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:

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

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

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

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{\ⅆ}{\ⅆt}z\left(t\right)\\ \text{•}& & \text{Separate variables}\\ & & \frac{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right){z\left(t\right)}^2}{z\left(t\right)+1}=-\frac{t^2}{t-1}\\ \text{•}& & \text{Integrate both sides with respect to}t\\ & & \int \frac{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right){z\left(t\right)}^2}{z\left(t\right)+1}\ⅆt=\int -\frac{t^2}{t-1}\ⅆt+\mathrm{\_C1}\\ \text{•}& & \text{Evaluate integral}\\ & & \frac{{z\left(t\right)}^2}{2}-z\left(t\right)+\ln \left(z\left(t\right)+1\right)=-\frac{t^2}{2}-t-\ln \left(t-1\right)+\mathrm{\_C1}\end{array}$

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

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

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

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

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

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

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

$\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}\\ & & \left[\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)=\left[\right]+\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)=\end{array}$