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Student[ODEs]

 ODESteps
 Show a step-by-step solution process for ODEs, IVPs, or systems

 Calling Sequence ODESteps(ODE) ODESteps(ODE, y(x)) ODESteps(sys)

Parameters

 ODE - an ordinary differential equation y - name ; the dependent variable x - name ; the independent variable sys - set ; an ODE system including initial values

Description

 • The ODESteps() command solves an ordinary differential equation (ODE) or system of ODEs.
 • The input may include a corresponding set of initial values, which would make it an initial value problem (IVP).
 • The output shows a series of steps in the solving process.
 • The following types of ODEs and ODE systems and/or solving methods are considered:

Examples

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

A first order ODE:

 > $\mathrm{ode1}≔{t}^{2}\left(z\left(t\right)+1\right)+{z\left(t\right)}^{2}\left(t-1\right)\left(\frac{ⅆ}{ⅆt}z\left(t\right)\right)=0$
 ${\mathrm{ode1}}{≔}{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}$ (1)
 > $\mathrm{ODESteps}\left(\mathrm{ode1}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}{=}{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}t\\ {}& {}& {\int }\frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{=}{\int }{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{\mathrm{C1}}\end{array}$ (2)

A first order IVP:

 > $\mathrm{ivp1}≔\left\{{t}^{2}\left(z\left(t\right)+1\right)+{z\left(t\right)}^{2}\left(t-1\right)\left(\frac{ⅆ}{ⅆt}z\left(t\right)\right)=0,z\left(3\right)=1\right\}$
 ${\mathrm{ivp1}}{≔}\left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}$ (3)
 > $\mathrm{ODESteps}\left(\mathrm{ivp1}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}{=}{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}t\\ {}& {}& {\int }\frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{=}{\int }{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}z{}\left(3\right)=1\\ {}& {}& {-}\frac{{1}}{{2}}{+}{\mathrm{ln}}{}\left({2}\right){=}{-}\frac{{15}}{{2}}{-}{\mathrm{ln}}{}\left({2}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=7+2{}\mathrm{ln}{}\left(2\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\end{array}$ (4)

A second order ODE:

 > $\mathrm{ode2}≔2x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)-9{x}^{2}+\left(2\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+{x}^{2}+1\right)\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)=0$
 ${\mathrm{ode2}}{≔}{2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}$ (5)
 > $\mathrm{ODESteps}\left(\mathrm{ode2}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\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{Make substitution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u=\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to reduce order of ODE}\\ {}& {}& {2}{}{x}{}{u}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{u}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({x}\right)\right){=}{0}\\ \text{▫}& {}& \text{Check if ODE is exact}\\ {}& \text{◦}& \text{ODE is exact if the lhs is the total derivative of a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{C}^{2}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{function}\\ {}& {}& \left[{}\right]{=}{0}\\ {}& \text{◦}& \text{Compute derivative of lhs}\\ {}& {}& \frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right){+}\left(\frac{{\partial }}{{\partial }{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Evaluate derivatives}\\ {}& {}& {2}{}{x}{=}{2}{}{x}\\ {}& \text{◦}& \text{Condition met, ODE is exact}\\ \text{•}& {}& \text{Exact ODE implies solution will be of this form}\\ {}& {}& \left[{F}{}\left({x}{,}{u}\right){=}{\mathrm{C1}}{,}{M}{}\left({x}{,}{u}\right){=}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right){,}{N}{}\left({x}{,}{u}\right){=}\frac{{\partial }}{{\partial }{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right)\right]\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{by integrating}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}M{}\left(x,u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {F}{}\left({x}{,}{u}\right){=}\left[{}\right]{+}{\mathrm{_F1}}{}\left({u}\right)\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& {F}{}\left({x}{,}{u}\right){=}{{x}}^{{2}}{}{u}{-}{3}{}{{x}}^{{3}}{+}{\mathrm{_F1}}{}\left({u}\right)\\ \text{•}& {}& \text{Take derivative of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u\\ {}& {}& {N}{}\left({x}{,}{u}\right){=}\frac{{\partial }}{{\partial }{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right)\\ \text{•}& {}& \text{Compute derivative}\\ {}& {}& {{x}}^{{2}}{+}{2}{}{u}{+}{1}{=}{{x}}^{{2}}{+}\frac{{ⅆ}}{{ⅆ}{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({u}\right)\\ \text{•}& {}& \text{Isolate for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆu}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{_F1}{}\left(u\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({u}\right){=}{2}{}{u}{+}{1}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(u\right)\\ {}& {}& {\mathrm{_F1}}{}\left({u}\right){=}{{u}}^{{2}}{+}{u}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into equation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,u\right)\\ {}& {}& {F}{}\left({x}{,}{u}\right){=}{{x}}^{{2}}{}{u}{-}{3}{}{{x}}^{{3}}{+}{{u}}^{{2}}{+}{u}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into the solution of the ODE}\\ {}& {}& {{x}}^{{2}}{}{u}{-}{3}{}{{x}}^{{3}}{+}{{u}}^{{2}}{+}{u}{=}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(x\right)\\ {}& {}& \left\{{u}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}{,}{u}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right\}\\ \text{•}& {}& \text{Solve 1st ODE for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(x\right)\\ {}& {}& {u}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Make substitution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u=\frac{ⅆ}{ⅆx}\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){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Integrate both sides to solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }\left({-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C2}}\\ \text{•}& {}& \text{Compute lhs}\\ {}& {}& {y}{}\left({x}\right){=}{\int }\left({-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C2}}\\ \text{•}& {}& \text{Solve 2nd ODE for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(x\right)\\ {}& {}& {u}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Make substitution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u=\frac{ⅆ}{ⅆx}\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){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Integrate both sides to solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }\left({-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C2}}\\ \text{•}& {}& \text{Compute lhs}\\ {}& {}& {y}{}\left({x}\right){=}{\int }\left({-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C2}}\end{array}$ (6)

