Solve the following system of equations using initial conditions \(x\left ( 0\right ) =3;y\left ( 0\right ) =1\)\begin{align*} x^{\prime } & =3x+y-2\sin \left ( t\right ) \\ y^{\prime } & =4x+3y+6\cos \left ( t\right ) \end{align*}
Solution:
Writing the above in matrix form\[\begin{pmatrix} x^{\prime }\\ y^{\prime }\end{pmatrix} =\begin{pmatrix} 3 & 1\\ 4 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix} +\begin{pmatrix} -2\sin \left ( t\right ) \\ 6\cos \left ( t\right ) \end{pmatrix} \] In vector/matrix notations it becomes\[ \mathbf{x}^{\prime }=A\mathbf{x+f}\] The eigenvalues of \(A\) are \(\{5,1\}\) and the matrix whose columns are the corresponding eigenvectors (in same order as the eigenvalues) is\[ P=\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\\ 1 & 1 \end{pmatrix} \] The fundamental matrix \(\Omega \) is given by\[ \Omega =\begin{pmatrix} \frac{1}{2}e^{5t} & -\frac{1}{2}e^{t}\\ e^{5t} & e^{t}\end{pmatrix} \] Since \(\left \vert P\right \vert =1\neq 0\), hence these solutions are linearly independent eigenvectors, (This check was not needed in this case, since the eigenvalues are distinct).\begin{align*} \mathbf{x}_{h} & =\Omega \mathbf{c}\\ & =\begin{pmatrix} \frac{1}{2}e^{5t} & -\frac{1}{2}e^{t}\\ e^{5t} & e^{t}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} \end{align*}
Now the particular solution \(\mathbf{x}_{p}\) is found using the variation of parameters method for systems of equations.\begin{equation} \mathbf{x}=\mathbf{x}_{h}+\mathbf{x}_{p} \tag{4} \end{equation} To find the particular solution, assume \[ \mathbf{x}_{p}=\Omega \mathbf{u}\] and \(\mathbf{u}\) is found\[ \mathbf{x}_{p}^{\prime }=\Omega ^{\prime }\mathbf{u+}\Omega \mathbf{u}^{\prime }\] Substituting the above in the ode \(\mathbf{x}^{\prime }=A\mathbf{x+f}\) gives\[ \Omega ^{\prime }\mathbf{u}+\Omega \mathbf{u}^{\prime }=\left ( A\Omega \right ) \mathbf{u+f}\] But \(\Omega ^{\prime }=A\Omega \) (since \(\Omega \) is a fundamental solution) then the above becomes\begin{align*} \Omega ^{\prime }\mathbf{u}+\Omega \mathbf{u}^{\prime } & =\Omega ^{\prime }\mathbf{u+f}\\ \Omega \mathbf{u}^{\prime } & =\mathbf{f}\\ \mathbf{u}^{\prime } & =\Omega ^{-1}\mathbf{f}\\ \mathbf{u} & =\int \Omega ^{-1}\mathbf{f}\,\mathrm{d}t \end{align*}
But \(\Omega =\begin{pmatrix} \frac{1}{2}e^{5t} & -\frac{1}{2}e^{t}\\ e^{5t} & e^{t}\end{pmatrix} \), hence \(\Omega ^{-1}=\begin{pmatrix} \frac{1}{e^{5t}} & \frac{1}{2e^{5t}}\\ -\frac{1}{e^{t}} & \frac{1}{2e^{t}}\end{pmatrix} \), and \(\mathbf{f}=\begin{pmatrix} -2\sin \left ( t\right ) \\ 6\cos \left ( t\right ) \end{pmatrix} \) hence the above becomes\begin{align*} \mathbf{u} & =\int \begin{pmatrix} \frac{1}{e^{5t}} & \frac{1}{2e^{5t}}\\ -\frac{1}{e^{t}} & \frac{1}{2e^{t}}\end{pmatrix}\begin{pmatrix} -2\sin \left ( t\right ) \\ 6\cos \left ( t\right ) \end{pmatrix} \,\mathrm{d}t\\ & =\int \begin{pmatrix} 3\frac{\cos t}{e^{5t}}-2\frac{\sin t}{e^{5t}}\\ 3\frac{\cos t}{e^{t}}+\frac{2}{e^{t}}\sin t \end{pmatrix} \,\mathrm{d}t\\ & =\begin{pmatrix} \frac{1}{2}e^{-5t}\left ( \sin t-\cos t\right ) \\ \frac{1}{2}e^{-t}\left ( \sin t-5\cos t\right ) \end{pmatrix} \end{align*}
Then 6 , since \(\mathbf{x}_{p}=\Omega \mathbf{u}\) then\begin{align*} \mathbf{x}_{p} & =\begin{pmatrix} \frac{1}{2}e^{5t} & -\frac{1}{2}e^{t}\\ e^{5t} & e^{t}\end{pmatrix}\begin{pmatrix} \frac{1}{2}e^{-5t}\left ( \sin t-\cos t\right ) \\ \frac{1}{2}e^{-t}\left ( \sin t-5\cos t\right ) \end{pmatrix} \\ & =\begin{pmatrix} -\frac{1}{4}e^{t}e^{-t}\left ( \sin t-5\cos t\right ) -\frac{1}{4}e^{-5t}e^{5t}\left ( \cos t-\sin t\right ) \\ \frac{1}{2}e^{t}e^{-t}\left ( \sin t-5\cos t\right ) -\frac{1}{2}e^{-5t}e^{5t}\left ( \cos t-\sin t\right ) \end{pmatrix} \allowbreak \\ & =\begin{pmatrix} \cos t\\ \sin t-3\cos t \end{pmatrix} \end{align*}
