This help page gives a few examples of using the command ODESteps to solve Cauchy-Euler equations.

See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.

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

$\mathrm{ode1}≔x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-4x\mathrm{diff}\left(y\left(x\right)\,x\right)+2y\left(x\right)=0$

$\mathrm{ode1}≔x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode1}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & 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{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 1st derivative of}y\text{with respect to}x\text{, using the chain rule}\\ & & \frac{\ⅆ}{\ⅆt}y\left(t\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{\ⅆ}{\ⅆt}y\left(t\right)=\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x}\\ & \text{◦}& \text{Calculate the 2nd derivative of}y\text{with respect to}x\text{, using the chain rule}\\ & & \frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)=\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)=-\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x^2}+\frac{\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)}{x^2}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & x^2\left(-\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x^2}+\frac{\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)}{x^2}\right)-4\frac{\ⅆ}{\ⅆt}y\left(t\right)+2y\left(t\right)=0\\ \text{•}& & \text{Simplify}\\ & & -5\frac{\ⅆ}{\ⅆt}y\left(t\right)+\frac{{\ⅆ}^2}{\ⅆt^2}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(\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}t=\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}}\end{array}$

$\mathrm{ode2}≔x^3\mathrm{diff}\left(y\left(x\right)\,x\,x\,x\right)+3x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-6x\mathrm{diff}\left(y\left(x\right)\,x\right)-6y\left(x\right)=0$

