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.

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

$\mathrm{ivp1}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-\mathrm{diff}\left(y\left(x\right)\,x\right)-x\exp \left(x\right)=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=0\right)=0\,y\left(0\right)=1\right\}$

$\mathrm{ivp1}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-x{\ⅇ}^x=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=0\right\}}=0\,y\left(0\right)=1\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp1}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-x{\ⅇ}^x=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}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}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)+x{\ⅇ}^x\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}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{c\_\_1}{y}_1\left(x\right)+\mathrm{c\_\_2}{y}_2\left(x\right)+y_p\left(x\right)\\ \text{•}& & \text{Substitute in solutions of the homogeneous ODE}\\ & & y\left(x\right)=\mathrm{c\_\_1}+\mathrm{c\_\_2}{\ⅇ}^x+y_p\left(x\right)\\ \text{▫}& & \text{Find a particular solution}{y}_p\left(x\right)\text{of the ODE}\\ & \text{◦}& \text{Use variation of parameters to find}{y}_p\text{here}f\left(x\right)\text{is the forcing function}\\ & & \left[y_p\left(x\right)=-y_1\left(x\right)\int \frac{y_2\left(x\right)f\left(x\right)}{W\left(y_1\left(x\right)\,y_2\left(x\right)\right)}\ⅆx+y_2\left(x\right)\int \frac{y_1\left(x\right)f\left(x\right)}{W\left(y_1\left(x\right)\,y_2\left(x\right)\right)}\ⅆx\,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}{y}_p\left(x\right)\\ & & y_p\left(x\right)=-\int x{\ⅇ}^x\ⅆx+{\ⅇ}^x\int x\ⅆx\\ & \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{c\_\_1}+\mathrm{c\_\_2}{\ⅇ}^x+{\ⅇ}^x\left(1-x+\frac{1}{2}{x}^2\right)\\ \text{▫}& & \text{Check validity of solution}y\left(x\right)=\mathrm{c\_\_1}+\mathrm{c\_\_2}{\ⅇ}^x+{\ⅇ}^x\left(1-x+\frac{1}{2}{x}^2\right)\\ & \text{◦}& \text{Use initial condition}y\left(0\right)=1\\ & & 1=\mathrm{c\_\_1}+\mathrm{c\_\_2}+1\\ & \text{◦}& \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\mathrm{c\_\_2}{\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}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=0\right\}}=0\\ & & 0=\mathrm{c\_\_2}\\ & \text{◦}& \text{Solve for}\mathrm{c\_\_1}\text{and}\mathrm{c\_\_2}\\ & & \left\{\mathrm{c\_\_1}=0\,\mathrm{c\_\_2}=0\right\}\\ & \text{◦}& \text{Substitute constant values into general solution and simplify}\\ & & y\left(x\right)={\ⅇ}^x\left(1-x+\frac{1}{2}{x}^2\right)\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)={\ⅇ}^x\left(1-x+\frac{1}{2}{x}^2\right)\end{array}$

$\mathrm{ivp2}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+\frac{5{\mathrm{diff}\left(y\left(x\right)\,x\right)}^2}{y\left(x\right)}=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=1\right)=-3\,y\left(1\right)=1\right\}$