Substituting the above into Eq. (4) gives\begin{align*} \mathbf{x} & =\Omega \mathbf{c+x}_{p}\\ & =\begin{pmatrix} \frac{1}{2}e^{5t} & -\frac{1}{2}e^{t}\\ e^{5t} & e^{t}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} +\begin{pmatrix} \cos t\\ \sin t-3\cos t \end{pmatrix} \end{align*}
Applying initial conditions \(x\left ( 0\right ) =3;y\left ( 0\right ) =1\) the constants are found by solving the following\begin{align*} \begin{pmatrix} 3\\ 1 \end{pmatrix} & =\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} +\begin{pmatrix} 1\\ -3 \end{pmatrix} \\\begin{pmatrix} 2\\ 4 \end{pmatrix} & =\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} \end{align*}
Hence\[\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} =\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\\ 1 & 1 \end{pmatrix} ^{-1}\begin{pmatrix} 2\\ 4 \end{pmatrix} =\begin{pmatrix} 4\\ 0 \end{pmatrix} \] And the final solution is\begin{align*} \mathbf{x} & =\Omega \mathbf{c+x}_{p}\\\begin{pmatrix} x\\ y \end{pmatrix} & =\begin{pmatrix} \frac{1}{2}e^{5t} & -\frac{1}{2}e^{t}\\ e^{5t} & e^{t}\end{pmatrix}\begin{pmatrix} 4\\ 0 \end{pmatrix} +\begin{pmatrix} \cos t\\ \sin t-3\cos t \end{pmatrix} \end{align*}
Or\begin{align*} x & =\cos t+2e^{5t}\\ y & =\sin t-3\cos t+4e^{5t} \end{align*}
Find the complete solution of the system with initial conditions \(x\left ( 0\right ) =-1,y\left ( 0\right ) =2,z\left ( 0\right ) =8\)\begin{align*} x^{\prime } & =3x-z\\ y^{\prime } & =-2x+2y+z\\ z^{\prime } & =8x-3z \end{align*}
solution:
\begin{align*} \begin{pmatrix} x^{\prime }\\ y^{\prime }\\ z^{\prime }\end{pmatrix} & =\begin{pmatrix} 3 & 0 & -1\\ -2 & 2 & 1\\ 8 & 0 & -3 \end{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix} \\ \mathbf{x}^{\prime } & =A\mathbf{x} \end{align*}
To diagonalize \(A\), its eigenvectors matrix \(P\) is found. Then the matrix \(D\) is \[ P^{-1}AP=D \] \(D\) is a matrix with the eigenvalues of \(A\) on its diagonal. \(P\) has the eigenvectors as its columns. In this problem it is found that\[ P=\begin{pmatrix} \frac{1}{4} & \frac{1}{2} & 0\\ -\frac{1}{6} & 0 & 1\\ 1 & 1 & 0 \end{pmatrix} \] And \[ D=\begin{pmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2 \end{pmatrix} \] Now to decouple \(A\), let
\[ \mathbf{x}=P\mathbf{z}\] Hence the problem becomes \begin{align*} \mathbf{x}^{\prime } & =P\mathbf{z}^{\prime }\\ & =A\mathbf{x} \end{align*}
Therefore\begin{align*} P\mathbf{z}^{\prime } & =A\mathbf{x}\\ & =AP\mathbf{z} \end{align*}
or\begin{align*} \mathbf{z}^{\prime } & =P^{-1}AP\mathbf{z}\\ & =D\mathbf{z} \end{align*}
The new system is which is diagonalizable (decoupled) hence easy to solve. Solving for \(\mathbf{z}\) gives \[\begin{pmatrix} z_{1}^{\prime }\\ z_{2}^{\prime }\\ z_{3}^{\prime }\end{pmatrix} =\begin{pmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} z_{1}\\ z_{2}\\ z_{3}\end{pmatrix} \] Hence\begin{align*} \begin{pmatrix} z_{1}\\ z_{2}\\ z_{3}\end{pmatrix} & =\begin{pmatrix} e^{-t} & 0 & 0\\ 0 & e^{t} & 0\\ 0 & 0 & e^{2t}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} \\ \mathbf{z} & =\Omega _{z}\mathbf{c} \end{align*}