A second order IVP:

 > $\mathrm{ivp2}≔\left\{\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)-\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)-x{ⅇ}^{x}=0,\genfrac{}{}{0}{}{\frac{ⅆ}{ⅆx}y\left(x\right)}{\phantom{x=0}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{\frac{ⅆ}{ⅆx}y\left(x\right)}}{x=0}=0,y\left(0\right)=1\right\}$
 ${\mathrm{ivp2}}{≔}\left\{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}\genfrac{}{}{0}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}{\phantom{\left\{{x}{=}{0}\right\}}}{|}\genfrac{}{}{0}{}{\phantom{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}}{\left\{{x}{=}{0}\right\}}{=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}$ (7)
 > $\mathrm{ODESteps}\left(\mathrm{ivp2}\right)$
 $\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{{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{c__1}+\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{c__1}{+}{\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{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}{=}{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}$ (8)

An Cauchy-Euler equation:

 > $\mathrm{EC}≔{x}^{2}\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)-4x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+2y\left(x\right)=0$
 ${\mathrm{EC}}{≔}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{4}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{2}{}{y}{}\left({x}\right){=}{0}$ (9)
 > $\mathrm{ODESteps}\left(\mathrm{EC}\right)$
 $\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(\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){}{\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}$ (10)

Finding a series solution:

 > $\mathrm{series_ode}≔\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)+x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+y\left(x\right)=0$
 ${\mathrm{series_ode}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{y}{}\left({x}\right){=}{0}$ (11)
 > $\mathrm{ODESteps}\left(\mathrm{series_ode}\right)$
 $\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 DE with series expansions}\\ {}& \text{◦}& \text{Convert}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\cdot \left(\frac{ⅆ}{ⅆx}\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}\\ {}& {}& \left[{}\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}{=}{2}}^{{\mathrm{\infty }}}{}{{a}}_{{k}}{}{k}{}\left({k}{-}{1}\right){}{{x}}^{{k}{-}{2}}\\ {}& \text{◦}& \text{Shift index using}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{->}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}k+2\\ {}& {}& \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{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}$ (12)

Solving an ODE with special function solution:

