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.

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

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

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

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)+3\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+4\frac{\ⅆ}{\ⅆx}y\left(x\right)+2y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}3\\ & & \frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\\ \text{•}& & \text{Characteristic polynomial of ODE}\\ & & r^3+3r^2+4r+2=0\\ \text{•}& & \text{Roots of the characteristic polynomial}\\ & & r=\left[\mathrm{-1}\,\mathrm{-1}-\mathrm{I}\,\mathrm{-1}+\mathrm{I}\right]\\ \text{•}& & \text{Solution from}r=\mathrm{-1}\\ & & y_1\left(x\right)={\ⅇ}^{-x}\\ \text{•}& & \text{Solutions from}r=\mathrm{-1}-\mathrm{I}\text{and}r=\mathrm{-1}+\mathrm{I}\\ & & \left[y_2\left(x\right)={\ⅇ}^{-x}\sin \left(x\right)\,y_3\left(x\right)={\ⅇ}^{-x}\cos \left(x\right)\right]\\ \text{•}& & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathrm{c\_\_1}{y}_1\left(x\right)+\mathrm{c\_\_2}{y}_2\left(x\right)+\mathrm{c\_\_3}{y}_3\left(x\right)\\ \text{•}& & \text{Substitute in solutions and simplify}\\ & & y\left(x\right)={\ⅇ}^{-x}\left(\mathrm{c\_\_1}+\mathrm{c\_\_2}\sin \left(x\right)+\mathrm{c\_\_3}\cos \left(x\right)\right)\end{array}$

$\mathrm{macro}\left(Y=⟨y\left[1\right]\left(x\right)\,y\left[2\right]\left(x\right)⟩\right)\:$

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

$\mathrm{sys2}≔\left[\begin{array}{c}\frac{\ⅆ}{\ⅆx}{y}_1\left(x\right)\\ \frac{\ⅆ}{\ⅆx}{y}_2\left(x\right)\end{array}\right]=\left[\begin{array}{cc}7& 1\\ \mathrm{-4}& 3\end{array}\right]·\left[\begin{array}{c}{y}_1\left(x\right)\\ y_2\left(x\right)\end{array}\right]$

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

$\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}{cc}7& 1\\ \mathrm{-4}& 3\end{array}\right]·\left[\begin{array}{c}{y}_1\left(x\right)\\ y_2\left(x\right)\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}7& 1\\ \mathrm{-4}& 3\end{array}\right]·\overrightarrow{y}\left(x\right)\\ \text{•}& & \text{Define the coefficient matrix}\\ & & A=\left[\begin{array}{cc}7& 1\\ \mathrm{-4}& 3\end{array}\right]\\ \text{•}& & \text{Rewrite the system as}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=A·\overrightarrow{y}\left(x\right)\\ \text{•}& & \text{To solve the system, find the eigenvalues and eigenvectors of}A\\ \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)={\ⅇ}^{5x}\cdot \left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\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}\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 homogeneous system}\\ & & \mathrm{\lambda }{\ⅇ}^{\mathrm{\lambda }x}\left(x\overrightarrow{v}+\overrightarrow{p}\right)+{\ⅇ}^{\mathrm{\lambda }x}\overrightarrow{v}=\left({\ⅇ}^{\mathrm{\lambda }x}A\right)·\left(x\overrightarrow{v}+\overrightarrow{p}\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}\right)+{\ⅇ}^{\mathrm{\lambda }x}\overrightarrow{v}={\ⅇ}^{\mathrm{\lambda }x}\left(\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}\mathrm{I}\\ & & \left(\mathrm{\lambda }\cdot 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 homogeneous system}\\ & & \left(A-\mathrm{\lambda }\cdot I\right)·\overrightarrow{p}=\overrightarrow{v}\\ \text{•}& & \text{Choose}\overrightarrow{p}\text{to use in the second solution to the homogeneous system from eigenvalue}5\\ & & \left(\left[\begin{array}{cc}7& 1\\ \mathrm{-4}& 3\end{array}\right]-5\cdot \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\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)={\ⅇ}^{5x}\cdot \left(x\cdot \left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right]+\left[\begin{array}{c}-\frac{1}{4}\\ 0\end{array}\right]\right)\\ \text{•}& & \text{General solution to the system of ODEs}\\ & & \overrightarrow{y}=\mathrm{c\_\_1}{\overrightarrow{y}}_1\left(x\right)+\mathrm{c\_\_2}{\overrightarrow{y}}_2\left(x\right)\\ \text{•}& & \text{Substitute solutions into the general solution}\\ & & \overrightarrow{y}=\mathrm{c\_\_1}{\ⅇ}^{5x}\cdot \left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right]+\mathrm{c\_\_2}{\ⅇ}^{5x}\cdot \left(x\cdot \left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right]+\left[\begin{array}{c}-\frac{1}{4}\\ 0\end{array}\right]\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{{\ⅇ}^{5x}\left(2\mathrm{c\_\_2}x+2\mathrm{c\_\_1}+\mathrm{c\_\_2}\right)}{4}\\ {\ⅇ}^{5x}\left(\mathrm{c\_\_2}x+\mathrm{c\_\_1}\right)\end{array}\right]\end{array}$