$\mathrm{ode2}≔x^3\left(\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\right)+3x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-6x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-6y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode2}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & x^3\left(\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\right)+3x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-6x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-6y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}3\\ & & \frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\\ \text{•}& & \text{Isolate 3rd derivative}\\ & & \frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)=\frac{6y\left(x\right)}{x^3}-\frac{3\left(\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)x-2\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x^2}\\ \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}^3}{\ⅆx^3}y\left(x\right)-\frac{6y\left(x\right)}{x^3}+\frac{3\left(\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)x-2\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x^2}=0\\ \text{•}& & \text{Multiply by denominators of the ODE}\\ & & x^3\left(\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\right)-6y\left(x\right)+3x\left(\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)x-2\frac{\ⅆ}{\ⅆx}y\left(x\right)\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 1st derivative of}y\text{with respect to}x\text{, using the chain rule}\\ & & \frac{\ⅆ}{\ⅆt}y\left(t\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{\ⅆ}{\ⅆt}y\left(t\right)=\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x}\\ & \text{◦}& \text{Calculate the 2nd derivative of}y\text{with respect to}x\text{, using the chain rule}\\ & & \frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)=\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)=-\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x^2}+\frac{\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)}{x^2}\\ & \text{◦}& \text{Calculate the 3rd derivative of}y\text{with respect to}x\text{, using the chain rule}\\ & & \frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)=\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)=\frac{2\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}{x^3}-\frac{3\left(\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\right)}{x^3}+\frac{\frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)}{x^3}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & x^3\left(\frac{2\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}{x^3}-\frac{3\left(\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\right)}{x^3}+\frac{\frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)}{x^3}\right)-6y\left(t\right)+3x\left(\left(-\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x^2}+\frac{\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)}{x^2}\right)x-\frac{2\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}{x}\right)=0\\ \text{•}& & \text{Simplify}\\ & & -7\frac{\ⅆ}{\ⅆt}y\left(t\right)+\frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)-6y\left(t\right)=0\\ \text{▫}& & \text{Convert linear ODE into a system of first order ODEs}\\ & \text{◦}& \text{Define new variable}{y}_1\left(t\right)\\ & & y_1\left(t\right)=y\left(t\right)\\ & \text{◦}& \text{Define new variable}{y}_2\left(t\right)\\ & & y_2\left(t\right)=\frac{\ⅆ}{\ⅆt}y\left(t\right)\\ & \text{◦}& \text{Define new variable}{y}_3\left(t\right)\\ & & y_3\left(t\right)=\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\\ & \text{◦}& \text{Isolate for}\frac{\ⅆ}{\ⅆt}{y}_3\left(t\right)\text{using original ODE}\\ & & \frac{\ⅆ}{\ⅆt}{y}_3\left(t\right)=7y_2\left(t\right)+6y_1\left(t\right)\\ & & \text{Convert linear ODE into a system of first order ODEs}\\ & & \left[y_2\left(t\right)=\frac{\ⅆ}{\ⅆt}{y}_1\left(t\right)\,y_3\left(t\right)=\frac{\ⅆ}{\ⅆt}{y}_2\left(t\right)\,\frac{\ⅆ}{\ⅆt}{y}_3\left(t\right)=7y_2\left(t\right)+6y_1\left(t\right)\right]\\ \text{•}& & \text{Define vector}\\ & & \overrightarrow{y}\left(t\right)=\left[\begin{array}{c}{y}_3\left(t\right)\\ y_1\left(t\right)\\ y_2\left(t\right)\end{array}\right]\\ \text{•}& & \text{System to solve}\\ & & \frac{\ⅆ}{\ⅆt}\overrightarrow{y}\left(t\right)=A·\overrightarrow{y}\left(t\right)\\ \text{•}& & \text{To solve the system find eigenvalues and eigenvectors of}A\\ & & A=\left[\begin{array}{ccc}0& 6& 7\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]\\ \text{•}& & \text{Eigenpairs of A}\\ & & \left[\left[\mathrm{-1}\,\left[\begin{array}{c}\mathrm{-1}\\ \mathrm{-1}\\ 1\end{array}\right]\right]\,\left[3\,\left[\begin{array}{c}3\\ \frac{1}{3}\\ 1\end{array}\right]\right]\,\left[\mathrm{-2}\,\left[\begin{array}{c}\mathrm{-2}\\ -\frac{1}{2}\\ 1\end{array}\right]\right]\right]\\ \text{•}& & \text{Consider eigenpair}\\ & & \left[\mathrm{-1}\,\left[\begin{array}{c}\mathrm{-1}\\ \mathrm{-1}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution to homogeneous system from eigenpair}\\ & & {\overrightarrow{y}}_1\left(t\right)=\left[\right]\\ \text{•}& & \text{Consider eigenpair}\\ & & \left[3\,\left[\begin{array}{c}3\\ \frac{1}{3}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution to homogeneous system from eigenpair}\\ & & {\overrightarrow{y}}_2\left(t\right)=\left[\right]\\ \text{•}& & \text{Consider eigenpair}\\ & & \left[\mathrm{-2}\,\left[\begin{array}{c}\mathrm{-2}\\ -\frac{1}{2}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution to homogeneous system from eigenpair}\\ & & {\overrightarrow{y}}_3\left(t\right)=\left[\right]\\ \text{•}& & \text{General solution to the system of ODEs}\\ & & \overrightarrow{y}\left(t\right)=\mathrm{\_C1}{\overrightarrow{y}}_1\left(t\right)+\mathrm{\_C2}{\overrightarrow{y}}_2\left(t\right)+\mathrm{\_C3}{\overrightarrow{y}}_3\left(t\right)\\ \text{•}& & \text{Substitute solutions into the general solution}\\ & & \overrightarrow{y}\left(t\right)=\left[\right]+\left[\right]+\left[\right]\\ \text{•}& & \text{First component of the vector is the solution to the ODE}\\ & & y\left(t\right)=-\left(-3\mathrm{\_C2}{\ⅇ}^{5t}+\mathrm{\_C1}{\ⅇ}^t+2\mathrm{\_C3}\right){\ⅇ}^{-2t}\\ \text{•}& & \text{Change variables back using}t=\ln \left(x\right)\\ & & y\left(x\right)=-\left(-3\mathrm{\_C2}{\ⅇ}^{5\left[\right]}+\mathrm{\_C1}{\ⅇ}^{\left[\right]}+2\mathrm{\_C3}\right){\ⅇ}^{-2\left[\right]}\\ \text{•}& & \text{Simplify}\\ & & y\left(x\right)=3\mathrm{\_C2}{x}^3-\frac{\mathrm{\_C1}}{x}-\frac{2\mathrm{\_C3}}{x^2}\end{array}$