Where \(c_{i}\) are the constants of integration and \(\Omega _{z}\) is the fundamental matrix in the \(z\) space. Now, since \(\mathbf{x}=P\mathbf{z}\) then the solution is converted back\begin{align*} \mathbf{x} & =P\Omega _{z}\mathbf{c}\\ & =\begin{pmatrix} \frac{1}{4} & \frac{1}{2} & 0\\ -\frac{1}{6} & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}\begin{pmatrix} e^{-t} & 0 & 0\\ 0 & e^{t} & 0\\ 0 & 0 & e^{2t}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} \\ & =\begin{pmatrix} \frac{1}{4}e^{-t} & \frac{1}{2}e^{t} & 0\\ -\frac{1}{6}e^{-t} & 0 & e^{2t}\\ e^{-t} & e^{t} & 0 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} \end{align*}
Applying initial conditions \(x\left ( 0\right ) =-1,y\left ( 0\right ) =2,z\left ( 0\right ) =8\)\[\begin{pmatrix} -1\\ 2\\ 8 \end{pmatrix} =\begin{pmatrix} \frac{1}{4} & \frac{1}{2} & 0\\ -\frac{1}{6} & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} \] Hence\[\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} \frac{1}{4} & \frac{1}{2} & 0\\ -\frac{1}{6} & 0 & 1\\ 1 & 1 & 0 \end{pmatrix} ^{-1}\begin{pmatrix} -1\\ 2\\ 8 \end{pmatrix} =\begin{pmatrix} 20\\ -12\\ \frac{16}{3}\end{pmatrix} \] Therefore, the final solution is\begin{align*} \begin{pmatrix} x\left ( t\right ) \\ y\left ( t\right ) \\ z\left ( t\right ) \end{pmatrix} & =\begin{pmatrix} \frac{1}{4}e^{-t} & \frac{1}{2}e^{t} & 0\\ -\frac{1}{6}e^{-t} & 0 & e^{2t}\\ e^{-t} & e^{t} & 0 \end{pmatrix}\begin{pmatrix} 20\\ -12\\ \frac{16}{3}\end{pmatrix} \\\begin{pmatrix} x\left ( t\right ) \\ y\left ( t\right ) \\ z\left ( t\right ) \end{pmatrix} & =\begin{pmatrix} 5e^{-t}-6e^{t}\\ \frac{16}{3}e^{2t}-\frac{10}{3}e^{-t}\\ 20e^{-t}-12e^{t}\end{pmatrix} \end{align*}
Produce a matrix that diagonalizes \(A=\begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 2\\ 0 & 1 & 3 \end{pmatrix} \), or show the matrix is not diagonalizable.
Solution:
The eigenvalues of \(A\) are \[\begin{pmatrix} \lambda _{1}\\ \lambda _{2}\\ \lambda 3 \end{pmatrix} =\begin{pmatrix} \frac{1}{2}\sqrt{17}+\frac{3}{2}\\ \frac{3}{2}-\frac{1}{2}\sqrt{17}\\ 0 \end{pmatrix} \] Since the eigenvalues are distinct then the matrix is diagonalizable. The corresponding eigenvectors are\[ P=\begin{pmatrix} 0 & 0 & -2\\ \frac{1}{2}\sqrt{17}-\frac{3}{2} & -\frac{1}{2}\sqrt{17}-\frac{3}{2} & -3\\ 1 & 1 & 1 \end{pmatrix} \] Where the first column is the first eigenvector associated with first eigenvalue \(\lambda _{1}\) and the second column is second first eigenvector associated with the second eigenvalue \(\lambda _{2}\). Therefore the diagonalized matrix of \(A\) is the matrix \(D\) given by\[ D=P^{-1}AP \] \(D\) does not need to be computed from the above, since it is given also by\begin{align*} D & =\begin{pmatrix} \lambda _{1} & 0 & 0\\ 0 & \lambda _{2} & 0\\ 0 & 0 & \lambda 3 \end{pmatrix} \\ & =\begin{pmatrix} \frac{1}{2}\sqrt{17}+\frac{3}{2} & 0 & 0\\ 0 & \frac{3}{2}-\frac{1}{2}\sqrt{17} & 0\\ 0 & 0 & 0 \end{pmatrix} \\ & =\begin{pmatrix} 3.5616 & 0 & 0\\ 0 & -0.5616 & 0\\ 0 & 0 & 0 \end{pmatrix} \end{align*}
Note that Jordan form of the matrix is its diagonalizable matrix (if it is diagonalizable).
Find the general solution to \begin{align*} x^{\prime } & =-x+3y\\ y^{\prime } & =3x-y\\ z^{\prime } & =-2x-2y+6z \end{align*}
solution:
\begin{align*} \begin{pmatrix} x^{\prime }\\ y^{\prime }\\ z^{\prime }\end{pmatrix} & =\begin{pmatrix} -1 & 3 & 0\\ 3 & -1 & 0\\ -2 & -2 & 6 \end{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix} \\ \mathbf{x}^{\prime } & =A\mathbf{x} \end{align*}
