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Let \(\alpha _{i}\) be the \(i^{th}\) eigenvalue of \(M_{1}\) and let \(v_{i}\) be an eigenvector associated with \(\alpha _{i}\). This implies\[ M_{1}v_{i}=\alpha _{i}v_{i}\] Similarly, let \(\beta _{i}\) be the \(i^{th}\) eigenvalue of \(M_{2}\) and let \(u_{i}\) be an eigenvector associated with \(\beta _{i}\). This implies\[ M_{2}u_{i}=\beta _{i}u_{i}\] We start by post multiplying \(M_{1}M_{2}\) with an eigenvector of \(M_{1}\) associated with eigenvalue \(\alpha _{i}\), this results in \[ M_{1}M_{2}v_{i}=M_{2}M_{1}v_{i}\] Where we just took advantage of commuting \(M_{1}M_{2}\) by changing the order in the RHS above. But \(M_{1}v_{i}=\alpha _{i}v_{i}\), hence the above becomes\[ M_{1}M_{2}v_{i}=M_{2}\alpha _{i}v_{i}\] Since \(\alpha _{i}\) is scalar, we can move it to the left and obtain\[ M_{1}\left ( M_{2}v_{i}\right ) =\alpha _{i}\left ( M_{2}v_{i}\right ) \] We see now that \(M_{2}v_{i}\) itself is an eigenvector of \(M_{1}\).
What the above means is that if \(v_{i}\) is an eigenvector of \(M_{1}\) associated with an eigenvalue \(\alpha _{i},\) then so will be \(M_{2}v_{i}\). Now an important point follows: Since the eigenvalues are distinct, then all the eigenvectors that belong to each eigenvalues are scalar multiple of each others. What this means, is that \(M_{2}v_{i}\) is some scaled version of \(v_{i}\) since both are in the same eigenspace associated with \(\alpha _{i}\). The eigenspace associated with an eigenvalue is just the space spanned by all the eigenvectors of this eigenvalue. This means this space is one dimensional in this case.
This is critical, since it then tells us that \(M_{2}v_{i}=\beta _{i}v_{i}\) where \(\beta _{i}\) is the above scalar, which is the eigenvalue of \(M_{2}\). Without this restriction, we could not say that \(M_{2}v_{i}=\beta _{i}v_{i}\).
Therefore, the above means that each eigenvector of \(M_{1}\) is also an eigenvector of \(M_{2}\). Or said in other way, the matrix \(M_{1}\) and \(M_{2}\) share the same eigenspaces.
But this complete the proof. Since the nonsingular matrix \(T\) which diagonalizes a matrix is made of up of the eigenvectors of the matrix\(.\) The columns of \(T\) are the eigenvectors of the matrix\(.\) And since \(M_{1},M_{2}\) share the same eigenvectors, hence the same \(T\) will diagonalize both of them at the same time.
QED
We want to show that \(\frac{d}{dt}\Psi \left ( t\right ) =A\left ( t\right ) \Psi \left ( t\right ) \). Where \(\Psi \left ( t\right ) =e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }\). To expand \(e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }\)we will use the definition of matrix exponential \[ e^{M}=I+M+\frac{1}{2}M^{2}+\frac{1}{3!}M^{3}+\cdots \] Therefore\[ e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }=I+\int _{0}^{t}A\left ( \tau \right ) d\tau +\frac{1}{2}\left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) \left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) +\frac{1}{3!}\left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) \left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) \left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) +\cdots \] To make it easier to see, we will expand only the first 2 terms in expansion:\begin{equation} e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }=I+\int _{0}^{t}A\left ( \tau \right ) d\tau +\frac{1}{2}\left [ \int _{0}^{t}A\left ( \tau \right ) d\tau \int _{0}^{t}A\left ( \tau \right ) d\tau \right ] +\cdots \tag{1} \end{equation} Taking the time derivative of the above and using the product rule \(\frac{d}{dt}\left ( XY\right ) =X\frac{d}{dt}Y+Y\frac{d}{dt}X\) gives\begin{align*} \frac{d}{dt}\Psi \left ( t\right ) & =\frac{d}{dt}\left ( e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }\right ) \\ & =\frac{d}{dt}\left ( I+\int _{0}^{t}A\left ( \tau \right ) d\tau +\frac{1}{2}\left [ \int _{0}^{t}A\left ( \tau \right ) d\tau \int _{0}^{t}A\left ( \tau \right ) d\tau \right ] +\cdots \right ) \\ & =\overset{0}{\overbrace{\frac{d}{dt}I}}+\frac{d}{dt}\int _{0}^{t}A\left ( \tau \right ) d\tau +\frac{1}{2}\frac{d}{dt}\left [ \int _{0}^{t}A\left ( \tau \right ) d\tau \int _{0}^{t}A\left ( \tau \right ) d\tau \right ] +\cdots \\ & =A\left ( t\right ) +\frac{1}{2}\left [ \left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) A\left ( t\right ) +A\left ( t\right ) \left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) \right ] +\cdots \end{align*}
