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

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

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

$\mathrm{high\_order\_ode1}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\,x\right)+3\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+4\mathrm{diff}\left(y\left(x\right)\,x\right)+2y\left(x\right)=0$

$\mathrm{high\_order\_ode1}≔\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)+3\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+4\frac{\ⅆ}{\ⅆx}y\left(x\right)+2y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{high\_order\_ode1}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)+3\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+4\frac{\ⅆ}{\ⅆx}y\left(x\right)+2y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}3\\ & & \frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\\ \text{▫}& & \text{Convert linear ODE into a system of first order ODEs}\\ & \text{◦}& \text{Define new variable}{y}_1\left(x\right)\\ & & y_1\left(x\right)=y\left(x\right)\\ & \text{◦}& \text{Define new variable}{y}_2\left(x\right)\\ & & y_2\left(x\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & \text{◦}& \text{Define new variable}{y}_3\left(x\right)\\ & & y_3\left(x\right)=\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ & \text{◦}& \text{Isolate for}\frac{\ⅆ}{\ⅆx}{y}_3\left(x\right)\text{using original ODE}\\ & & \frac{\ⅆ}{\ⅆx}{y}_3\left(x\right)=-3y_3\left(x\right)-4y_2\left(x\right)-2y_1\left(x\right)\\ & & \text{Convert linear ODE into a system of first order ODEs}\\ & & \left[y_2\left(x\right)=\frac{\ⅆ}{\ⅆx}{y}_1\left(x\right)\,y_3\left(x\right)=\frac{\ⅆ}{\ⅆx}{y}_2\left(x\right)\,\frac{\ⅆ}{\ⅆx}{y}_3\left(x\right)=-3y_3\left(x\right)-4y_2\left(x\right)-2y_1\left(x\right)\right]\\ \text{•}& & \text{Define vector}\\ & & \overrightarrow{y}\left(x\right)=\left[\begin{array}{c}{y}_3\left(x\right)\\ y_1\left(x\right)\\ y_2\left(x\right)\end{array}\right]\\ \text{•}& & \text{System to solve}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=A·\overrightarrow{y}\left(x\right)\\ \text{•}& & \text{To solve the system find eigenvalues and eigenvectors of}A\\ & & A=\left[\begin{array}{ccc}\mathrm{-3}& \mathrm{-2}& \mathrm{-4}\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]\\ \text{•}& & \text{Eigenpairs of A}\\ & & \left[\left[\mathrm{-1}\,\left[\begin{array}{c}\mathrm{-1}\\ \mathrm{-1}\\ 1\end{array}\right]\right]\,\left[\mathrm{-1}+\mathrm{I}\,\left[\begin{array}{c}\mathrm{-1}+\mathrm{I}\\ -\frac{1}{2}-\frac{\mathrm{I}}{2}\\ 1\end{array}\right]\right]\,\left[\mathrm{-1}-\mathrm{I}\,\left[\begin{array}{c}\mathrm{-1}-\mathrm{I}\\ -\frac{1}{2}+\frac{\mathrm{I}}{2}\\ 1\end{array}\right]\right]\right]\\ \text{•}& & \text{Consider eigenpair}\\ & & \left[\mathrm{-1}\,\left[\begin{array}{c}\mathrm{-1}\\ \mathrm{-1}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution to homogeneous system from eigenpair}\\ & & {\overrightarrow{y}}_1\left(x\right)=\left[\right]\\ \text{•}& & \text{Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\\ & & \left[\mathrm{-1}+\mathrm{I}\,\left[\begin{array}{c}\mathrm{-1}+\mathrm{I}\\ -\frac{1}{2}-\frac{\mathrm{I}}{2}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution from eigenpair}\\ & & \left[\right]\\ \text{•}& & \text{Use Euler identity to write solution in terms of sin and cos}\\ & & \left[\right]\\ \text{•}& & \text{Simplify expression}\\ & & \left[\right]\\ \text{•}& & \text{Both real and imaginary parts are solutions to the homogeneous system}\\ & & \left[{\overrightarrow{y}}_2\left(x\right)=\left[\right]\,{\overrightarrow{y}}_3\left(x\right)=\left[\right]\right]\\ \text{•}& & \text{General solution to the system of ODEs}\\ & & \overrightarrow{y}\left(x\right)=\mathrm{\_C1}{\overrightarrow{y}}_1\left(x\right)+\mathrm{\_C2}{\overrightarrow{y}}_2\left(x\right)+\mathrm{\_C3}{\overrightarrow{y}}_3\left(x\right)\\ \text{•}& & \text{Substitute solutions into the general solution}\\ & & \overrightarrow{y}\left(x\right)=\left[\right]+\left[\right]+\left[\right]\\ \text{•}& & \text{First component of the vector is the solution to the ODE}\\ & & y\left(x\right)=-{\ⅇ}^{-x}\left(\left(\mathrm{\_C2}+\mathrm{\_C3}\right)\sin \left(x\right)+\left(\mathrm{\_C2}-\mathrm{\_C3}\right)\cos \left(x\right)+\mathrm{\_C1}\right)\end{array}$

