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:

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

$\mathrm{ode1}≔t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\mathrm{diff}\left(z\left(t\right)\,t\right)=0$

$\mathrm{ode1}≔t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=0$

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{\ⅆ}{\ⅆt}z\left(t\right)\\ \text{•}& & \text{Separate variables}\\ & & \frac{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right){z\left(t\right)}^2}{z\left(t\right)+1}=-\frac{t^2}{t-1}\\ \text{•}& & \text{Integrate both sides with respect to}t\\ & & \int \frac{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right){z\left(t\right)}^2}{z\left(t\right)+1}\ⅆt=\int -\frac{t^2}{t-1}\ⅆt+\mathrm{C1}\\ \text{•}& & \text{Evaluate integral}\\ & & \frac{{z\left(t\right)}^2}{2}-z\left(t\right)+\ln \left(z\left(t\right)+1\right)=-\frac{t^2}{2}-t-\ln \left(t-1\right)+\mathrm{C1}\end{array}$

$\mathrm{ivp1}≔\left\{t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\mathrm{diff}\left(z\left(t\right)\,t\right)=0\,z\left(3\right)=1\right\}$

$\mathrm{ivp1}≔\left\{t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=0\,z\left(3\right)=1\right\}$

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

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

$\mathrm{ode2}≔2x\mathrm{diff}\left(y\left(x\right)\,x\right)-9x^2+\left(2\mathrm{diff}\left(y\left(x\right)\,x\right)+x^2+1\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)=0$

$\mathrm{ode2}≔2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-9x^2+\left(2\frac{\ⅆ}{\ⅆx}y\left(x\right)+x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)=0$

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & 2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-9x^2+\left(2\frac{\ⅆ}{\ⅆx}y\left(x\right)+x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\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{Make substitution}u=\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{to reduce order of ODE}\\ & & 2xu\left(x\right)-9x^2+\left(2u\left(x\right)+x^2+1\right)\left(\frac{\ⅆ}{\ⅆx}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}{C}^2\text{function}\\ & & \frac{\ⅆ}{\ⅆx}F\left(x\,u\left(x\right)\right)=0\\ & \text{◦}& \text{Compute derivative of lhs}\\ & & \frac{\partial }{\partial x}F\left(x\,u\right)+\left(\frac{\partial }{\partial u}F\left(x\,u\right)\right)\left(\frac{\ⅆ}{\ⅆx}u\left(x\right)\right)=0\\ & \text{◦}& \text{Evaluate derivatives}\\ & & 2x=2x\\ & \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}F\left(x\,u\right)\,N\left(x\,u\right)=\frac{\partial }{\partial u}F\left(x\,u\right)\right]\\ \text{•}& & \text{Solve for}F\left(x\,u\right)\text{by integrating}M\left(x\,u\right)\text{with respect to}x\\ & & F\left(x\,u\right)=\int \left(2xu-9x^2\right)\ⅆx+\mathrm{\_F1}\left(u\right)\\ \text{•}& & \text{Evaluate integral}\\ & & F\left(x\,u\right)=x^{2}u-3x^3+\mathrm{\_F1}\left(u\right)\\ \text{•}& & \text{Take derivative of}F\left(x\,u\right)\text{with respect to}u\\ & & N\left(x\,u\right)=\frac{\partial }{\partial u}F\left(x\,u\right)\\ \text{•}& & \text{Compute derivative}\\ & & x^2+2u+1=x^2+\frac{\ⅆ}{\ⅆu}\mathrm{\_F1}\left(u\right)\\ \text{•}& & \text{Isolate for}\frac{\ⅆ}{\ⅆu}\mathrm{\_F1}\left(u\right)\\ & & \frac{\ⅆ}{\ⅆu}\mathrm{\_F1}\left(u\right)=2u+1\\ \text{•}& & \text{Solve for}\mathrm{\_F1}\left(u\right)\\ & & \mathrm{\_F1}\left(u\right)=u^2+u\\ \text{•}& & \text{Substitute}\mathrm{\_F1}\left(u\right)\text{into equation for}F\left(x\,u\right)\\ & & F\left(x\,u\right)=x^{2}u-3x^3+u^2+u\\ \text{•}& & \text{Substitute}F\left(x\,u\right)\text{into the solution of the ODE}\\ & & x^{2}u-3x^3+u^2+u=\mathrm{C1}\\ \text{•}& & \text{Solve for}u\left(x\right)\\ & & \left\{u\left(x\right)=-\frac{x^2}{2}-\frac{1}{2}-\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\,u\left(x\right)=-\frac{x^2}{2}-\frac{1}{2}+\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\right\}\\ \text{•}& & \text{Solve 1st ODE for}u\left(x\right)\\ & & u\left(x\right)=-\frac{x^2}{2}-\frac{1}{2}-\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\\ \text{•}& & \text{Make substitution}u=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=-\frac{x^2}{2}-\frac{1}{2}-\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\\ \text{•}& & \text{Integrate both sides to solve for}y\left(x\right)\\ & & \int \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\ⅆx=\int \left(-\frac{x^2}{2}-\frac{1}{2}-\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\right)\ⅆx+\mathrm{C2}\\ \text{•}& & \text{Compute lhs}\\ & & y\left(x\right)=\int \left(-\frac{x^2}{2}-\frac{1}{2}-\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\right)\ⅆx+\mathrm{C2}\\ \text{•}& & \text{Solve 2nd ODE for}u\left(x\right)\\ & & u\left(x\right)=-\frac{x^2}{2}-\frac{1}{2}+\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\\ \text{•}& & \text{Make substitution}u=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=-\frac{x^2}{2}-\frac{1}{2}+\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\\ \text{•}& & \text{Integrate both sides to solve for}y\left(x\right)\\ & & \int \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\ⅆx=\int \left(-\frac{x^2}{2}-\frac{1}{2}+\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\right)\ⅆx+\mathrm{C2}\\ \text{•}& & \text{Compute lhs}\\ & & y\left(x\right)=\int \left(-\frac{x^2}{2}-\frac{1}{2}+\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{C1}+1}}{2}\right)\ⅆx+\mathrm{C2}\end{array}$