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

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

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

$\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 the forcing function}\\ & & \overrightarrow{f}\left(x\right)=\left[\begin{array}{c}1\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{Define the coefficient matrix}\\ & & A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\\ \text{•}& & \text{Rewrite the system as}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{y}\left(x\right)=A·\overrightarrow{y}\left(x\right)+\overrightarrow{f}\\ \text{•}& & \text{To solve the system, find the eigenvalues and eigenvectors of}A\\ \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={\ⅇ}^{-x}\cdot \left[\begin{array}{c}\mathrm{-1}\\ 1\end{array}\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={\ⅇ}^{4x}\cdot \left[\begin{array}{c}\frac{2}{3}\\ 1\end{array}\right]\\ \text{•}& & \text{General solution of the system of ODEs can be written in terms of the particular solution}{\overrightarrow{y}}_p\left(x\right)\\ & & \overrightarrow{y}\left(x\right)=\mathrm{c\_\_1}{\overrightarrow{y}}_1+\mathrm{c\_\_2}{\overrightarrow{y}}_2+{\overrightarrow{y}}_p\left(x\right)\\ \text{▫}& & \text{Fundamental matrix}\\ & \text{◦}& \text{Let}\mathrm{\phi }\left(x\right)\text{be the matrix whose columns are the independent solutions of the homogeneous system.}\\ & & \mathrm{\phi }\left(x\right)=\left[\begin{array}{cc}-{\ⅇ}^{-x}& \frac{2{\ⅇ}^{4x}}{3}\\ {\ⅇ}^{-x}& {\ⅇ}^{4x}\end{array}\right]\\ & \text{◦}& \text{The fundamental matrix,}\mathrm{\Phi }\left(x\right)\text{is a normalized version of}\mathrm{\phi }\left(x\right)\text{satisfying}\mathrm{\Phi }\left(0\right)=I\text{where}I\text{is the identity matrix}\\ & & \mathrm{\Phi }\left(x\right)=\mathrm{\phi }\left(x\right)·{\mathrm{\phi }\left(0\right)}^{\mathrm{-1}}\\ & \text{◦}& \text{Substitute the value of}\mathrm{\phi }\left(x\right)\text{and}\mathrm{\phi }\left(0\right)\\ & & \mathrm{\Phi }\left(x\right)=\left[\begin{array}{cc}-{\ⅇ}^{-x}& \frac{2{\ⅇ}^{4x}}{3}\\ {\ⅇ}^{-x}& {\ⅇ}^{4x}\end{array}\right]·{\left[\begin{array}{cc}\mathrm{-1}& \frac{2}{3}\\ 1& 1\end{array}\right]}^{\mathrm{-1}}\\ & \text{◦}& \text{Evaluate and simplify to get the 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 of the system of ODEs using variation of parameters}\\ & \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}}_p\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}}_p\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 its 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 its 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)={\mathrm{\Phi }\left(x\right)}^{\mathrm{-1}}·\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{Integrate to solve for}\overrightarrow{v}\left(x\right)\\ & & \overrightarrow{v}\left(x\right)={\int }_0^x{\mathrm{\Phi }\left(s\right)}^{\mathrm{-1}}·\overrightarrow{f}\left(s\right)\ⅆs\\ & \text{◦}& \text{Plug}\overrightarrow{v}\left(x\right)\text{into the equation for the particular solution}\\ & & {\overrightarrow{y}}_p\left(x\right)=\mathrm{\Phi }\left(x\right)·{\int }_0^x{\mathrm{\Phi }\left(s\right)}^{\mathrm{-1}}·\overrightarrow{f}\left(s\right)\ⅆs\\ & \text{◦}& \text{Plug in the fundamental matrix and the forcing function and compute}\\ & & {\overrightarrow{y}}_p\left(x\right)=\left[\begin{array}{c}\frac{1}{2}-\frac{2{\ⅇ}^{-x}}{5}-\frac{{\ⅇ}^x}{3}+\frac{7{\ⅇ}^{4x}}{30}\\ -\frac{3}{4}+\frac{2{\ⅇ}^{-x}}{5}+\frac{7{\ⅇ}^{4x}}{20}\end{array}\right]\\ \text{•}& & \text{Plug particular solution back into general solution}\\ & & \overrightarrow{y}\left(x\right)=\mathrm{c\_\_1}{\overrightarrow{y}}_1+\mathrm{c\_\_2}{\overrightarrow{y}}_2+\left[\begin{array}{c}\frac{1}{2}-\frac{2{\ⅇ}^{-x}}{5}-\frac{{\ⅇ}^x}{3}+\frac{7{\ⅇ}^{4x}}{30}\\ -\frac{3}{4}+\frac{2{\ⅇ}^{-x}}{5}+\frac{7{\ⅇ}^{4x}}{20}\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{{\ⅇ}^{-x}\left(-30\mathrm{c\_\_1}-12\right)}{30}+\frac{\left(20\mathrm{c\_\_2}+7\right){\ⅇ}^{4x}}{30}-\frac{{\ⅇ}^x}{3}+\frac{1}{2}\\ \frac{{\ⅇ}^{-x}\left(20\mathrm{c\_\_1}+8\right)}{20}-\frac{3}{4}+\frac{\left(20\mathrm{c\_\_2}+7\right){\ⅇ}^{4x}}{20}\end{array}\right]\end{array}$