$\mathrm{ivp2}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{5{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2}{y\left(x\right)}=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=\mathrm{-3}\,y\left(1\right)=1\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp2}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{5{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2}{y\left(x\right)}=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=\mathrm{-3}\,y\left(1\right)=1\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Define new dependent variable}u\\ & & u\left(x\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ \text{•}& & \text{Compute}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}u\left(x\right)=\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Use chain rule on the lhs}\\ & & \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\left(\frac{\ⅆ}{\ⅆy}u\left(y\right)\right)=\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Substitute in the definition of}u\\ & & u\left(y\right)\left(\frac{\ⅆ}{\ⅆy}u\left(y\right)\right)=\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Make substitutions}\frac{\ⅆ}{\ⅆx}y\left(x\right)=u\left(y\right)\text{,}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=u\left(y\right)\left(\frac{\ⅆ}{\ⅆy}u\left(y\right)\right)\text{to reduce order of ODE}\\ & & u\left(y\right)\left(\frac{\ⅆ}{\ⅆy}u\left(y\right)\right)+\frac{5{u\left(y\right)}^2}{y}=0\\ \text{•}& & \text{Solve for the highest derivative}\\ & & \frac{\ⅆ}{\ⅆy}u\left(y\right)=-\frac{5u\left(y\right)}{y}\\ \text{•}& & \text{Separate variables}\\ & & \frac{\frac{\ⅆ}{\ⅆy}u\left(y\right)}{u\left(y\right)}=-\frac{5}{y}\\ \text{•}& & \text{Integrate both sides with respect to}y\\ & & \int \frac{\frac{\ⅆ}{\ⅆy}u\left(y\right)}{u\left(y\right)}\ⅆy=\int -\frac{5}{y}\ⅆy+\mathrm{c\_\_1}\\ \text{•}& & \text{Evaluate integral}\\ & & \ln \left(u\left(y\right)\right)=-5\ln \left(y\right)+\mathrm{c\_\_1}\\ \text{•}& & \text{Solve for}u\left(y\right)\\ & & u\left(y\right)=\frac{{\ⅇ}^{\mathrm{c\_\_1}}}{y^5}\\ \text{•}& & \text{Redefine the integration constant(s)}\\ & & u\left(y\right)=\frac{\mathrm{c\_\_1}}{y^5}\\ \text{•}& & \text{Solve 1st ODE for}u\left(y\right)\\ & & u\left(y\right)=\frac{\mathrm{c\_\_1}}{y^5}\\ \text{•}& & \text{Revert to original variables with substitution}u\left(y\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{,}y=y\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{\mathrm{c\_\_1}}{{y\left(x\right)}^5}\\ \text{•}& & \text{Solve for the highest derivative}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{\mathrm{c\_\_1}}{{y\left(x\right)}^5}\\ \text{•}& & \text{Separate variables}\\ & & \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right){y\left(x\right)}^5=\mathrm{c\_\_1}\\ \text{•}& & \text{Integrate both sides with respect to}x\\ & & \int \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right){y\left(x\right)}^5\ⅆx=\int \mathrm{c\_\_1}\ⅆx+\mathrm{c\_\_2}\\ \text{•}& & \text{Evaluate integral}\\ & & \frac{{y\left(x\right)}^6}{6}=\mathrm{c\_\_1}x+\mathrm{c\_\_2}\\ \text{•}& & \text{Solve for}y\left(x\right)\\ & & \left\{y\left(x\right)={\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=\left(-\frac{1}{2}-\frac{\mathrm{I}\sqrt{3}}{2}\right){\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=\left(-\frac{1}{2}+\frac{\mathrm{I}\sqrt{3}}{2}\right){\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=\left(\frac{1}{2}-\frac{\mathrm{I}\sqrt{3}}{2}\right){\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=\left(\frac{1}{2}+\frac{\mathrm{I}\sqrt{3}}{2}\right){\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=-{\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right\}\\ \text{•}& & \text{Simplify}\\ & & \left\{y\left(x\right)={\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=-{\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=-\frac{\left(1+\mathrm{I}\sqrt{3}\right){\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{2}\,y\left(x\right)=\frac{\left(1+\mathrm{I}\sqrt{3}\right){\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{2}\,y\left(x\right)=-\frac{\left(\mathrm{I}\sqrt{3}-1\right){\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{2}\,y\left(x\right)=\frac{\left(\mathrm{I}\sqrt{3}-1\right){\left(6\mathrm{c\_\_1}x+6\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{2}\right\}\\ \text{•}& & \text{Redefine the integration constant(s)}\\ & & \left\{y\left(x\right)={\left(6\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=-{\left(6\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=-\frac{\left(1+\mathrm{I}\sqrt{3}\right){\left(6\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{2}\,y\left(x\right)=\frac{\left(1+\mathrm{I}\sqrt{3}\right){\left(6\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{2}\,y\left(x\right)=-\frac{\left(\mathrm{I}\sqrt{3}-1\right){\left(6\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{2}\,y\left(x\right)=\frac{\left(\mathrm{I}\sqrt{3}-1\right){\left(6\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{2}\right\}\\ \text{•}& & \text{Redefine the integration constant(s)}\\ & & \left\{y\left(x\right)={\left(\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=\mathrm{c\_\_1}{\left(x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=\mathrm{c\_\_2}{\left(\mathrm{c\_\_1}x+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right\}\\ \text{▫}& & \text{Check validity of solution}y\left(x\right)={\left(\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ & \text{◦}& \text{Use initial condition}y\left(1\right)=1\\ & & 1={\left(\mathrm{c\_\_1}+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ & \text{◦}& \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{\mathrm{c\_\_1}}{6{\left(\mathrm{c\_\_1}x+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\\ & \text{◦}& \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=\mathrm{-3}\\ & & \mathrm{-3}=\frac{\mathrm{c\_\_1}}{6{\left(\mathrm{c\_\_1}+\mathrm{c\_\_2}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\\ & \text{◦}& \text{Solve for}\mathrm{c\_\_1}\text{and}\mathrm{c\_\_2}\\ & & \left\{\mathrm{c\_\_1}=\mathrm{-18}\,\mathrm{c\_\_2}=19\right\}\\ & \text{◦}& \text{Substitute constant values into general solution and simplify}\\ & & y\left(x\right)={\left(-18x+19\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ & \text{◦}& \text{Solution has no non-real coefficients. Ignore remaining solutions.}\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)={\left(-18x+19\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\end{array}$