The eigenvalues of \(A\) are \(\left \{ -4,2,6\right \} \), hence the matrix is diagonalizable (since all its eigenvalues are distinct). Therefore \begin{align*} P^{-1}AP & =D\\ & =\begin{pmatrix} -4 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 6 \end{pmatrix} \end{align*}
The eigenvector matrix \(P\) is\[ P=\begin{pmatrix} -1 & 1 & 0\\ 1 & 1 & 0\\ 0 & 1 & 1 \end{pmatrix} \] Where the first column if the eigenvector associated with the first eigenvalues \(-4\) and so on. Now, let\[ \mathbf{x}=P\mathbf{z}\] Which leads to (as was done in problem 2) in order to decouple the system of equations in the \(z\) space\[ \mathbf{z}^{\prime }=D\mathbf{z}\] Now the new system is decoupled. Solving for \(\mathbf{z}\)\[\begin{pmatrix} z_{1}^{\prime }\\ z_{2}^{\prime }\\ z_{3}^{\prime }\end{pmatrix} =\begin{pmatrix} -4 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 6 \end{pmatrix}\begin{pmatrix} z_{1}\\ z_{2}\\ z_{3}\end{pmatrix} \] Hence\begin{align*} \begin{pmatrix} z_{1}\\ z_{2}\\ z_{3}\end{pmatrix} & =\begin{pmatrix} e^{-4t} & 0 & 0\\ 0 & e^{2t} & 0\\ 0 & 0 & e^{6t}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} \\ \mathbf{z} & =\Omega _{z}\mathbf{c} \end{align*}
Where \(c\) are the constants of integration and \(\Omega _{z}\) is the fundamental matrix in the \(z\) space. Since \(\mathbf{x}=P\mathbf{z}\) then\begin{align*} \mathbf{x} & =P\Omega _{z}\mathbf{c}\\ & =\begin{pmatrix} -1 & 1 & 0\\ 1 & 1 & 0\\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} e^{-4t} & 0 & 0\\ 0 & e^{2t} & 0\\ 0 & 0 & e^{6t}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} \\ & =\begin{pmatrix} -e^{-4t} & e^{2t} & 0\\ e^{-4t} & e^{2t} & 0\\ 0 & e^{2t} & e^{6t}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} \\ & =\begin{pmatrix} c_{2}e^{2t}-c_{1}e^{-4t}\\ c_{2}e^{2t}+c_{1}e^{-4t}\\ c_{2}e^{2t}+c_{3}e^{6t}\end{pmatrix} \end{align*}
Find the general solution for
\begin{align*} 2x^{\prime }+x+y^{\prime }+2y & =e^{t}\\ 3x^{\prime }-7x+3y^{\prime }+y & =0 \end{align*}
solution:
First the equations are transformed such that each equation contains only \(x^{\prime }\) or \(y^{\prime }\) on its own. This is to allow the system to be written as \(\mathbf{x}^{\prime }=A\mathbf{x}\). Solving for \(x^{\prime },y^{\prime }\) gives\begin{align*} x^{\prime } & =-\frac{10}{3}x-\frac{5}{3}y+e^{t}\\ y^{\prime } & =\frac{17}{3}x+\frac{4}{3}y-e^{t} \end{align*}
Hence\begin{align*} \begin{pmatrix} x^{\prime }\\ y^{\prime }\end{pmatrix} & =\begin{pmatrix} -\frac{10}{3} & -\frac{5}{3}\\ \frac{17}{3} & \frac{4}{3}\end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix} +\begin{pmatrix} e^{t}\\ -e^{t}\end{pmatrix} \\ \mathbf{x}^{\prime } & =A\mathbf{x}+\mathbf{f} \end{align*}
The eigenvalues of \(A\) are \(\left \{ -1+2i,-1-2i\right \} \). Since they are distinct, the matrix is diagonalizable. The \(P\) matrix of eigenvectors is\begin{align*} P & =\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix} \\ D & =\begin{pmatrix} -1+2i & 0\\ 0 & -1-2i \end{pmatrix} \end{align*}
Let \[ \mathbf{x}=P\mathbf{z}\] Hence \(\mathbf{x}^{\prime }=A\mathbf{x}+\mathbf{f}\) becomes decoupled as follows\begin{align*} \mathbf{z}^{\prime } & =D\mathbf{z}+P^{-1}\mathbf{f}\\ & =D\mathbf{z}+\mathbf{G} \end{align*}
Where \(\mathbf{G}=P^{-1}\mathbf{f=}\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix} ^{-1}\begin{pmatrix} e^{t}\\ -e^{t}\end{pmatrix} =\begin{pmatrix} -\left ( \frac{1}{2}+\frac{5}{6}i\right ) e^{t}\\ -\left ( \frac{1}{2}-\frac{5}{6}i\right ) e^{t}\end{pmatrix} \)