$\mathrm{macro}\left(Y=\overrightarrow{y}\right)\:$

$\mathrm{sys2}≔\mathrm{diff}\left(Y\left(x\right)\,x\right)=\mathrm{Matrix}\left(\left[\left[7\,1\right]\,\left[-4\,3\right]\right]\right)·Y\left(x\right)$

$\mathrm{sys2}≔\frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=\left[\begin{array}{cc}7& 1\\ \mathrm{-4}& 3\end{array}\right]·\overrightarrow{y}\left(x\right)$

$\mathrm{ODESteps}\left(\mathrm{sys2}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=\left[\begin{array}{cc}7& 1\\ \mathrm{-4}& 3\end{array}\right]·\overrightarrow{y}\left(x\right)\\ \text{•}& & \text{System to solve}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=A·\overrightarrow{y}\left(x\right)\\ \text{•}& & \text{To solve the system find eigenvalues and eigenvectors of}A\\ & & A=\left[\begin{array}{cc}7& 1\\ \mathrm{-4}& 3\end{array}\right]\\ \text{•}& & \text{Eigenpairs of A}\\ & & \left[\left[5\,\left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right]\right]\,\left[5\,\left[\begin{array}{c}0\\ 0\end{array}\right]\right]\right]\\ \text{•}& & \text{Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\\ & & \left[5\,\left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{First solution from eigenvalue}5\\ & & {\overrightarrow{y}}_1\left(x\right)=\left[\right]\\ \text{•}& & \text{Form of the 2nd homogeneous solution where}\overrightarrow{p}\text{is to be solved for,}\mathrm{\lambda }=5\text{is the eigenvalue, and}\overrightarrow{v}\text{is the eigenvector}\\ & & {\overrightarrow{y}}_2\left(x\right)={\ⅇ}^{\mathrm{\lambda }x}\left(x\overrightarrow{v}+\overrightarrow{p}+\overrightarrow{v}\right)\\ \text{•}& & \text{Note that the}x\text{multiplying}\overrightarrow{v}\text{makes this solution linearly independent to the 1st solution obtained from}\mathrm{\lambda }=5\\ \text{•}& & \text{substitute}{\overrightarrow{y}}_2\left(x\right)\text{into the system}\\ & & \mathrm{\lambda }{\ⅇ}^{\mathrm{\lambda }x}\left(x\overrightarrow{v}+\overrightarrow{p}+\overrightarrow{v}\right)+{\ⅇ}^{\mathrm{\lambda }x}\overrightarrow{v}=\left({\ⅇ}^{\mathrm{\lambda }x}A\right)·\left(x\overrightarrow{v}+\overrightarrow{p}+\overrightarrow{v}\right)\\ \text{•}& & \text{Use the fact that}\overrightarrow{v}\text{is an eigenvector of}A\\ & & \mathrm{\lambda }{\ⅇ}^{\mathrm{\lambda }x}\left(x\overrightarrow{v}+\overrightarrow{p}+\overrightarrow{v}\right)+{\ⅇ}^{\mathrm{\lambda }x}\overrightarrow{v}={\ⅇ}^{\mathrm{\lambda }x}\left(\mathrm{\lambda }\overrightarrow{v}+\mathrm{\lambda }x\overrightarrow{v}+A·\overrightarrow{p}\right)\\ \text{•}& & \text{Simplify equation}\\ & & \mathrm{\lambda }\overrightarrow{p}+\overrightarrow{v}=A·\overrightarrow{p}\\ \text{•}& & \text{Make use of the identity matrix}I\\ & & \left(\mathrm{\lambda }I\right)·\overrightarrow{p}+\overrightarrow{v}=A·\overrightarrow{p}\\ \text{•}& & \text{Condition}\overrightarrow{p}\text{must meet for}{\overrightarrow{y}}_2\left(x\right)\text{to be a solution to the system}\\ & & \left(-\mathrm{\lambda }I+A\right)·\overrightarrow{p}=\overrightarrow{v}\\ \text{•}& & \text{Choose}\overrightarrow{p}\text{to use in the second solution to the system from eigenvalue}5\\ & & \left(\left[\right]\right)·\overrightarrow{p}=\left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right]\\ \text{•}& & \text{Choice of}\overrightarrow{p}\\ & & \overrightarrow{p}=\left[\begin{array}{c}-\frac{1}{4}\\ 0\end{array}\right]\\ \text{•}& & \text{second solution from eigenvalue}5\\ & & {\overrightarrow{y}}_2\left(x\right)=\left[\right]\\ \text{•}& & \text{General solution to the system of ODEs}\\ & & \overrightarrow{y}\left(x\right)=\mathrm{\_C1}{\overrightarrow{y}}_1\left(x\right)+\mathrm{\_C2}{\overrightarrow{y}}_2\left(x\right)\\ \text{•}& & \text{Substitute solutions into the general solution}\\ & & \overrightarrow{y}\left(x\right)=\left[\right]+\left[\right]\end{array}$