$\mathrm{sys4}≔\left\{\mathrm{diff}\left(w\left(x\right)\,x\right)=w\left(x\right)+2z\left(x\right)\,\mathrm{diff}\left(z\left(x\right)\,x\right)=3w\left(x\right)+2z\left(x\right)+\exp \left(x\right)\right\}$

$\mathrm{sys4}≔\left\{\frac{\ⅆ}{\ⅆx}w\left(x\right)=w\left(x\right)+2z\left(x\right)\,\frac{\ⅆ}{\ⅆx}z\left(x\right)=3w\left(x\right)+2z\left(x\right)+{\ⅇ}^x\right\}$

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

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{\ⅆ}{\ⅆx}w\left(x\right)=w\left(x\right)+2z\left(x\right)\,\frac{\ⅆ}{\ⅆx}z\left(x\right)=3w\left(x\right)+2z\left(x\right)+{\ⅇ}^x\right\}\\ \text{•}& & \text{Define vector}\\ & & \overrightarrow{w}\left(x\right)=\left[\begin{array}{c}w\left(x\right)\\ z\left(x\right)\end{array}\right]\\ \text{•}& & \text{Convert system into a vector equation}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{w}\left(x\right)=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]·\overrightarrow{w}\left(x\right)+\left[\begin{array}{c}0\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{System to solve}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{w}\left(x\right)=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]·\overrightarrow{w}\left(x\right)+\left[\begin{array}{c}0\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{Define the forcing function}\\ & & \overrightarrow{f}\left(x\right)=\left[\begin{array}{c}0\\ {\ⅇ}^x\end{array}\right]\\ \text{•}& & \text{Define the coefficient matrix}\\ & & A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\\ \text{•}& & \text{Rewrite the system as}\\ & & \frac{\ⅆ}{\ⅆx}\overrightarrow{w}\left(x\right)=A·\overrightarrow{w}\left(x\right)+\overrightarrow{f}\\ \text{•}& & \text{To solve the system, find the eigenvalues and eigenvectors of}A\\ \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{w}}_1={\ⅇ}^{-x}\cdot \left[\begin{array}{c}\mathrm{-1}\\ 1\end{array}\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{w}}_2={\ⅇ}^{4x}\cdot \left[\begin{array}{c}\frac{2}{3}\\ 1\end{array}\right]\\ \text{•}& & \text{General solution of the system of ODEs can be written in terms of the particular solution}{\overrightarrow{w}}_p\left(x\right)\\ & & \overrightarrow{w}\left(x\right)=\mathrm{c\_\_1}{\overrightarrow{w}}_1+\mathrm{c\_\_2}{\overrightarrow{w}}_2+{\overrightarrow{w}}_p\left(x\right)\\ \text{▫}& & \text{Fundamental matrix}\\ & \text{◦}& \text{Let}\mathrm{\phi }\left(x\right)\text{be the matrix whose columns are the independent solutions of the homogeneous system.