Solving for \(\mathbf{z}\)\[\begin{pmatrix} z_{1}^{\prime }\\ z_{2}^{\prime }\end{pmatrix} =\begin{pmatrix} -1+2i & 0\\ 0 & -1-2i \end{pmatrix}\begin{pmatrix} z_{1}\\ z_{2}\end{pmatrix} +\begin{pmatrix} -\left ( \frac{1}{2}+\frac{5}{6}i\right ) e^{t}\\ -\left ( \frac{1}{2}-\frac{5}{6}i\right ) e^{t}\end{pmatrix} \] Since the system is decoupled it can be solved as each equation on its own as follows\begin{align*} z_{1}^{\prime } & =\left ( -1+2i\right ) z_{1}-\left ( \frac{1}{2}+\frac{5}{6}i\right ) e^{t}\\ z_{2}^{\prime } & =\left ( -1-2i\right ) z_{2}-\left ( \frac{1}{2}-\frac{5}{6}i\right ) e^{t} \end{align*}
The solution is\begin{align*} z_{1}^{\prime }+\left ( 1-2i\right ) z_{1} & =\left ( -\frac{1}{2}-\frac{5}{6}i\right ) e^{t}\\ z_{2}^{\prime }+\left ( 1+2i\right ) z_{2} & =\left ( -\frac{1}{2}+\frac{5}{6}i\right ) e^{t} \end{align*}
Hence, using integrating factor\begin{align*} d\left ( e^{\int \left ( 1-2i\right ) \,\mathrm{d}t}z_{1}\right ) & =e^{\int \left ( 1-2i\right ) \,d\tau }\left ( -\frac{1}{2}-\frac{5}{6}i\right ) e^{t}\\ d\left ( e^{\int \left ( 1+2i\right ) \,\mathrm{d}t}z_{2}\right ) & =e^{\int \left ( 1+2i\right ) \,d\tau }\left ( -\frac{1}{2}+\frac{5}{6}i\right ) e^{t} \end{align*}
Therefore\begin{align*} d\left ( e^{\left ( 1-2i\right ) t}z_{1}\right ) & =e^{\left ( 1-2i\right ) t}\left ( -\frac{1}{2}-\frac{5}{6}i\right ) e^{t}\\ d\left ( e^{\left ( 1+2i\right ) t}z_{2}\right ) & =e^{\left ( 1+2i\right ) t}\left ( -\frac{1}{2}+\frac{5}{6}i\right ) e^{t} \end{align*}
Hence\begin{align*} e^{\left ( 1-2i\right ) t}z_{1} & =\left ( -\frac{1}{2}-\frac{5}{6}i\right ) \int e^{\left ( 1-2i\right ) t+t}\,\mathrm{d}t+c_{1}\\ e^{\left ( 1+2i\right ) t}z_{2} & =\left ( -\frac{1}{2}+\frac{5}{6}i\right ) \int e^{\left ( 1+2i\right ) t+t}\,\mathrm{d}t+c_{2} \end{align*}
or\begin{align*} e^{\left ( 1-2i\right ) t}z_{1} & =\left ( -\frac{1}{2}-\frac{5}{6}i\right ) \left ( \frac{1}{4}+\frac{i}{4}\right ) e^{-2it+2t}+c_{1}\\ e^{\left ( 1+2i\right ) t}z_{2} & =\left ( -\frac{1}{2}+\frac{5}{6}i\right ) \left ( \frac{1}{4}-\frac{i}{4}\right ) e^{2it+2t}+c_{2} \end{align*}
or\begin{align*} z_{1} & =\left ( -\frac{1}{2}-\frac{5}{6}i\right ) \left ( \frac{1}{4}+\frac{i}{4}\right ) e^{-2it+2t-(1-2i)t}+c_{1}e^{-\left ( 1-2i\right ) t}\\ z_{2} & =\left ( -\frac{1}{2}+\frac{5}{6}i\right ) \left ( \frac{1}{4}-\frac{i}{4}\right ) e^{2it+2t-\left ( 1+2i\right ) t}+c_{2}e^{-\left ( 1+2i\right ) t} \end{align*}
Hence\begin{align*} z_{1} & =\left ( \frac{1}{12}-\frac{1}{3}i\right ) e^{t}+c_{1}e^{-\left ( 1-2i\right ) t}\\ z_{2} & =\left ( \frac{1}{12}+\frac{1}{3}i\right ) e^{t}+c_{2}e^{-\left ( 1+2i\right ) t} \end{align*}
But \(\mathbf{x}=P\mathbf{z}\) therefore \begin{align*} \mathbf{x} & =\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix}\begin{pmatrix} \left ( \frac{1}{12}-\frac{1}{3}i\right ) e^{t}+c_{1}e^{-\left ( 1-2i\right ) t}\\ \left ( \frac{1}{12}+\frac{1}{3}i\right ) e^{t}+c_{2}e^{-\left ( 1+2i\right ) t}\end{pmatrix} \\ & =\begin{pmatrix} \frac{1}{6}e^{t}-\left ( \frac{7}{17}-\frac{6}{17}i\right ) c_{1}e^{-\left ( 1-2i\right ) t}-\left ( \frac{7}{17}+\frac{6}{17}i\right ) c_{2}e^{-\left ( 1+2i\right ) t}\\ \frac{1}{6}e^{t}+c_{1}e^{-\left ( 1-2i\right ) t}+c_{2}e^{-\left ( 1+2i\right ) t}\end{pmatrix} \end{align*}
