This help page gives a few examples of using the command ODESteps to solve ordinary differential equations in terms of special functions.

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

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

$\mathrm{ode1}≔x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+4x\mathrm{diff}\left(y\left(x\right)\,x\right)+\left(25x^2-9\right)y\left(x\right)=0$

$\mathrm{ode1}≔x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+\left(25x^2-9\right)y\left(x\right)=0$

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

$\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)+\left(25x^2-9\right)y\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{\left(25x^2-9\right)y\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{\left(25x^2-9\right)y\left(x\right)}{x^2}=0\\ \text{•}& & \text{Simplify ODE}\\ & & x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+25y\left(x\right)x^2+4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-9y\left(x\right)=0\\ \text{•}& & \text{Make a change of variables}\\ & & t=5x\\ \text{•}& & \text{Compute}\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=5\frac{\ⅆ}{\ⅆt}y\left(t\right)\\ \text{•}& & \text{Compute second derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=25\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\\ \text{•}& & \text{Apply change of variables to the ODE}\\ & & t^2\left(\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\right)+y\left(t\right)t^2+4t\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)-9y\left(t\right)=0\\ \text{•}& & \text{Make a change of variables}\\ & & y\left(t\right)=\frac{u\left(t\right)}{t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ \text{•}& & \text{Compute}\frac{\ⅆ}{\ⅆt}y\left(t\right)\\ & & \frac{\ⅆ}{\ⅆt}y\left(t\right)=-\frac{3u\left(t\right)}{2t^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{\frac{\ⅆ}{\ⅆt}u\left(t\right)}{t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ \text{•}& & \text{Compute}\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\\ & & \frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)=\frac{15u\left(t\right)}{4t^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{3\left(\frac{\ⅆ}{\ⅆt}u\left(t\right)\right)}{t^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{\frac{{\ⅆ}^2}{\ⅆt^2}u\left(t\right)}{t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ \text{•}& & \text{Apply change of variables to the ODE}\\ & & u\left(t\right)t^2+\left(\frac{{\ⅆ}^2}{\ⅆt^2}u\left(t\right)\right)t^2+\left(\frac{\ⅆ}{\ⅆt}u\left(t\right)\right)t-\frac{45u\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}y\left(x\right)\text{back to}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^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ \text{•}& & \text{Make the change from}t\text{back to}x\\ & & y\left(x\right)=\frac{\left(\mathrm{C1}\mathrm{BesselJ}\left(\frac{3\sqrt{5}}{2}\,5x\right)+\mathrm{C2}\mathrm{BesselY}\left(\frac{3\sqrt{5}}{2}\,5x\right)\right)\sqrt{5}}{25x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$

$\mathrm{ode2}≔\left(-x^2+1\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-x\mathrm{diff}\left(y\left(x\right)\,x\right)+y\left(x\right)=0$

$\mathrm{ode2}≔\left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0$

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & \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{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{C1}\sin \left(\mathrm{\theta }\right)+\mathrm{C2}\cos \left(\mathrm{\theta }\right)\\ \text{•}& & \text{Revert back to}x\\ & & y\left(x\right)=\mathrm{C1}\sin \left(\arccos \left(x\right)\right)+\mathrm{C2}\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{C1}\sqrt{-x^2+1}+\mathrm{C2}x\end{array}$

$\mathrm{ode3}≔\left(-x^2+1\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-x\mathrm{diff}\left(y\left(x\right)\,x\right)+4y\left(x\right)=0$

$\mathrm{ode3}≔\left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+4y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode3}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \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)+4y\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{4y\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{4y\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)+4y\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}}+4y\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}}+4y\left(x\right)=0\\ \text{•}& & \text{Simplify ODE}\\ & & \frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)+4y\left(x\right)=0\\ \text{•}& & \text{ODE is that of a harmonic oscillator with given general solution}\\ & & y\left(\mathrm{\theta }\right)=\mathrm{C1}\sin \left(2\mathrm{\theta }\right)+\mathrm{C2}\cos \left(2\mathrm{\theta }\right)\\ \text{•}& & \text{Revert back to}x\\ & & y\left(x\right)=\mathrm{C1}\sin \left(2\arccos \left(x\right)\right)+\mathrm{C2}\cos \left(2\arccos \left(x\right)\right)\\ \text{•}& & \text{Apply double angle identities to solution}\\ & & y\left(x\right)=\mathrm{C1}\sin \left(\arccos \left(x\right)\right)\cos \left(\arccos \left(x\right)\right)+\mathrm{C2}\left(2{\cos \left(\arccos \left(x\right)\right)}^2-1\right)\\ \text{•}& & \text{Use trig identity to simplify sin}\\ & & \sin \left(\arccos \left(x\right)\right)=\sqrt{-x^2+1}\\ \text{•}& & \text{Simplify solution to the ODE}\\ & & y\left(x\right)=\mathrm{C1}x\sqrt{-x^2+1}+\mathrm{C2}\left(2x^2-1\right)\end{array}$

$\mathrm{ode4}≔\left(-x^2+1\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-x\mathrm{diff}\left(y\left(x\right)\,x\right)+9y\left(x\right)=0$

$\mathrm{ode4}≔\left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+9y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode4}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \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)+9y\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{9y\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{9y\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)+9y\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}}+9y\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}}+9y\left(x\right)=0\\ \text{•}& & \text{Simplify ODE}\\ & & \frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)+9y\left(x\right)=0\\ \text{•}& & \text{ODE is that of a harmonic oscillator with given general solution}\\ & & y\left(\mathrm{\theta }\right)=\mathrm{C1}\sin \left(3\mathrm{\theta }\right)+\mathrm{C2}\cos \left(3\mathrm{\theta }\right)\\ \text{•}& & \text{Revert back to}x\\ & & y\left(x\right)=\mathrm{C1}\sin \left(3\arccos \left(x\right)\right)+\mathrm{C2}\cos \left(3\arccos \left(x\right)\right)\end{array}$

$\mathrm{ode5}≔\left(-x^2+1\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-x\mathrm{diff}\left(y\left(x\right)\,x\right)-4y\left(x\right)=0$

$\mathrm{ode5}≔\left(-x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-4y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode5}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \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)-4y\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{4y\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{4y\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)-4y\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}}-4y\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}}-4y\left(x\right)=0\\ \text{•}& & \text{Simplify ODE}\\ & & -4y\left(x\right)+\frac{{\ⅆ}^2}{\ⅆ{\mathrm{\theta }}^2}y\left(\mathrm{\theta }\right)=0\\ \text{•}& & \text{ODE is second order linear with characteristic polynomial that is the difference of squares with given general solution}\\ & & y\left(\mathrm{\theta }\right)=\mathrm{C1}{\ⅇ}^{2\mathrm{\theta }}+\mathrm{C2}{\ⅇ}^{-2\mathrm{\theta }}\\ \text{•}& & \text{Revert back to}x\\ & & y\left(x\right)=\mathrm{C1}{\ⅇ}^{2\arccos \left(x\right)}+\mathrm{C2}{\ⅇ}^{-2\arccos \left(x\right)}\end{array}$