}\\ & & \mathrm{\phi }\left(x\right)=\left[\begin{array}{cc}-{\ⅇ}^{-x}& \frac{2{\ⅇ}^{4x}}{3}\\ {\ⅇ}^{-x}& {\ⅇ}^{4x}\end{array}\right]\\ & \text{◦}& \text{The fundamental matrix,}\mathrm{\Phi }\left(x\right)\text{is a normalized version of}\mathrm{\phi }\left(x\right)\text{satisfying}\mathrm{\Phi }\left(0\right)=I\text{where}I\text{is the identity matrix}\\ & & \mathrm{\Phi }\left(x\right)=\mathrm{\phi }\left(x\right)·{\mathrm{\phi }\left(0\right)}^{\mathrm{-1}}\\ & \text{◦}& \text{Substitute the value of}\mathrm{\phi }\left(x\right)\text{and}\mathrm{\phi }\left(0\right)\\ & & \mathrm{\Phi }\left(x\right)=\left[\begin{array}{cc}-{\ⅇ}^{-x}& \frac{2{\ⅇ}^{4x}}{3}\\ {\ⅇ}^{-x}& {\ⅇ}^{4x}\end{array}\right]·{\left[\begin{array}{cc}\mathrm{-1}& \frac{2}{3}\\ 1& 1\end{array}\right]}^{\mathrm{-1}}\\ & \text{◦}& \text{Evaluate and simplify to get the 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 of the system of ODEs using variation of parameters}\\ & \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{w}}_p\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{w}}_p\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 its 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 its 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)={\mathrm{\Phi }\left(x\right)}^{\mathrm{-1}}·\overrightarrow{f}\left(x\right)\\ & \text{◦}& \text{Integrate to solve for}\overrightarrow{v}\left(x\right)\\ & & \overrightarrow{v}\left(x\right)={\int }_0^x{\mathrm{\Phi }\left(s\right)}^{\mathrm{-1}}·\overrightarrow{f}\left(s\right)\ⅆs\\ & \text{◦}& \text{Plug}\overrightarrow{v}\left(x\right)\text{into the equation for the particular solution}\\ & & {\overrightarrow{w}}_p\left(x\right)=\mathrm{\Phi }\left(x\right)·{\int }_0^x{\mathrm{\Phi }\left(s\right)}^{\mathrm{-1}}·\overrightarrow{f}\left(s\right)\ⅆs\\ & \text{◦}& \text{Plug in the fundamental matrix and the forcing function and compute}\\ & & {\overrightarrow{w}}_p\left(x\right)=\left[\begin{array}{c}\frac{{\ⅇ}^{-x}}{5}-\frac{{\ⅇ}^x}{3}+\frac{2{\ⅇ}^{4x}}{15}\\ -\frac{{\ⅇ}^{-x}}{5}+\frac{{\ⅇ}^{4x}}{5}\end{array}\right]\\ \text{•}& & \text{Plug particular solution back into general solution}\\ & & \overrightarrow{w}\left(x\right)=\mathrm{c\_\_1}{\overrightarrow{w}}_1+\mathrm{c\_\_2}{\overrightarrow{w}}_2+\left[\begin{array}{c}\frac{{\ⅇ}^{-x}}{5}-\frac{{\ⅇ}^x}{3}+\frac{2{\ⅇ}^{4x}}{15}\\ -\frac{{\ⅇ}^{-x}}{5}+\frac{{\ⅇ}^{4x}}{5}\end{array}\right]\\ \text{•}& & \text{Substitute in vector of dependent variables}\\ & & \left[\begin{array}{c}w\left(x\right)\\ z\left(x\right)\end{array}\right]=\left[\begin{array}{c}\frac{{\ⅇ}^{-x}\left(-15\mathrm{c\_\_1}+3\right)}{15}+\frac{\left(10\mathrm{c\_\_2}+2\right){\ⅇ}^{4x}}{15}-\frac{{\ⅇ}^x}{3}\\ \frac{{\ⅇ}^{-x}\left(\left(1+5\mathrm{c\_\_2}\right){\ⅇ}^{5x}-1+5\mathrm{c\_\_1}\right)}{5}\end{array}\right]\\ \text{•}& & \text{Solution to the system of ODEs}\\ & & \left\{w\left(x\right)=\frac{{\ⅇ}^{-x}\left(-15\mathrm{c\_\_1}+3\right)}{15}+\frac{\left(10\mathrm{c\_\_2}+2\right){\ⅇ}^{4x}}{15}-\frac{{\ⅇ}^x}{3}\,z\left(x\right)=\frac{{\ⅇ}^{-x}\left(\left(1+5\mathrm{c\_\_2}\right){\ⅇ}^{5x}-1+5\mathrm{c\_\_1}\right)}{5}\right\}\end{array}$