The final solution is\begin{align*} x\left ( t\right ) & =\frac{1}{6}e^{t}-\left ( \frac{7}{17}-\frac{6}{17}i\right ) c_{1}e^{-\left ( 1-2i\right ) t}-\left ( \frac{7}{17}+\frac{6}{17}i\right ) c_{2}e^{-\left ( 1+2i\right ) t}\\ y\left ( t\right ) & =\frac{1}{6}e^{t}+c_{1}e^{-\left ( 1-2i\right ) t}+c_{2}e^{-\left ( 1+2i\right ) t} \end{align*}
or\begin{align*} x\left ( t\right ) & =-\left ( \frac{7}{17}c_{1}e^{-\left ( 1-2i\right ) t}-\frac{6}{17}ic_{1}e^{-\left ( 1-2i\right ) t}\right ) -\left ( \frac{7}{17}c_{2}e^{-\left ( 1+2i\right ) t}+\frac{6}{17}ic_{2}e^{-\left ( 1+2i\right ) t}\right ) +\frac{1}{6}e^{t}\\ y\left ( t\right ) & =c_{1}e^{-t}e^{2it}+c_{2}e^{-t}e^{-2it}+\frac{1}{6}e^{t} \end{align*}
or\begin{align*} x\left ( t\right ) & =\frac{1}{6}e^{t}-e^{-t}\left ( \frac{7}{17}c_{1}e^{2it}-\frac{6}{17}ic_{1}e^{2it}\right ) -e^{-t}\left ( \frac{7}{17}c_{2}e^{-2it}+\frac{6}{17}ic_{2}e^{-2it}\right ) \\ y\left ( t\right ) & =e^{-t}\left ( c_{1}\left ( \cos 2t+i\sin 2t\right ) +c_{2}\left ( \cos 2t-i\sin 2t\right ) \right ) +\frac{1}{6}e^{t} \end{align*}
or\begin{align*} x\left ( t\right ) & =-e^{-t}\left ( \frac{7}{17}c_{1}\left ( \cos 2t+i\sin 2t\right ) -\frac{6}{17}ic_{1}\left ( \cos 2t+i\sin 2t\right ) \right ) \\ & -e^{-t}\left ( \frac{7}{17}c_{2}\left ( \cos 2t-i\sin 2t\right ) +\frac{6}{17}ic_{2}\left ( \cos 2t-i\sin 2t\right ) \right ) +\frac{1}{6}e^{t}\\ y\left ( t\right ) & =e^{-t}\left ( c_{1}\cos 2t+c_{1}i\sin 2t+c_{2}\cos 2t-ic_{2}\sin 2t\right ) +\frac{1}{6}e^{t} \end{align*}
or\begin{align*} x\left ( t\right ) & =-e^{-t}\left ( \frac{7}{17}c_{1}\cos 2t+\frac{7}{17}ic_{1}\sin 2t-ic_{1}\frac{6}{17}\cos 2t+c_{1}\frac{6}{17}\sin 2t\right ) \\ & -e^{-t}\left ( \frac{7}{17}c_{2}\cos 2t-i\frac{7}{17}c_{2}\sin 2t+\frac{6}{17}ic_{2}\cos 2t+\frac{6}{17}c_{2}\sin 2t\right ) +\frac{1}{6}e^{t}\\ y\left ( t\right ) & =e^{-t}\left ( \left ( c_{1}+c_{2}\right ) \cos 2t+i\left ( c_{1}-c_{2}\right ) \sin 2t\right ) +\frac{1}{6}e^{t} \end{align*}
or\begin{align*} x\left ( t\right ) & =\frac{1}{6}e^{t}-e^{-t}\left ( \begin{array} [c]{c}\frac{7}{17}c_{1}\cos 2t+\frac{7}{17}ic_{1}\sin 2t-ic_{1}\frac{6}{17}\cos 2t+c_{1}\frac{6}{17}\sin 2t\\ -\frac{7}{17}c_{2}\cos 2t+i\frac{7}{17}c_{2}\sin 2t-\frac{6}{17}ic_{2}\cos 2t-\frac{6}{17}c_{2}\sin 2t \end{array} \right ) \\ y\left ( t\right ) & =e^{-t}\left ( \left ( c_{1}+c_{2}\right ) \cos 2t+i\left ( c_{1}-c_{2}\right ) \sin 2t\right ) +\frac{1}{6}e^{t} \end{align*}
or\begin{align*} x\left ( t\right ) & =\frac{1}{6}e^{t}-e^{-t}\left ( \frac{7}{17}\left ( c_{1}-c_{2}\right ) \cos 2t+i\frac{7}{17}\left ( c_{1}+c_{2}\right ) \sin 2t-i\frac{6}{17}\left ( c_{1}+c_{2}\right ) \cos 2t+\frac{6}{17}\left ( c_{1}-c_{2}\right ) \sin 2t\right ) \\ y\left ( t\right ) & =e^{-t}\left ( \left ( c_{1}+c_{2}\right ) \cos 2t-i\left ( c_{1}-c_{2}\right ) \sin 2t\right ) +\frac{1}{6}e^{t} \end{align*}
Let \(\left ( c_{1}+c_{2}\right ) i=A\) and let \(\left ( c_{1}-c_{2}\right ) =B\) then above becomes\begin{align*} x\left ( t\right ) & =\frac{1}{6}e^{t}-e^{-t}\left ( \frac{7}{17}B\cos 2t-\frac{7}{17}A\sin 2t+\frac{6}{17}A\cos 2t+\frac{6}{17}B\sin 2t\right ) \\ y\left ( t\right ) & =\frac{1}{6}e^{t}-ie^{-t}\left ( B\sin 2t+A\cos 2t\right ) \end{align*}
I am not sure how to move the remaining complex number into the constants over both solutions. According to CAS, the answer should be real\begin{align*} y\left ( t\right ) & =\frac{1}{6}e^{t}-\frac{1}{5}e^{-t}\left ( 7B\cos 2t+7A\sin 2t+6A\cos 2t-6B\sin 2t\right ) \\ x\left ( t\right ) & =\frac{1}{6}e^{t}+e^{-t}\left ( B\sin 2t+A\cos 2t\right ) \end{align*}
May be I am close, but do not see it now. So will stop here.