 > $\mathrm{Bessel_ode}≔{x}^{2}\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)+4x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+\left(25{x}^{2}-9\right)y\left(x\right)=0$
 ${\mathrm{Bessel_ode}}{≔}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{4}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({25}{}{{x}}^{{2}}{-}{9}\right){}{y}{}\left({x}\right){=}{0}$ (13)
 > $\mathrm{ODESteps}\left(\mathrm{Bessel_ode}\right)$
 $\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){+}\left({25}{}{{x}}^{{2}}{-}{9}\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{Isolate 2nd derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{\left({25}{}{{x}}^{{2}}{-}{9}\right){}{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{\left({25}{}{{x}}^{{2}}{-}{9}\right){}{y}{}\left({x}\right)}{{{x}}^{{2}}}{=}{0}\\ \text{•}& {}& \text{Simplify ODE}\\ {}& {}& {{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{25}{}{y}{}\left({x}\right){}{{x}}^{{2}}{+}{4}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{9}{}{y}{}\left({x}\right){=}{0}\\ \text{•}& {}& \text{Make a change of variables}\\ {}& {}& {t}{=}{5}{}{x}\\ \text{•}& {}& \text{Compute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\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){=}{5}{}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\\ \text{•}& {}& \text{Compute second derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{25}{}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\\ \text{•}& {}& \text{Apply change of variables to the ODE}\\ {}& {}& {{t}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right){+}{y}{}\left({t}\right){}{{t}}^{{2}}{+}{4}{}{t}{}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right){-}{9}{}{y}{}\left({t}\right){=}{0}\\ \text{•}& {}& \text{Make a change of variables}\\ {}& {}& {y}{}\left({t}\right){=}\frac{{u}{}\left({t}\right)}{{{t}}^{{3}}{{2}}}}\\ \text{•}& {}& \text{Compute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆt}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(t\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}{-}\frac{{3}{}{u}{}\left({t}\right)}{{2}{}{{t}}^{{5}}{{2}}}}{+}\frac{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)}{{{t}}^{{3}}{{2}}}}\\ \text{•}& {}& \text{Compute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{ⅆ}^{2}}{ⅆ{t}^{2}}\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){=}\frac{{15}{}{u}{}\left({t}\right)}{{4}{}{{t}}^{{7}}{{2}}}}{-}\frac{{3}{}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right)}{{{t}}^{{5}}{{2}}}}{+}\frac{\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)}{{{t}}^{{3}}{{2}}}}\\ \text{•}& {}& \text{Apply change of variables to the ODE}\\ {}& {}& {u}{}\left({t}\right){}{{t}}^{{2}}{+}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{{t}}^{{2}}{+}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{t}{-}\frac{{45}{}{u}{}\left({t}\right)}{{4}}{=}{0}\\ \text{•}& {}& \text{ODE is now of the Bessel form}\\ \text{•}& {}& \text{Solution to Bessel ODE}\\ {}& {}& {u}{}\left({t}\right){=}{\mathrm{C1}}{}{\mathrm{BesselJ}}{}\left(\frac{{3}{}\sqrt{{5}}}{{2}}{,}{t}\right){+}{\mathrm{C2}}{}{\mathrm{BesselY}}{}\left(\frac{{3}{}\sqrt{{5}}}{{2}}{,}{t}\right)\\ \text{•}& {}& \text{Make the change from}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{back to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(t\right)\\ {}& {}& {y}{}\left({t}\right){=}\frac{{\mathrm{C1}}{}{\mathrm{BesselJ}}{}\left(\frac{{3}{}\sqrt{{5}}}{{2}}{,}{t}\right){+}{\mathrm{C2}}{}{\mathrm{BesselY}}{}\left(\frac{{3}{}\sqrt{{5}}}{{2}}{,}{t}\right)}{{{t}}^{{3}}{{2}}}}\\ \text{•}& {}& \text{Make the change from}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}t\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{back to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {y}{}\left({x}\right){=}\frac{\left({\mathrm{C1}}{}{\mathrm{BesselJ}}{}\left(\frac{{3}{}\sqrt{{5}}}{{2}}{,}{5}{}{x}\right){+}{\mathrm{C2}}{}{\mathrm{BesselY}}{}\left(\frac{{3}{}\sqrt{{5}}}{{2}}{,}{5}{}{x}\right)\right){}\sqrt{{5}}}{{25}{}{{x}}^{{3}}{{2}}}}\end{array}$ (14)

Solving a third order ODE by writing it as a system:

 > $\mathrm{ode3}≔\frac{{ⅆ}^{3}}{ⅆ{x}^{3}}y\left(x\right)+3\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)+4\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+2y\left(x\right)=0$
 ${\mathrm{ode3}}{≔}\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{x}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{3}{}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{4}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{2}{}{y}{}\left({x}\right){=}{0}$ (15)
 > $\mathrm{ODESteps}\left(\mathrm{ode3}\right)$