$\mathrm{ivp3}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-\mathrm{diff}\left(y\left(x\right)\,x\right)-6y\left(x\right)=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=1\right)=a\,y\left(1\right)=0\right\}$

$\mathrm{ivp3}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-6y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=a\,y\left(1\right)=0\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp3}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-6y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}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}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}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(\mathrm{-2}\,3\right)\\ \text{•}& & \text{1st solution of the ODE}\\ & & y_1\left(x\right)={\ⅇ}^{-2x}\\ \text{•}& & \text{2nd solution of the ODE}\\ & & y_2\left(x\right)={\ⅇ}^{3x}\\ \text{•}& & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathrm{c\_\_1}{y}_1\left(x\right)+\mathrm{c\_\_2}{y}_2\left(x\right)\\ \text{•}& & \text{Substitute in solutions}\\ & & y\left(x\right)=\mathrm{c\_\_1}{\ⅇ}^{-2x}+\mathrm{c\_\_2}{\ⅇ}^{3x}\\ \text{▫}& & \text{Check validity of solution}y\left(x\right)=\mathrm{c\_\_1}{\ⅇ}^{-2x}+\mathrm{c\_\_2}{\ⅇ}^{3x}\\ & \text{◦}& \text{Use initial condition}y\left(1\right)=0\\ & & 0=\mathrm{c\_\_1}{\ⅇ}^{\mathrm{-2}}+\mathrm{c\_\_2}{\ⅇ}^3\\ & \text{◦}& \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=-2\mathrm{c\_\_1}{\ⅇ}^{-2x}+3\mathrm{c\_\_2}{\ⅇ}^{3x}\\ & \text{◦}& \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=a\\ & & a=-2\mathrm{c\_\_1}{\ⅇ}^{\mathrm{-2}}+3\mathrm{c\_\_2}{\ⅇ}^3\\ & \text{◦}& \text{Solve for}\mathrm{c\_\_1}\text{and}\mathrm{c\_\_2}\\ & & \left\{\mathrm{c\_\_1}=-\frac{a}{5{\ⅇ}^{\mathrm{-2}}}\,\mathrm{c\_\_2}=\frac{a}{5{\ⅇ}^3}\right\}\\ & \text{◦}& \text{Substitute constant values into general solution and simplify}\\ & & y\left(x\right)=-\frac{a\left({\ⅇ}^{2-2x}-{\ⅇ}^{-3+3x}\right)}{5}\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=-\frac{a\left({\ⅇ}^{2-2x}-{\ⅇ}^{-3+3x}\right)}{5}\end{array}$

$\mathrm{ivp4}≔\left\{x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-4x\mathrm{diff}\left(y\left(x\right)\,x\right)+2y\left(x\right)=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=1\right)=10\,y\left(1\right)=-1\right\}$