$\mathrm{sys3}≔⟨\mathrm{diff}\left(y\left[1\right]\left(x\right)\,x\right)\,\mathrm{diff}\left(y\left[2\right]\left(x\right)\,x\right)⟩=\mathrm{Matrix}\left(\left[\left[1\,2\right]\,\left[3\,2\right]\right]\right)·⟨y\left[1\right]\left(x\right)\,y\left[2\right]\left(x\right)⟩+⟨1\,\exp \left(x\right)⟩$

$\mathrm{sys3}≔\left[\begin{array}{c}\frac{\ⅆ}{\ⅆx}{y}_1\left(x\right)\\ \frac{\ⅆ}{\ⅆx}{y}_2\left(x\right)\end{array}\right]=\left[\begin{array}{c}{y}_1\left(x\right)+2y_2\left(x\right)+1\\ 3y_1\left(x\right)+2y_2\left(x\right)+{\ⅇ}^x\end{array}\right]$

$\mathrm{ODESteps}\left(\mathrm{sys3}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left[\begin{array}{c}\frac{\ⅆ}{\ⅆx}{y}_1\left(x\right)\\ \frac{\ⅆ}{\ⅆx}{y}_2\left(x\right)\end{array}\right]=\left[\begin{array}{c}{y}_1\left(x\right)+2y_2\left(x\right)+1\\ 3y_1\left(x\right)+2y_2\left(x\right)+{\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{Define vector}\\ & & \overrightarrow{y}\left(x\right)=\left[\begin{array}{c}{y}_1\left(x\right)\\ y_2\left(x\right)\end{array}\right]\\ \text{•}& & \text{System to solve}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]·\overrightarrow{y}\left(x\right)+\left[\begin{array}{c}1\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{Define forcing function}\\ & & \overrightarrow{f}\left(x\right)=\left[\begin{array}{c}1\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{To solve the system find eigenvalues and eigenvectors of}A\\ & & A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\\ \text{•}& & \text{Eigenpairs of A}\\ & & \left[\left[4\,\left[\begin{array}{c}\frac{2}{3}\\ 1\end{array}\right]\right]\,\left[\mathrm{-1}\,\left[\begin{array}{c}\mathrm{-1}\\ 1\end{array}\right]\right]\right]\\ \text{•}& & \text{Consider eigenpair}\\ & & \left[4\,\left[\begin{array}{c}\frac{2}{3}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution to homogeneous system from eigenpair}\\ & & {\overrightarrow{y}}_1\left(x\right)=\left[\right]\\ \text{•}& & \text{Consider eigenpair}\\ & & \left[\mathrm{-1}\,\left[\begin{array}{c}\mathrm{-1}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution to homogeneous system from eigenpair}\\ & & {\overrightarrow{y}}_2\left(x\right)=\left[\right]\\ \text{•}& & \text{General solution to the system of ODEs where}{\overrightarrow{y}}_p\left(x\right)\text{is a particular solution to the system}\\ & & \overrightarrow{y}\left(x\right)=\mathrm{\_C1}{\overrightarrow{y}}_1\left(x\right)+\mathrm{\_C2}{\overrightarrow{y}}_2\left(x\right)+{\overrightarrow{y}}_p\left(x\right)\\ \text{▫}& & \text{Fundamental matrix}\\ & \text{◦}& \text{The fundamental matrix is defined as the matrix with columns that are the solutions to the homogeneous system with initial condition being the basis vector in the standard basis at}x=0\\ & & \mathrm{\Phi }\left(x\right)\\ & \text{◦}& \text{Compute the solution to the homogeneous system for each basis vector of the standard basis being the inital condition at}x=0\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=A·\overrightarrow{y}\left(x\right)\\ & \text{◦}& \text{Let}B\text{be the matrix with solutions to the homogeneous system as columns evaluated at}x=0\\ & & B=\left[\begin{array}{cc}\frac{2}{3}& \mathrm{-1}\\ 1& 1\end{array}\right]\\ & \text{◦}& \text{For each basis vector we need to solve the system of linear equations for}{\overrightarrow{C}}_j\text{which contains the values to multiply each homogeneous solution by to get the basis vector as the intital condition}\\ & & B·{\overrightarrow{C}}_j={\hat{e}}_j\\ & \text{◦}& \text{Equation which must be executed for each basis vector}\\ & & {\overrightarrow{C}}_j=\left[\right]·{\hat{e}}_j\\ & \text{◦}& \text{Compute}{\overrightarrow{C}}_j\text{for j =}1\\ & & {\overrightarrow{C}}_1=\left[\begin{array}{c}\frac{3}{5}\\ -\frac{3}{5}\end{array}\right]\\ & \text{◦}& \text{1st column of the fundamental matrix}\\ & & \left[\right]+\left[\right]=\left[\begin{array}{c}\frac{3{\ⅇ}^{-x}}{5}+\frac{2{\ⅇ}^{4x}}{5}\\ -\frac{3{\ⅇ}^{-x}}{5}+\frac{3{\ⅇ}^{4x}}{5}\end{array}\right]\\ & \text{◦}& \text{Compute}{\overrightarrow{C}}_j\text{for j =}2\\ & & {\overrightarrow{C}}_2=\left[\begin{array}{c}\frac{3}{5}\\ \frac{2}{5}\end{array}\right]\\ & \text{◦}& \text{2nd column of the fundamental matrix}\\ & & \left[\right]+\left[\right]=\left[\begin{array}{c}-\frac{2{\ⅇ}^{-x}}{5}+\frac{2{\ⅇ}^{4x}}{5}\\ \frac{2{\ⅇ}^{-x}}{5}+\frac{3{\ⅇ}^{4x}}{5}\end{array}\right]\\ & & \text{Fundamental matrix}\\ & & \mathrm{\Phi }\left(x\right)=\left[\begin{array}{cc}\frac{3{\ⅇ}^{-x}}{5}+\frac{2{\ⅇ}^{4x}}{5}& -\frac{2{\ⅇ}^{-x}}{5}+\frac{2{\ⅇ}^{4x}}{5}\\ -\frac{3{\ⅇ}^{-x}}{5}+\frac{3{\ⅇ}^{4x}}{5}& \frac{2{\ⅇ}^{-x}}{5}+\frac{3{\ⅇ}^{4x}}{5}\end{array}\right]\\ \text{▫}& & \text{Find a particular solution to the system of ODEs using variation of paramaters}\\ & \text{◦}& \text{Let the particular solution be the fundamental matrix multiplied by}\overrightarrow{v}\left(x\right)\text{and solve for}\overrightarrow{v}\left(x\right)\\ & & {\overrightarrow{y}}_x\left(x\right)=\mathrm{\Phi }\left(x\right)·\overrightarrow{v}\left(x\right)\\ & \text{◦}& \text{Take the derivative of the particular solution}\\ & & \frac{\ⅆ}{\ⅆx}{\overrightarrow{y}}_x\left(x\right)=\left(\frac{\ⅆ}{\ⅆx}\mathrm{\Phi }\left(x\right)\right)·\overrightarrow{v}\left(x\right)+\mathrm{\Phi }\left(x\right)·\left(\frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)\right)\\ & \text{◦}& \text{Substitute particular solution and it's derivative into the system of ODEs}\\ & & \left(\frac{\ⅆ}{\ⅆx}\mathrm{\Phi }\left(x\right)\right)·\overrightarrow{v}\left(x\right)+\mathrm{\Phi }\left(x\right)·\left(\frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)\right)=A·\mathrm{\Phi }\left(x\right)·\overrightarrow{v}\left(x\right)+\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{The fundamental matrix has columns that are solutions to the homogeneous system so it's derivative follows that of the homogeneous system}\\ & & A·\mathrm{\Phi }\left(x\right)·\overrightarrow{v}\left(x\right)+\mathrm{\Phi }\left(x\right)·\left(\frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)\right)=A·\mathrm{\Phi }\left(x\right)·\overrightarrow{v}\left(x\right)+\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{Cancel like terms}\\ & & \mathrm{\Phi }\left(x\right)·\left(\frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)\right)=\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{Multiply by the inverse of the fundamental matrix}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)=\left[\right]·\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{Integrate to solve for}\overrightarrow{v}\left(x\right)\\ & & \overrightarrow{v}\left(x\right)={\int }_0^x\left[\right]·\overrightarrow{f}\left(s\right)\ⅆs\\ & \text{◦}& \text{Plug}\overrightarrow{v}\left(x\right)\text{into the equation for the particular solution}\\ & & {\overrightarrow{y}}_x\left(x\right)=\mathrm{\Phi }\left(x\right)·\left({\int }_0^x\left[\right]·\overrightarrow{f}\left(s\right)\ⅆs\right)\\ & \text{◦}& \text{Plug in the fundamental matrix and the forcing function and compute}\\ & & {\overrightarrow{y}}_x\left(x\right)=\left[\begin{array}{c}\frac{7{\ⅇ}^{4x}}{30}-\frac{{\ⅇ}^x}{3}+\frac{1}{2}-\frac{2{\ⅇ}^{-x}}{5}\\ \frac{7{\ⅇ}^{4x}}{20}-\frac{3}{4}+\frac{2{\ⅇ}^{-x}}{5}\end{array}\right]\\ & & \text{Find a particular solution to the system of ODEs using variation of paramaters}\\ & & {\overrightarrow{y}}_x\left(x\right)=\left[\begin{array}{c}\frac{7{\ⅇ}^{4x}}{30}-\frac{{\ⅇ}^x}{3}+\frac{1}{2}-\frac{2{\ⅇ}^{-x}}{5}\\ \frac{7{\ⅇ}^{4x}}{20}-\frac{3}{4}+\frac{2{\ⅇ}^{-x}}{5}\end{array}\right]\\ \text{•}& & \text{Plug particular solution back into general solution}\\ & & \overrightarrow{y}\left(x\right)=\mathrm{\_C1}{\overrightarrow{y}}_1\left(x\right)+\mathrm{\_C2}{\overrightarrow{y}}_2\left(x\right)+\left[\begin{array}{c}\frac{7{\ⅇ}^{4x}}{30}-\frac{{\ⅇ}^x}{3}+\frac{1}{2}-\frac{2{\ⅇ}^{-x}}{5}\\ \frac{7{\ⅇ}^{4x}}{20}-\frac{3}{4}+\frac{2{\ⅇ}^{-x}}{5}\end{array}\right]\\ \text{•}& & \text{Solution to the system of ODEs}\\ & & \left[\begin{array}{c}{y}_1\left(x\right)\\ y_2\left(x\right)\end{array}\right]=\left[\begin{array}{c}\frac{\left(-30\mathrm{\_C2}-12\right){\ⅇ}^{-x}}{30}+\frac{\left(20\mathrm{\_C1}+7\right){\ⅇ}^{4x}}{30}-\frac{{\ⅇ}^x}{3}+\frac{1}{2}\\ \frac{\left(20\mathrm{\_C2}+8\right){\ⅇ}^{-x}}{20}-\frac{3}{4}+\frac{\left(20\mathrm{\_C1}+7\right){\ⅇ}^{4x}}{20}\end{array}\right]\end{array}$