Using \(\int e^{At}dt=e^{At}A^{-1}\) determine general solution of the following matrix equation \(\frac{d\mathbf{N}\left ( t\right ) }{dt}=A\mathbf{N}\left ( t\right ) +\mathbf{F}\left ( t\right ) \) where \(\mathbf{F}\left ( t\right ) =\mathbf{B}t^{2}\) and \(\mathbf{B}\) is constant vector.
solution:
\begin{align*} \frac{d\mathbf{N}\left ( t\right ) }{dt} & =A\mathbf{N}\left ( t\right ) +\mathbf{F}\left ( t\right ) \\ \frac{d\mathbf{N}\left ( t\right ) }{dt}-A\mathbf{N}\left ( t\right ) & =\mathbf{F}\left ( t\right ) \\ \int _{0}^{t}d\left ( e^{-A\tau }\mathbf{N}\left ( \tau \right ) \right ) & =e^{-At}\mathbf{F}\left ( t\right ) \\ e^{-At}\mathbf{N}\left ( t\right ) -\mathbf{N}\left ( 0\right ) & =\int e^{-At}\mathbf{F}\left ( t\right ) \,\mathrm{d}\tau \\ e^{-At}\mathbf{N}\left ( t\right ) & =\mathbf{N}\left ( 0\right ) +\int e^{-At}\mathbf{F}\left ( t\right ) \,\mathrm{d}\tau \\ \mathbf{N}\left ( t\right ) & =e^{At}\mathbf{N}\left ( 0\right ) +e^{At}\int e^{-At}\mathbf{F}\left ( t\right ) \,\mathrm{d}\tau \\ \mathbf{N}\left ( t\right ) & =e^{At}\mathbf{N}\left ( 0\right ) +e^{At}\int e^{-At}\mathbf{B}t^{2}\,\mathrm{d}\tau \end{align*}
But \(e^{At}=Pe^{Dt}P^{-1},\)where \(P\) be the matrix of eigenvectors of \(A\) and \(D\) be the matrix of eigenvalues of \(A\) on its diagonal. then\begin{align*} \mathbf{N}\left ( t\right ) & =Pe^{Dt}P^{-1}\mathbf{N}\left ( 0\right ) +Pe^{Dt}P^{-1}\int Pe^{-Dt}P^{-1}\mathbf{B}t^{2}\,\mathrm{d}\tau \\ & =Pe^{Dt}P^{-1}\mathbf{N}\left ( 0\right ) +Pe^{Dt}\int e^{-Dt}P^{-1}\mathbf{B}t^{2}\,\mathrm{d}\tau \end{align*}
The solution due to the forcing function is contained in \(Pe^{Dt}\int e^{-Dt}P^{-1}\mathbf{B}t^{2}\,\mathrm{d}\tau \)
Solve problem 5 above using the matrix exponential method.
solution:
\begin{align*} 2x^{\prime }+x+y^{\prime }+2y & =e^{t}\\ 3x^{\prime }-7x+3y^{\prime }+y & =0 \end{align*}
solution:
First we solve the above such that each equation contains only \(x^{\prime }\) or \(y^{\prime }\) on its own. This is to allow us to write the system as \(\mathbf{x}^{\prime }=A\mathbf{x}\). By solving for \(x^{\prime },y^{\prime }\) we find\begin{align*} x^{\prime } & =-\frac{10}{3}x-\frac{5}{3}y+e^{t}\\ y^{\prime } & =\frac{17}{3}x+\frac{4}{3}y-e^{t} \end{align*}
Hence\begin{align*} \begin{pmatrix} x^{\prime }\\ y^{\prime }\end{pmatrix} & =\begin{pmatrix} -\frac{10}{3} & -\frac{5}{3}\\ \frac{17}{3} & \frac{4}{3}\end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix} +\begin{pmatrix} e^{t}\\ -e^{t}\end{pmatrix} \\ \mathbf{x}^{\prime } & =A\mathbf{x}+\mathbf{f} \end{align*}
The general solution is\begin{equation} \mathbf{x=}Pe^{Dt}P^{-1}\mathbf{x}\left ( 0\right ) +Pe^{Dt}\int _{0}^{t}e^{-D\tau }P^{-1}\mathbf{F}\,\mathrm{d}t \tag{1} \end{equation} Where the eigenvalues of \(A\) are \(\left \{ -1+2i,-1-2i\right \} \), hence, as was found in problem 5\begin{align*} P & =\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix} \\ D & =\begin{pmatrix} -1+2i & 0\\ 0 & -1-2i \end{pmatrix} \\ \Phi & =\begin{pmatrix} \left ( -\frac{7}{17}+\frac{6}{17}i\right ) e^{\left ( -1+2i\right ) t} & \left ( -\frac{7}{17}-\frac{6}{17}i\right ) e^{\left ( -1-2i\right ) t}\\ e^{\left ( -1+2i\right ) t} & e^{\left ( -1-2i\right ) t}\end{pmatrix} \end{align*}
