This help page gives a few examples of using the command ODESteps to solve first order initial value problems.

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

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

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

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

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

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

$\mathrm{ivp2}≔\left\{2xy\left(x\right)-9x^2+\left(2y\left(x\right)+x^2+1\right)\mathrm{diff}\left(y\left(x\right)\,x\right)=0\,y\left(0\right)=1\right\}$

$\mathrm{ivp2}≔\left\{2xy\left(x\right)-9x^2+\left(2y\left(x\right)+x^2+1\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=0\,y\left(0\right)=1\right\}$

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{2xy\left(x\right)-9x^2+\left(2y\left(x\right)+x^2+1\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=0\,y\left(0\right)=1\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)\\ \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\,y\right)+\left(\frac{\partial }{\partial y}F\left(x\,y\right)\right)\left(\frac{\ⅆ}{\ⅆx}y\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\,y\right)=\mathrm{\_C1}\,M\left(x\,y\right)=\frac{\partial }{\partial x}F\left(x\,y\right)\,N\left(x\,y\right)=\frac{\partial }{\partial y}F\left(x\,y\right)\right]\\ \text{•}& & \text{Solve for}F\left(x\,y\right)\text{by integrating}M\left(x\,y\right)\text{with respect to}x\\ & & F\left(x\,y\right)=\left[\right]+\mathrm{\_F1}\left(y\right)\\ \text{•}& & \text{Evaluate integral}\\ & & F\left(x\,y\right)=-3x^3+x^{2}y+\mathrm{\_F1}\left(y\right)\\ \text{•}& & \text{Take derivative of}F\left(x\,y\right)\text{with respect to}y\\ & & N\left(x\,y\right)=\frac{\partial }{\partial y}F\left(x\,y\right)\\ \text{•}& & \text{Compute derivative}\\ & & x^2+2y+1=x^2+\frac{\ⅆ}{\ⅆy}\mathrm{\_F1}\left(y\right)\\ \text{•}& & \text{Isolate for}\frac{\ⅆ}{\ⅆy}\mathrm{\_F1}\left(y\right)\\ & & \frac{\ⅆ}{\ⅆy}\mathrm{\_F1}\left(y\right)=2y+1\\ \text{•}& & \text{Solve for}\mathrm{\_F1}\left(y\right)\\ & & \mathrm{\_F1}\left(y\right)=y^2+y\\ \text{•}& & \text{Substitute}\mathrm{\_F1}\left(y\right)\text{into equation for}F\left(x\,y\right)\\ & & F\left(x\,y\right)=-3x^3+x^{2}y+y^2+y\\ \text{•}& & \text{Substitute}F\left(x\,y\right)\text{into the solution of the ODE}\\ & & -3x^3+x^{2}y+y^2+y=\mathrm{\_C1}\\ \text{•}& & \text{Solve for}y\left(x\right)\\ & & \left\{y\left(x\right)=-\frac{x^2}{2}-\frac{1}{2}-\frac{\sqrt{x^4+12x^3+2x^2+4\mathrm{\_C1}+1}}{2}\,y\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{Use initial condition}y\left(0\right)=1\\ & & 1=-\frac{1}{2}-\frac{\sqrt{4\mathrm{\_C1}+1}}{2}\\ \text{•}& & \text{Solution does not satisfy initial condition}\\ \text{•}& & \text{Use initial condition}y\left(0\right)=1\\ & & 1=-\frac{1}{2}+\frac{\sqrt{4\mathrm{\_C1}+1}}{2}\\ \text{•}& & \text{Solve for}\mathrm{\_C1}\\ & & \mathrm{\_C1}=2\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=-\frac{x^2}{2}-\frac{1}{2}+\frac{\sqrt{x^4+12x^3+2x^2+9}}{2}\end{array}$

$\mathrm{ivp3}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\right)-y\left(x\right)-x\exp \left(x\right)=0\,y\left(a\right)=b\right\}$

$\mathrm{ivp3}≔\left\{\frac{\ⅆ}{\ⅆx}y\left(x\right)-y\left(x\right)-x{\ⅇ}^x=0\,y\left(a\right)=b\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp3}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{\ⅆ}{\ⅆx}y\left(x\right)-y\left(x\right)-x{\ⅇ}^x=0\,y\left(a\right)=b\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)\\ \text{•}& & \text{Isolate the derivative}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=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}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)-y\left(x\right)=x{\ⅇ}^x\\ \text{•}& & \text{The ODE is linear; multiply by an integrating factor}\mathrm{\mu }\left(x\right)\\ & & \mathrm{\mu }\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)-y\left(x\right)\right)=\mathrm{\mu }\left(x\right)x{\ⅇ}^x\\ \text{•}& & \text{Assume the lhs of the ODE is the total derivative}\left[\right]\\ & & \mathrm{\mu }\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)-y\left(x\right)\right)=\left(\frac{\ⅆ}{\ⅆx}\mathrm{\mu }\left(x\right)\right)y\left(x\right)+\mathrm{\mu }\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\\ \text{•}& & \text{Isolate}\frac{\ⅆ}{\ⅆx}\mathrm{\mu }\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}\mathrm{\mu }\left(x\right)=-\mathrm{\mu }\left(x\right)\\ \text{•}& & \text{Solve to find the integrating factor}\\ & & \mathrm{\mu }\left(x\right)={\ⅇ}^{-x}\\ \text{•}& & \text{Integrate both sides with respect to}x\\ & & \left[\right]=\int \mathrm{\mu }\left(x\right)x{\ⅇ}^x\ⅆx+\mathrm{\_C1}\\ \text{•}& & \text{Evaluate the integral on the lhs}\\ & & \mathrm{\mu }\left(x\right)y\left(x\right)=\int \mathrm{\mu }\left(x\right)x{\ⅇ}^x\ⅆx+\mathrm{\_C1}\\ \text{•}& & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)=\frac{\int \mathrm{\mu }\left(x\right)x{\ⅇ}^x\ⅆx+\mathrm{\_C1}}{\mathrm{\mu }\left(x\right)}\\ \text{•}& & \text{Substitute}\mathrm{\mu }\left(x\right)={\ⅇ}^{-x}\\ & & y\left(x\right)=\frac{\int {\ⅇ}^{-x}x{\ⅇ}^x\ⅆx+\mathrm{\_C1}}{{\ⅇ}^{-x}}\\ \text{•}& & \text{Evaluate the integrals on the rhs}\\ & & y\left(x\right)=\frac{\frac{x^2}{2}+\mathrm{\_C1}}{{\ⅇ}^{-x}}\\ \text{•}& & \text{Simplify}\\ & & y\left(x\right)=\frac{{\ⅇ}^x\left(x^2+2\mathrm{\_C1}\right)}{2}\\ \text{•}& & \text{Use initial condition}y\left(a\right)=b\\ & & b=\frac{{\ⅇ}^a\left(a^2+2\mathrm{\_C1}\right)}{2}\\ \text{•}& & \text{Solve for}\mathrm{\_C1}\\ & & \mathrm{\_C1}=-\frac{{\ⅇ}^a{a}^2-2b}{2{\ⅇ}^a}\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=\frac{{\ⅇ}^x\left(x^2-\frac{{\ⅇ}^a{a}^2-2b}{{\ⅇ}^a}\right)}{2}\end{array}$