$\mathrm{sys4}≔\left\{\mathrm{diff}\left(y\left[1\right]\left(x\right)\,x\right)=y\left[1\right]\left(x\right)+2y\left[2\right]\left(x\right)\,\mathrm{diff}\left(y\left[2\right]\left(x\right)\,x\right)=3y\left[1\right]\left(x\right)+2y\left[2\right]\left(x\right)+\exp \left(x\right)\right\}$

$\mathrm{sys4}≔\left\{\frac{\ⅆ}{\ⅆx}{y}_1\left(x\right)=y_1\left(x\right)+2y_2\left(x\right)\,\frac{\ⅆ}{\ⅆx}{y}_2\left(x\right)=3y_1\left(x\right)+2y_2\left(x\right)+{\ⅇ}^x\right\}$

$\mathrm{ODESteps}\left(\mathrm{sys4}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{\ⅆ}{\ⅆx}{y}_1\left(x\right)=y_1\left(x\right)+2y_2\left(x\right)\,\frac{\ⅆ}{\ⅆx}{y}_2\left(x\right)=3y_1\left(x\right)+2y_2\left(x\right)+{\ⅇ}^x\right\}\\ \text{•}& & \text{Define vector}\\ & & \overrightarrow{y}\left(x\right)=\left[\begin{array}{c}{y}_1\left(x\right)\\ y_2\left(x\right)\end{array}\right]\\ \text{•}& & \text{Convert system into a vector equation}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=\left[\right]+\left[\begin{array}{c}0\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{System to solve}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]·\overrightarrow{y}\left(x\right)+\left[\begin{array}{c}0\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{Define forcing function}\\ & & \overrightarrow{f}\left(x\right)=\left[\begin{array}{c}0\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{To solve the system find eigenvalues and eigenvectors of}A\\ & & A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\\ \text{•}& & \text{Eigenpairs of A}\\ & & \left[\left[\mathrm{-1}\,\left[\begin{array}{c}\mathrm{-1}\\ 1\end{array}\right]\right]\,\left[4\,\left[\begin{array}{c}\frac{2}{3}\\ 1\end{array}\right]\right]\right]\\ \text{•}& & \text{Consider eigenpair}\\ & & \left[\mathrm{-1}\,\left[\begin{array}{c}\mathrm{-1}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution to homogeneous system from eigenpair}\\ & & {\overrightarrow{y}}_1\left(x\right)=\left[\right]\\ \text{•}& & \text{Consider eigenpair}\\ & & \left[4\,\left[\begin{array}{c}\frac{2}{3}\\ 1\end{array}\right]\right]\\ \text{•}& & \text{Solution to homogeneous system from eigenpair}\\ & & {\overrightarrow{y}}_2\left(x\right)=\left[\right]\\ \text{•}& & \text{General solution to the system of ODEs where}{\overrightarrow{y}}_p\left(x\right)\text{is a particular solution to the system}\\ & & \overrightarrow{y}\left(x\right)=\mathrm{\_C1}{\overrightarrow{y}}_1\left(x\right)+\mathrm{\_C2}{\overrightarrow{y}}_2\left(x\right)+{\overrightarrow{y}}_p\left(x\right)\\ \text{▫}& & \text{Fundamental matrix}\\ & \text{◦}& \text{The fundamental matrix is defined as the matrix with columns that are the solutions to the homogeneous system with initial condition being the basis vector in the standard basis at}x=0\\ & & \mathrm{\Phi }\left(x\right)\\ & \text{◦}& \text{Compute the solution to the homogeneous system for each basis vector of the standard basis being the inital condition at}x=0\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=A·\overrightarrow{y}\left(x\right)\\ & \text{◦}& \text{Let}B\text{be the matrix with solutions to the homogeneous system as columns evaluated at}x=0\\ & & B=\left[\begin{array}{cc}\mathrm{-1}& \frac{2}{3}\\ 1& 1\end{array}\right]\\ & \text{◦}& \text{For each basis vector we need to solve the system of linear equations for}{\overrightarrow{C}}_j\text{which