$\mathrm{ivp4}≔\left\{x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=10\,y\left(1\right)=\mathrm{-1}\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp4}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=10\,y\left(1\right)=\mathrm{-1}\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=-\frac{2y\left(x\right)}{x^2}+\frac{4\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x}\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{4\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x}+\frac{2y\left(x\right)}{x^2}=0\\ \text{•}& & \text{Multiply by denominators of the ODE}\\ & & x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2y\left(x\right)=0\\ \text{•}& & \text{Make a change of variables}\\ & & t=\ln \left(x\right)\\ \text{▫}& & \text{Substitute the change of variables back into the ODE}\\ & \text{◦}& \text{Calculate the}\text{1st}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)\left(\frac{\ⅆ}{\ⅆx}t\left(x\right)\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x}\\ & \text{◦}& \text{Calculate the}\text{2nd}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\left(\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\right){\left(\frac{\ⅆ}{\ⅆx}t\left(x\right)\right)}^2+\left(\frac{{\ⅆ}^2}{\ⅆx^2}t\left(x\right)\right)\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)}{x^2}-\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x^2}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & x^2\left(\frac{\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)}{x^2}-\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x^2}\right)-4\frac{\ⅆ}{\ⅆt}y\left(t\right)+2y\left(t\right)=0\\ \text{•}& & \text{Simplify}\\ & & \frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)-5\frac{\ⅆ}{\ⅆt}y\left(t\right)+2y\left(t\right)=0\\ \text{•}& & \text{Characteristic polynomial of ODE}\\ & & r^2-5r+2=0\\ \text{•}& & \text{Use quadratic formula to solve for}r\\ & & r=\frac{5\pm \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{c\_\_1}{y}_1\left(t\right)+\mathrm{c\_\_2}{y}_2\left(t\right)\\ \text{•}& & \text{Substitute in solutions}\\ & & y\left(t\right)=\mathrm{c\_\_1}{\ⅇ}^{\left(\frac{5}{2}-\frac{\sqrt{17}}{2}\right)t}+\mathrm{c\_\_2}{\ⅇ}^{\left(\frac{5}{2}+\frac{\sqrt{17}}{2}\right)t}\\ \text{•}& & \text{Change variables back using}t=\ln \left(x\right)\\ & & y\left(x\right)=\mathrm{c\_\_1}{\ⅇ}^{\left(\frac{5}{2}-\frac{\sqrt{17}}{2}\right)\ln \left(x\right)}+\mathrm{c\_\_2}{\ⅇ}^{\left(\frac{5}{2}+\frac{\sqrt{17}}{2}\right)\ln \left(x\right)}\\ \text{•}& & \text{Simplify}\\ & & y\left(x\right)=x^{\frac{5}{2}-\frac{\sqrt{17}}{2}}\left(x^{\sqrt{17}}\mathrm{c\_\_2}+\mathrm{c\_\_1}\right)\\ \text{▫}& & \text{Check validity of solution}y\left(x\right)=x^{\frac{5}{2}-\frac{\sqrt{17}}{2}}\left(x^{\sqrt{17}}\mathrm{c\_\_2}+\mathrm{c\_\_1}\right)\\ & \text{◦}& \text{Use initial condition}y\left(1\right)=\mathrm{-1}\\ & & \mathrm{-1}=\mathrm{c\_\_1}+\mathrm{c\_\_2}\\ & \text{◦}& \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{x^{\frac{5}{2}-\frac{\sqrt{17}}{2}}\left(\frac{5}{2}-\frac{\sqrt{17}}{2}\right)\left(x^{\sqrt{17}}\mathrm{c\_\_2}+\mathrm{c\_\_1}\right)}{x}+\frac{x^{\frac{5}{2}-\frac{\sqrt{17}}{2}}{x}^{\sqrt{17}}\sqrt{17}\mathrm{c\_\_2}}{x}\\ & \text{◦}& \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=10\\ & & 10=\left(\frac{5}{2}-\frac{\sqrt{17}}{2}\right)\left(\mathrm{c\_\_1}+\mathrm{c\_\_2}\right)+\sqrt{17}\mathrm{c\_\_2}\\ & \text{◦}& \text{Solve for}\mathrm{c\_\_1}\text{and}\mathrm{c\_\_2}\\ & & \left\{\mathrm{c\_\_1}=-\frac{1}{2}-\frac{25\sqrt{17}}{34}\,\mathrm{c\_\_2}=-\frac{1}{2}+\frac{25\sqrt{17}}{34}\right\}\\ & \text{◦}& \text{Substitute constant values into general solution and simplify}\\ & & y\left(x\right)=\frac{\left(\left(25\sqrt{17}-17\right)x^{\sqrt{17}}-25\sqrt{17}-17\right)x^{\frac{5}{2}-\frac{\sqrt{17}}{2}}}{34}\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=\frac{\left(\left(25\sqrt{17}-17\right)x^{\sqrt{17}}-25\sqrt{17}-17\right)x^{\frac{5}{2}-\frac{\sqrt{17}}{2}}}{34}\end{array}$

$\mathrm{ivp5}≔\left\{\left(-x^2+1\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-x\mathrm{diff}\left(y\left(x\right)\,x\right)+y\left(x\right)=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=2\right)=-1\,y\left(2\right)=1\right\}$