$\mathrm{ivp2}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-\mathrm{diff}\left(y\left(x\right)\,x\right)-x\exp \left(x\right)=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=0\right)=0\,y\left(0\right)=1\right\}$

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

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

$\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{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}{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)\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)}\ⅆx\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)}\ⅆx\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}{y}_p\left(x\right)\\ & & y_p\left(x\right)=-\left(\int x{\ⅇ}^x\ⅆx\right)+{\ⅇ}^x\left(\int x\ⅆx\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}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}y\left(0\right)=1\\ & & 1=\mathrm{c\_\_1}+\mathrm{\_C2}+1\\ & \text{◦}& \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}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}\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{\_C2}\\ & \text{◦}& \text{Solve for}\mathrm{c\_\_1}\text{and}\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^2-2x+2\right)}{2}\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}\end{array}$

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

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

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

$\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{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)\ln \left(x\right)}+\mathrm{C2}{\ⅇ}^{\left(\frac{5}{2}+\frac{\sqrt{17}}{2}\right)\ln \left(x\right)}\\ \text{•}& & \text{Simplify}\\ & & y\left(x\right)=x^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(x^{-\frac{\sqrt{17}}{2}}\mathrm{C1}+x^{\frac{\sqrt{17}}{2}}\mathrm{C2}\right)\end{array}$

$\mathrm{series\_ode}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x\mathrm{diff}\left(y\left(x\right)\,x\right)+y\left(x\right)=0$

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

$\mathrm{ODESteps}\left(\mathrm{series\_ode}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\\ \text{▫}& & \text{Rewrite DE with series expansions}\\ & \text{◦}& \text{Convert}x\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}k{x}^k\\ & \text{◦}& \text{Convert}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\text{to series expansion}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\sum _{k=2}^{\mathrm{\infty }}{a}_{k}k\left(k-1\right)x^{k-2}\\ & \text{◦}& \text{Shift index using}k\text{->}k+2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k+2}\left(k+2\right)\left(k+1\right)x^k\\ & & \text{Rewrite DE with series expansions}\\ & & \sum _{k=0}^{\mathrm{\infty }}\left(a_{k+2}\left(k+2\right)\left(k+1\right)+a_k\left(k+1\right)\right)x^k=0\\ \text{•}& & \text{Each term in the series must be 0, giving the recursion relation}\\ & & \left(k+1\right)\left(a_{k+2}\left(k+2\right)+a_k\right)=0\\ \text{•}& & \text{Recursion relation that defines the series solution to the ODE}\\ & & \left[y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\,a_{k+2}=-\frac{a_k}{k+2}\right]\end{array}$

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

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

$\mathrm{ODESteps}\left(\mathrm{Bessel\_ode}\right)$

$\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}$