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}\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)\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{3\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)}{x}-\frac{6\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x^2}-\frac{6y\left(x\right)}{x^3}=0\\ \text{•}& & \text{Multiply by denominators of the ODE}\\ & & 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{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{◦}& \text{Calculate the}\text{3rd}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)=\left(\frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)\right){\left(\frac{\ⅆ}{\ⅆx}t\left(x\right)\right)}^3+3\left(\frac{\ⅆ}{\ⅆx}t\left(x\right)\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}t\left(x\right)\right)\left(\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\right)+\left(\frac{{\ⅆ}^3}{\ⅆx^3}t\left(x\right)\right)\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)=\frac{\frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)}{x^3}-\frac{3\left(\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\right)}{x^3}+\frac{2\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}{x^3}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & x^3\left(\frac{\frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)}{x^3}-\frac{3\left(\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\right)}{x^3}+\frac{2\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}{x^3}\right)+3x^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)-6\frac{\ⅆ}{\ⅆt}y\left(t\right)-6y\left(t\right)=0\\ \text{•}& & \text{Simplify}\\ & & \frac{{\ⅆ}^3}{\ⅆt^3}y\left(t\right)-7\frac{\ⅆ}{\ⅆt}y\left(t\right)-6y\left(t\right)=0\\ \text{•}& & \text{Characteristic polynomial of ODE}\\ & & r^3-7r-6=0\\ \text{•}& & \text{Roots of the characteristic polynomial}\\ & & r=\left[3\,\mathrm{-2}\,\mathrm{-1}\right]\\ \text{•}& & \text{Solution from}r=3\\ & & y_1\left(t\right)={\ⅇ}^{3t}\\ \text{•}& & \text{Solution from}r=\mathrm{-2}\\ & & y_2\left(t\right)={\ⅇ}^{-2t}\\ \text{•}& & \text{Solution from}r=\mathrm{-1}\\ & & y_3\left(t\right)={\ⅇ}^{-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)+\mathrm{c\_\_3}{y}_3\left(t\right)\\ \text{•}& & \text{Substitute in solutions and simplify}\\ & & y\left(t\right)=\left(\mathrm{c\_\_1}{\ⅇ}^{5t}+\mathrm{c\_\_3}{\ⅇ}^t+\mathrm{c\_\_2}\right){\ⅇ}^{-2t}\\ \text{•}& & \text{Change variables back using}t=\ln \left(x\right)\\ & & y\left(x\right)=\frac{\mathrm{c\_\_1}{x}^5+\mathrm{c\_\_3}x+\mathrm{c\_\_2}}{x^2}\end{array}$