$\mathrm{ivp5}≔\left\{\left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=2\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=2\right\}}=\mathrm{-1}\,y\left(2\right)=1\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp5}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=2\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=2\right\}}=\mathrm{-1}\,y\left(2\right)=1\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{y\left(x\right)}{x^2-1}-\frac{x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x^2-1}\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x^2-1}-\frac{y\left(x\right)}{x^2-1}=0\\ \text{•}& & \text{Multiply by denominators of ODE}\\ & & \left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0\\ \text{•}& & \text{Make a change of variables}\\ & & \mathrm{\theta }=\arccos \left(x\right)\\ \text{•}& & \text{Calculate}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{with change of variables}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\left(\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)\right)\left(\frac{\ⅆ}{\ⅆx}\mathrm{\theta }\left(x\right)\right)\\ \text{•}& & \text{Compute}\text{1st}\text{derivative}\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=-\frac{\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)}{\sqrt{-x^2+1}}\\ \text{•}& & \text{Calculate}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\text{with change of variables}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\left(\frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)\right){\left(\frac{\ⅆ}{\ⅆx}\mathrm{\theta }\left(x\right)\right)}^2+\left(\frac{{\ⅆ}^2}{\ⅆx^2}\mathrm{\theta }\left(x\right)\right)\left(\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)\right)\\ \text{•}& & \text{Compute}\text{2nd}\text{derivative}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{\frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)}{-x^2+1}-\frac{x\left(\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)\right)}{{\left(-x^2+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ \text{•}& & \text{Apply the change of variables to the ODE}\\ & & \left(-x^2+1\right)\left(\frac{\frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)}{-x^2+1}-\frac{x\left(\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)\right)}{{\left(-x^2+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)+\frac{x\left(\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)\right)}{\sqrt{-x^2+1}}+y\left(x\right)=0\\ \text{•}& & \text{Multiply through}\\ & & -\frac{\left(\frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)\right)x^2}{-x^2+1}+\frac{\frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)}{-x^2+1}+\frac{x^3\left(\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)\right)}{{\left(-x^2+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{x\left(\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)\right)}{{\left(-x^2+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{x\left(\frac{\ⅆ}{\ⅆ\mathrm{\theta }}y\left(\mathrm{\theta }\right)\right)}{\sqrt{-x^2+1}}+y\left(x\right)=0\\ \text{•}& & \text{Simplify ODE}\\ & & \frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)+y\left(x\right)=0\\ \text{•}& & \text{ODE is that of a harmonic oscillator with given general solution}\\ & & y\left(\mathrm{\theta }\right)=\mathrm{c\_\_1}\sin \left(\mathrm{\theta }\right)+\mathrm{c\_\_2}\cos \left(\mathrm{\theta }\right)\\ \text{•}& & \text{Revert back to}x\\ & & y\left(x\right)=\mathrm{c\_\_1}\sin \left(\arccos \left(x\right)\right)+\mathrm{c\_\_2}\cos \left(\arccos \left(x\right)\right)\\ \text{•}& & \text{Use trig identity to simplify}\sin \left(\arccos \left(x\right)\right)\\ & & \sin \left(\arccos \left(x\right)\right)=\sqrt{-x^2+1}\\ \text{•}& & \text{Simplify solution to the ODE}\\ & & y\left(x\right)=\mathrm{c\_\_1}\sqrt{-x^2+1}+\mathrm{c\_\_2}x\\ \text{▫}& & \text{Check validity of solution}y\left(x\right)=\mathrm{c\_\_1}\sqrt{-x^2+1}+\mathrm{c\_\_2}x\\ & \text{◦}& \text{Use initial condition}y\left(2\right)=1\\ & & 1=\mathrm{c\_\_1}\sqrt{\mathrm{-3}}+2\mathrm{c\_\_2}\\ & \text{◦}& \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=-\frac{\mathrm{c\_\_1}x}{\sqrt{-x^2+1}}+\mathrm{c\_\_2}\\ & \text{◦}& \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=2\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=2\right\}}=\mathrm{-1}\\ & & \mathrm{-1}=\frac{2\mathrm{c\_\_1}\sqrt{\mathrm{-3}}}{3}+\mathrm{c\_\_2}\\ & \text{◦}& \text{Solve for}\mathrm{c\_\_1}\text{and}\mathrm{c\_\_2}\\ & & \left\{\mathrm{c\_\_1}=3\sqrt{\mathrm{-3}}\,\mathrm{c\_\_2}=5\right\}\\ & \text{◦}& \text{Substitute constant values into general solution and simplify}\\ & & y\left(x\right)=3\mathrm{I}\sqrt{3}\sqrt{-x^2+1}+5x\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=3\mathrm{I}\sqrt{3}\sqrt{-x^2+1}+5x\end{array}$