contains the values to multiply each homogeneous solution by to get the basis vector as the intital condition}\\ & & B·{\overrightarrow{C}}_j={\hat{e}}_j\\ & \text{◦}& \text{Equation which must be executed for each basis vector}\\ & & {\overrightarrow{C}}_j=\left[\right]·{\hat{e}}_j\\ & \text{◦}& \text{Compute}{\overrightarrow{C}}_j\text{for j =}1\\ & & {\overrightarrow{C}}_1=\left[\begin{array}{c}-\frac{3}{5}\\ \frac{3}{5}\end{array}\right]\\ & \text{◦}& \text{1st column of the fundamental matrix}\\ & & \left[\right]+\left[\right]=\left[\begin{array}{c}\frac{3{\ⅇ}^{-x}}{5}+\frac{2{\ⅇ}^{4x}}{5}\\ -\frac{3{\ⅇ}^{-x}}{5}+\frac{3{\ⅇ}^{4x}}{5}\end{array}\right]\\ & \text{◦}& \text{Compute}{\overrightarrow{C}}_j\text{for j =}2\\ & & {\overrightarrow{C}}_2=\left[\begin{array}{c}\frac{2}{5}\\ \frac{3}{5}\end{array}\right]\\ & \text{◦}& \text{2nd column of the fundamental matrix}\\ & & \left[\right]+\left[\right]=\left[\begin{array}{c}-\frac{2{\ⅇ}^{-x}}{5}+\frac{2{\ⅇ}^{4x}}{5}\\ \frac{2{\ⅇ}^{-x}}{5}+\frac{3{\ⅇ}^{4x}}{5}\end{array}\right]\\ & & \text{Fundamental matrix}\\ & & \mathrm{\Phi }\left(x\right)=\left[\begin{array}{cc}\frac{3{\ⅇ}^{-x}}{5}+\frac{2{\ⅇ}^{4x}}{5}& -\frac{2{\ⅇ}^{-x}}{5}+\frac{2{\ⅇ}^{4x}}{5}\\ -\frac{3{\ⅇ}^{-x}}{5}+\frac{3{\ⅇ}^{4x}}{5}& \frac{2{\ⅇ}^{-x}}{5}+\frac{3{\ⅇ}^{4x}}{5}\end{array}\right]\\ \text{▫}& & \text{Find a particular solution to the system of ODEs using variation of paramaters}\\ & \text{◦}& \text{Let the particular solution be the fundamental matrix multiplied by}\overrightarrow{v}\left(x\right)\text{and solve for}\overrightarrow{v}\left(x\right)\\ & & {\overrightarrow{y}}_x\left(x\right)=\mathrm{\Phi }\left(x\right)·\overrightarrow{v}\left(x\right)\\ & \text{◦}& \text{Take the derivative of the particular solution}\\ & & \frac{\ⅆ}{\ⅆx}{\overrightarrow{y}}_x\left(x\right)=\left(\frac{\ⅆ}{\ⅆx}\mathrm{\Phi }\left(x\right)\right)·\overrightarrow{v}\left(x\right)+\mathrm{\Phi }\left(x\right)·\left(\frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)\right)\\ & \text{◦}& \text{Substitute particular solution and it's derivative into the system of ODEs}\\ & & \left(\frac{\ⅆ}{\ⅆx}\mathrm{\Phi }\left(x\right)\right)·\overrightarrow{v}\left(x\right)+\mathrm{\Phi }\left(x\right)·\left(\frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)\right)=A·\mathrm{\Phi }\left(x\right)·\overrightarrow{v}\left(x\right)+\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{The fundamental matrix has columns that are solutions to the homogeneous system so it's derivative follows that of the homogeneous system}\\ & & A·\mathrm{\Phi }\left(x\right)·\overrightarrow{v}\left(x\right)+\mathrm{\Phi }\left(x\right)·\left(\frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)\right)=A·\mathrm{\Phi }\left(x\right)·\overrightarrow{v}\left(x\right)+\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{Cancel like terms}\\ & & \mathrm{\Phi }\left(x\right)·\left(\frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)\right)=\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{Multiply by the inverse of the fundamental matrix}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{v}\left(x\right)=\left[\right]·\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{Integrate to solve for}\overrightarrow{v}\left(x\right)\\ & & \overrightarrow{v}\left(x\right)={\int }_0^x\left[\right]\end{array}$