This help page gives a few examples of using the command ODESteps to solve second order ordinary differential equations.

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

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

$\mathrm{ode1}≔2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-9x^2\+\left(2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\+x^2\+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)\=0$

$\mathrm{ode1}≔2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-9x^2+\left(2\frac{\ⅆ}{\ⅆx}y\left(x\right)+x^2+1\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)=0$

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

$\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)=-\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}$

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

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

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-x{\ⅇ}^x=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{\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)\end{array}$

$\mathrm{ode3}≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\+\frac{5{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2}{y\left(x\right)}\=0$

$\mathrm{ode3}≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{5{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2}{y\left(x\right)}=0$

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{5{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2}{y\left(x\right)}=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Define new dependent variable}u\\ & & u\left(x\right)=\left[\right]\\ \text{•}& & \text{Compute}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ & & \left[\right]=\left[\right]\\ \text{•}& & \text{Use chain rule on the lhs}\\ & & \left(\left[\right]\right)\left(\left[\right]\right)=\left[\right]\\ \text{•}& & \text{Substitute in the definition of}u\\ & & u\left(y\right)\left(\left[\right]\right)=\left[\right]\\ \text{•}& & \text{Make substitutions}\frac{\ⅆ}{\ⅆx}y\left(x\right)=u\left(y\right)\text{,}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=u\left(y\right)\left(\frac{\ⅆ}{\ⅆy}u\left(y\right)\right)\text{to reduce order of ODE}\\ & & u\left(y\right)\left(\frac{\ⅆ}{\ⅆy}u\left(y\right)\right)+\frac{5{u\left(y\right)}^2}{y}=0\\ \text{•}& & \text{Separate variables}\\ & & \frac{\frac{\ⅆ}{\ⅆy}u\left(y\right)}{u\left(y\right)}=-\frac{5}{y}\\ \text{•}& & \text{Integrate both sides with respect to}y\\ & & \int \frac{\frac{\ⅆ}{\ⅆy}u\left(y\right)}{u\left(y\right)}\ⅆy=\int -\frac{5}{y}\ⅆy+\mathrm{C1}\\ \text{•}& & \text{Evaluate integral}\\ & & \ln \left(u\left(y\right)\right)=-5\ln \left(y\right)+\mathrm{C1}\\ \text{•}& & \text{Solve for}u\left(y\right)\\ & & u\left(y\right)=\frac{{\ⅇ}^{\mathrm{C1}}}{y^5}\\ \text{•}& & \text{Solve 1st ODE for}u\left(y\right)\\ & & u\left(y\right)=\frac{{\ⅇ}^{\mathrm{C1}}}{y^5}\\ \text{•}& & \text{Revert to original variables with substitution}u\left(y\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{,}y=y\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{{\ⅇ}^{\mathrm{C1}}}{{y\left(x\right)}^5}\\ \text{•}& & \text{Separate variables}\\ & & \left(\frac{\ⅆ}{\ⅆ}\right)\end{array}$