Taking advantage of the commute property we write the second term above as\[ \frac{d}{dt}\Psi \left ( t\right ) =A\left ( t\right ) +\frac{1}{2}\left [ A\left ( t\right ) \left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) +A\left ( t\right ) \left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) \right ] +\cdots \] Therefore\begin{align*} \frac{d}{dt}\Psi \left ( t\right ) & =A\left ( t\right ) +\frac{1}{2}\left [ 2A\left ( t\right ) \left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) \right ] +\cdots \\ & =A\left ( t\right ) +A\left ( t\right ) \int _{0}^{t}A\left ( \tau \right ) d\tau +\cdots \end{align*}
Since all the \(A\left ( t\right ) \) are on the left side, we can now factor \(A\left ( t\right ) \) out and obtain\[ \frac{d}{dt}\Psi \left ( t\right ) =A\left ( t\right ) \overset{\Psi \left ( t\right ) }{\overbrace{\left ( I+\int _{0}^{t}A\left ( \tau \right ) d\tau +\cdots \right ) }}\] Comparing the term inside \(\left ( \cdot \right ) \) in the above expression above with equation (1) we see it is \(\Psi \left ( t\right ) \). (If we have expanded more terms, it would be more clear, but the idea is the same as shown above). Therefore we conclude that\[ \frac{d}{dt}e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }=A\left ( t\right ) e^{^{\int _{0}^{t}A\left ( \tau \right ) d\tau }}\] Or \[ \fbox{$\frac{d}{dt}\Psi \left ( t\right ) =A\left ( t\right ) \Psi \left ( t\right ) $}\] Hence \(\Psi \left ( t\right ) \) satisfies the state equation. Now we need to show that \(x\left ( t\right ) =e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }x\left ( 0\right ) \) is the state solution. Since \(\Psi \left ( t\right ) \) is the fundamental matrix, then each of its columns is an independent solution to \(x^{\prime }=A\left ( t\right ) x\) by definition. Hence a linear combinations of the columns of \(\Psi \left ( t\right ) \) gives the solution \(x\left ( t\right ) \). As shown in class, we now obtain the general solution by assuming \(x\left ( t\right ) =\Psi \left ( t\right ) \theta \left ( t\right ) \) and then from this end up with the fundamental solution \(x\left ( t\right ) \) as\[ \vec{x}\left ( t\right ) =\Psi \left ( t\right ) \Psi ^{-1}\left ( 0\right ) x\left ( 0\right ) +{\displaystyle \int \limits _{0}^{t}} \Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \] But since this is free system, so there is no input \(u\left ( t\right ) \) and since \(\Psi \left ( 0\right ) =I\) then \(\Psi ^{-1}\left ( 0\right ) =I\) and the above reduces to\[ \vec{x}\left ( t\right ) =\Psi \left ( t\right ) x\left ( 0\right ) \] But \(\Psi \left ( t\right ) =e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }\) hence\[ \vec{x}\left ( t\right ) =e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }x\left ( 0\right ) \]
We are told that \[ A\left ( t\right ) A\left ( \tau \right ) =A\left ( \tau \right ) A\left ( t\right ) \] Lets integrate both sides from \(0\) to \(t\) w.r.t to \(\tau \). The equality will remain since we are integrating over the same interval of equal quantities, hence\[{\displaystyle \int \limits _{0}^{t}} A\left ( t\right ) A\left ( \tau \right ) d\tau ={\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) A\left ( t\right ) d\tau \] Now the integral on the LHS has \(A\left ( t\right ) \) which can be taken out of the integral, keeping the order to the left, and the integral on RHS has \(A\left ( t\right ) \) which can now be taken out of the integral, keeping the order to the right, which results in\[ A\left ( t\right ) \left ({\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) d\tau \right ) =\left ({\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) d\tau \right ) A\left ( t\right ) \] But the above is the assumptions we used in part (a). Therefore, \[ A\left ( t\right ) A\left ( \tau \right ) =A\left ( \tau \right ) A\left ( t\right ) \overset{\text{implies}}{\Rightarrow }A\left ( t\right ) \left ({\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) d\tau \right ) =\left ({\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) d\tau \right ) A\left ( t\right ) \] Therefore we can use the same solution found in (a)\[ \vec{x}\left ( t\right ) =e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }x\left ( 0\right ) \]
Since \(A\left ( t\right ) A\left ( \tau \right ) =A\left ( \tau \right ) A\left ( t\right ) \), then from problem (2) we know that \[ \Psi \left ( t\right ) =e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }\] is the fundamental matrix for \(x^{\prime }\left ( t\right ) =A\left ( t\right ) x\left ( t\right ) \). We now need to show that, given that \(A\left ( t\right ) \) has distinct eigenvalues for each \(t\), the fundamental matrix can be written as \[ \Psi \left ( t\right ) =T^{-1}e^{\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau }T \] For some constant matrix \(T\). The important point is that \(T\) must be constant in the above. In addition, we need to show that the above \(\Psi \left ( t\right ) \) satisfies \(\frac{d}{dt}\Psi \left ( t\right ) =A\left ( t\right ) \Psi \left ( t\right ) \).