Now \[ e^{Dt}=\begin{pmatrix} e^{\lambda _{1}t} & 0\\ 0 & e^{\lambda _{2}t}\end{pmatrix} =\begin{pmatrix} e^{\left ( -1+2i\right ) t} & 0\\ 0 & e^{\left ( -1-2i\right ) t}\end{pmatrix} \] and\[ e^{-D\tau }=\begin{pmatrix} e^{-\lambda _{1}t} & 0\\ 0 & e^{-\lambda _{2}t}\end{pmatrix} =\begin{pmatrix} e^{\left ( 1-2i\right ) t} & 0\\ 0 & e^{\left ( 1+2i\right ) t}\end{pmatrix} \] and \[ P^{-1}=\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix} ^{-1}=\begin{pmatrix} -\frac{17}{12}i & \frac{1}{2}-\frac{7}{12}i\\ \frac{17}{12}i & \frac{1}{2}+\frac{7}{12}i \end{pmatrix} \] Hence Eq. (1) becomes\begin{align*} \mathbf{x} & =Pe^{Dt}P^{-1}\mathbf{x}\left ( 0\right ) +Pe^{Dt}\int _{0}^{t}e^{-D\tau }P^{-1}\mathbf{F}\,d\tau \\ & =\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix}\begin{pmatrix} e^{\left ( -1+2i\right ) t} & 0\\ 0 & e^{\left ( -1-2i\right ) t}\end{pmatrix}\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix} ^{-1}\mathbf{x}\left ( 0\right ) \\ & +\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix}\begin{pmatrix} e^{\left ( -1+2i\right ) t} & 0\\ 0 & e^{\left ( -1-2i\right ) t}\end{pmatrix} \\ & \int _{0}^{t}\begin{pmatrix} e^{\left ( 1-2i\right ) t} & 0\\ 0 & e^{\left ( 1+2i\right ) t}\end{pmatrix}\begin{pmatrix} -\frac{7}{17}+\frac{6}{17}i & -\frac{7}{17}-\frac{6}{17}i\\ 1 & 1 \end{pmatrix} ^{-1}\begin{pmatrix} e^{t}\\ -e^{t}\end{pmatrix} \,\mathrm{d}t \end{align*}
Simplifying gives\begin{align*} \mathbf{x} & =\begin{pmatrix} \left ( \frac{1}{2}-\frac{7}{12}i\right ) e^{-\left ( 1+2i\right ) t}+\left ( \frac{1}{2}+\frac{7}{12}i\right ) e^{-\left ( 1-2i\right ) t} & \frac{5}{12}ie^{-\left ( 1-2i\right ) t}-\frac{5}{12}ie^{-\left ( 1+2i\right ) t}\\ \frac{17}{12}ie^{-\left ( 1+2i\right ) t}-\frac{17}{12}ie^{-\left ( 1-2i\right ) t} & \left ( \frac{1}{2}+\frac{7}{12}i\right ) e^{-\left ( 1+2i\right ) t}+\left ( \frac{1}{2}-\frac{7}{12}i\right ) e^{-\left ( 1-2i\right ) t}\end{pmatrix} \mathbf{x}\left ( 0\right ) \\ & +\begin{pmatrix} -\left ( \frac{7}{17}-\frac{6}{17}i\right ) e^{-\left ( 1-2i\right ) t} & -\left ( \frac{7}{17}+\frac{6}{17}i\right ) e^{-\left ( 1+2i\right ) t}\\ e^{-\left ( 1-2i\right ) t} & e^{-\left ( 1+2i\right ) t}\end{pmatrix} \int _{0}^{t}\begin{pmatrix} -\left ( \frac{1}{2}+\frac{5}{6}i\right ) e^{\left ( 2-2i\right ) t}\\ -\left ( \frac{1}{2}-\frac{5}{6}i\right ) e^{\left ( 2+2i\right ) t}\end{pmatrix} \allowbreak \,\mathrm{d}t \end{align*}
The integration yields\[\begin{pmatrix} -\frac{1}{12}+\frac{i}{3}+\frac{e^{\left ( 2-2i\right ) t}}{12}-\frac{ie^{\left ( 2-2i\right ) t}}{3}\\ -\frac{1}{12}+\frac{i}{3}+\frac{e^{\left ( 2+2i\right ) t}}{12}+\frac{ie^{\left ( 2+2i\right ) t}}{3}\end{pmatrix} \] Hence the above simplifies to\begin{align*} \mathbf{x} & =\begin{pmatrix} \left ( \frac{1}{2}-\frac{7}{12}i\right ) e^{-\left ( 1+2i\right ) t}+\left ( \frac{1}{2}+\frac{7}{12}i\right ) e^{-\left ( 1-2i\right ) t} & \frac{5}{12}ie^{-\left ( 1-2i\right ) t}-\frac{5}{12}ie^{-\left ( 1+2i\right ) t}\\ \frac{17}{12}ie^{-\left ( 1+2i\right ) t}-\frac{17}{12}ie^{-\left ( 1-2i\right ) t} & \left ( \frac{1}{2}+\frac{7}{12}i\right ) e^{-\left ( 1+2i\right ) t}+\left ( \frac{1}{2}-\frac{7}{12}i\right ) e^{-\left ( 1-2i\right ) t}\end{pmatrix} \mathbf{x}\left ( 0\right ) \\ & +\begin{pmatrix} \left ( \frac{31}{204}-\frac{11}{102}i\right ) e^{-\left ( 1+2i\right ) t}-\left ( \frac{1}{12}+\frac{1}{6}i\right ) e^{-\left ( 1-2i\right ) t}+\left ( \frac{1}{12}-\frac{1}{6}i\right ) e^{-\left ( 1+2i\right ) t}e^{\left ( 2+2i\right ) t}+\left ( \frac{1}{12}+\frac{1}{6}i\right ) e^{-\left ( 1-2i\right ) t}e^{\left ( 2-2i\right ) t}\\ \left ( \frac{1}{12}+\frac{1}{3}i\right ) e^{-\left ( 1+2i\right ) t}e^{\left ( 2+2i\right ) t}-\left ( \frac{1}{12}-\frac{1}{3}i\right ) e^{-\left ( 1-2i\right ) t}-\left ( \frac{1}{12}-\frac{1}{3}i\right ) e^{-\left ( 1+2i\right ) t}+\left ( \frac{1}{12}-\frac{1}{3}i\right ) e^{-\left ( 1-2i\right ) t}e^{\left ( 2-2i\right ) t}\end{pmatrix} \end{align*}