{\mathrm{c__1}}{}x+6{}\mathrm{_C2}\right)}^{1}{6}}\\ {}& \text{◦}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(1\right)=1\\ {}& {}& {1}{=}{\left({6}{}{{ⅇ}}^{\mathrm{c__1}}{+}{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{c__1}}}{{\left({6}{}{{ⅇ}}^{\mathrm{c__1}}{}{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}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0}{}{\phantom{\left(\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)}}{\left\{x=1\right\}}=-3\\ {}& {}& {-3}{=}\frac{{{ⅇ}}^{\mathrm{c__1}}}{{\left({6}{}{{ⅇ}}^{\mathrm{c__1}}{+}{6}{}{\mathrm{_C2}}\right)}^{{5}}{{6}}}}\\ {}& \text{◦}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{c__1}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{and}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C2}\\ {}& {}& \left\{\mathrm{c__1}{=}{\mathrm{ln}}{}\left({3}\right){+}{I}{}{\mathrm{\pi }}{,}{\mathrm{_C2}}{=}\frac{{19}}{{6}}\right\}\\ {}& \text{◦}& \text{Substitute constant values into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}{\left({-}{18}{}{x}{+}{19}\right)}^{{1}}{{6}}}\\ \text{▫}& {}& \text{Check validity of solution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)=-{\left(6{}{ⅇ}^{\mathrm{c__1}}{}x+6{}\mathrm{_C2}\right)}^{1}{6}}\\ {}& \text{◦}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(1\right)=1\\ {}& {}& {1}{=}{-}{\left({6}{}{{ⅇ}}^{\mathrm{c__1}}{+}{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{c__1}}}{{\left({6}{}{{ⅇ}}^{\mathrm{c__1}}{}{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}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0}{}{\phantom{\left(\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)}}{\left\{x=1\right\}}=-3\\ {}& {}& {-3}{=}{-}\frac{{{ⅇ}}^{\mathrm{c__1}}}{{\left({6}{}{{ⅇ}}^{\mathrm{c__1}}{+}{6}{}{\mathrm{_C2}}\right)}^{{5}}{{6}}}}\\ {}& \text{◦}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{c__1}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{and}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C2}\\ {}& \text{◦}& \text{The solution does not satisfy the initial conditions}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}{\left({-}{18}{}{x}{+}{19}\right)}^{{1}}{{6}}}\end{array}$ (4)
 > $\mathrm{ivp3}≔\left\{\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)-\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)-6y\left(x\right)=0,\genfrac{}{}{0}{}{\frac{ⅆ}{ⅆx}y\left(x\right)}{\phantom{x=1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{\frac{ⅆ}{ⅆx}y\left(x\right)}}{x=1}=a,y\left(1\right)=0\right\}$
 ${\mathrm{ivp3}}{≔}\left\{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{6}{}{y}{}\left({x}\right){=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{a}{,}{y}{}\left({1}\right){=}{0}\right\}$ (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{Check validity of solution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)=\mathrm{c__1}{}{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{c__1}{}{{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{c__1}{}{{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}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0}{}{\phantom{\left(\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\right)}}{\left\{x=1\right\}}=a\\ {}& {}& {a}{=}{-}{2}{}\mathrm{c__1}{}{{ⅇ}}^{{-2}}{+}{3}{}{\mathrm{_C2}}{}{{ⅇ}}^{{3}}\\ {}& \text{◦}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{c__1}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{and}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C2}\\ {}& {}& \left\{\mathrm{c__1}{=}{-}\frac{{a}}{{5}{}{{ⅇ}}^{{-2}}}{,}{\mathrm{_C2}}{=}\frac{{a}}{{5}{}{{ⅇ}}^{{3}}}\right\}\\ {}& \text{◦}& \text{Substitute constant values into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{a}{}\left({{ⅇ}}^{{2}{-}{2}{}{x}}{-}{{ⅇ}}^{{-}{3}{+}{3}{}{x}}\right)}{{5}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{a}{}\left({{ⅇ}}^{{2}{-}{2}{}{x}}{-}{{ⅇ}}^{{-}{3}{+}{3}{}{x}}\right)}{{5}}\end{array}$ (6)
 > $\mathrm{ivp4}≔\left\{{x}^{2}\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)-4x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+2y\left(x\right)=0,\genfrac{}{}{0}{}{\frac{ⅆ}{ⅆx}y\left(x\right)}{\phantom{x=1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{\frac{ⅆ}{ⅆx}y\left(x\right)}}{x=1}=10,y\left(1\right)=-1\right\}$
 ${\mathrm{ivp4}}{≔}\left\{{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{4}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{2}{}{y}{}\left({x}\right){=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{1}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{1}\right\}}{=}{10}{,}{y}{}\left({1}\right){=}{-1}\right\}$ (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}{}\end{array}$