The first step is to find the constant \(T\) matrix. Since \(A\left ( t\right ) A\left ( \tau \right ) =A\left ( \tau \right ) A\left ( t\right ) \,,\) then by selecting \(\tau =0\), which is the initial time, then \(A\left ( t\right ) A\left ( 0\right ) =A\left ( 0\right ) A\left ( t\right ) \). Therefore, each \(A\left ( t\right ) \) commutes with the same matrix \(A\left ( 0\right ) \). i.e. \(A\left ( t_{1}\right ) \) will commute with \(A\left ( 0\right ) \) and \(A\left ( t_{2}\right ) \) will commute with \(A\left ( 0\right ) \) and so on. But by problem 1, we showed that when two matrices commute, then they have the same eigenvectors. Therefore, we can select the eigenvectors of \(A\left ( 0\right ) \) to use to construct the \(T\) matrix from, by using the \(n\) linearly independent eigenvectors of \(A\left ( 0\right ) \) as the columns of \(T\). Lets call it \(T_{0}\). Therefore, \(T_{0}\) is now constant and do not change. Now that we found a constant \(T_{0}\) matrix to use for diagonalization of each \(A\left ( t\right ) \) matrix, we will show the rest of the solution using \(T_{0}\,.\) Since\[ e^{M}=I+M+\frac{1}{2}M^{2}+\frac{1}{3!}M^{3}+\cdots ={\displaystyle \sum \limits _{i=0}^{\infty }} \frac{M^{i}}{i!}\] Therefore, applying the above to \begin{align*} \Psi \left ( t\right ) & =e^{\int _{0}^{t}A\left ( \tau \right ) d\tau }\\ & ={\displaystyle \sum \limits _{i=0}^{\infty }} \frac{1}{i!}\left ( \int _{0}^{t}A\left ( \tau \right ) d\tau \right ) ^{i} \end{align*}
Since \(A\) has distinct eigenvalues at all time, we can diagonalize it using the constant \(T_{0}\), hence\begin{align*} \Psi \left ( t\right ) & ={\displaystyle \sum \limits _{i=0}^{\infty }} \frac{1}{i!}\left ( \int _{0}^{t}T_{0}^{-1}\Lambda \left ( \tau \right ) T_{0}d\tau \right ) ^{i}\\ & =I+\int _{0}^{t}T_{0}^{-1}\Lambda \left ( \tau \right ) T_{0}d\tau +\frac{1}{2}\int _{0}^{t}T_{0}^{-1}\Lambda \left ( \tau \right ) T_{0}d\tau \int _{0}^{t}T_{0}^{-1}\Lambda \left ( \tau \right ) T_{0}d\tau +\cdots \\ & =I+T_{0}^{-1}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) T_{0}+\frac{1}{2}T_{0}^{-1}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) \left ( T_{0}T_{0}^{-1}\right ) \left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) T_{0}+\cdots \end{align*}
All the inner \(T_{0}T_{0}^{-1}\) result in \(I\) since \(T_{0}\) is invertible, therefore the above become\[ \Psi \left ( t\right ) =I+T_{0}^{-1}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) T_{0}+\frac{1}{2!}T_{0}^{-1}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) ^{2}T_{0}+\frac{1}{3!}T_{0}^{-1}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) ^{3}T_{0}+\cdots \] Pre-multiply both sides by \(T_{0}\)\[ T_{0}\Psi \left ( t\right ) =T_{0}+\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) T_{0}+\frac{1}{2!}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) ^{2}T_{0}+\frac{1}{3!}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) ^{3}T_{0}+\cdots \] Post multiply both sides by \(T_{0}^{-1}\), and again replacing all of the \(T_{0}T_{0}^{-1}\) products with \(I\) gives\begin{align*} T_{0}\Psi \left ( t\right ) T_{0}^{-1} & =I+\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) T_{0}T_{0}^{-1}+\frac{1}{2!}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) ^{2}T_{0}T_{0}^{-1}+\frac{1}{3!}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) ^{3}T_{0}T_{0}^{-1}+\cdots \\ & =I+\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) +\frac{1}{2!}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) ^{2}+\frac{1}{3!}\left ( \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) ^{3}+\cdots \\ & =e^{\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau } \end{align*}
Therefore\begin{equation} \Psi \left ( t\right ) =T_{0}^{-1}e^{\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau }T_{0} \tag{1} \end{equation} Given equation (1), we need now to show that it leads to \(\frac{d}{dt}\Psi \left ( t\right ) =A\left ( t\right ) \Psi \left ( t\right ) \). \begin{align} \frac{d}{dt}\Psi \left ( t\right ) & =\frac{d}{dt}\left ( T_{0}^{-1}e^{\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau }T_{0}\right ) \nonumber \\ & =T_{0}^{-1}\left ( \frac{d}{dt}e^{\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau }\right ) T_{0} \tag{2} \end{align}
Since \(\Lambda \left ( \tau \right ) \) is a diagonal matrix (by definition, it has the eigenvalues on the diagonal), therefore it commutes with another \(\Lambda \left ( t\right ) \) (any diagonal matrix commutes with another diagonal matrix). Hence \begin{equation} \fbox{$\Lambda \left ( \tau \right ) \Lambda \left ( t\right ) =\Lambda \left ( t\right ) \Lambda \left ( \tau \right ) $} \tag{3} \end{equation} What this means is that we can expand \(e^{\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau }\) in power series and simplified as follows\[ e^{\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau }=I+\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\frac{1}{2}\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\frac{1}{3!}\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\cdots \] Substituting this into (2)\begin{align*} \frac{d}{dt}\Psi \left ( t\right ) & =T_{0}^{-1}\left ( \frac{d}{dt}\left [ I+\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\frac{1}{2}\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\frac{1}{3!}\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\cdots \right ] \right ) T_{0}\\ & =T_{0}^{-1}\left ( \left [ \Lambda \left ( t\right ) +\frac{1}{2}\left ( \Lambda \left ( t\right ) \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \Lambda \left ( t\right ) \right ) +\cdots \right ] \right ) T_{0} \end{align*}
Since \(\Lambda \left ( \tau \right ) \) commute, then using (3)\begin{align*} \frac{d}{dt}\Psi \left ( t\right ) & =T_{0}^{-1}\left ( \left [ \Lambda \left ( t\right ) +\frac{1}{2}\left ( \Lambda \left ( t\right ) \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\Lambda \left ( t\right ) \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \right ) +\cdots \right ] \right ) T_{0}\\ & =T_{0}^{-1}\left ( \left [ \Lambda \left ( t\right ) +\Lambda \left ( t\right ) \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\cdots \right ] \right ) T_{0}\\ & =T_{0}^{-1}\left ( \Lambda \left ( t\right ) \left [ I+\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\frac{1}{2}\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\cdots \right ] \right ) T_{0}\\ & =\overset{A(t)}{\overbrace{\left [ T_{0}^{-1}\Lambda \left ( t\right ) T_{0}\right ] }}\overset{\Psi \left ( t\right ) \text{ from (1)}}{\overbrace{T^{-1}\left ( I+\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\frac{1}{2}\int _{0}^{t}\Lambda \left ( \tau \right ) d\tau \int _{0}^{t}\Lambda \left ( \tau \right ) d\tau +\cdots \right ) T_{0}}} \end{align*}
Hence\[ \frac{d}{dt}\Psi \left ( t\right ) =A\left ( t\right ) \Psi \left ( t\right ) \]
For \(A\left ( t\right ) =\begin{pmatrix} -\frac{4}{t} & -\frac{2}{t^{2}}\\ 1 & 0 \end{pmatrix} \), we first need to find the fundamental matrix \(\Psi \left ( t\right ) \) and then \(\Phi \left ( t,\tau \right ) =\Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) \). Let the 2 linearly independent initial conditions be \[ X^{01}=\begin{pmatrix} 1\\ 0 \end{pmatrix} ,X^{02}=\begin{pmatrix} 0\\ 1 \end{pmatrix} \] We know solve \(x^{\prime }=A\left ( t\right ) x\) using both of these initial conditions and obtain two linearly independent solutions to use to construct \(\Psi \left ( t\right ) \) with. Using the first initial conditions \(x_{1}\left ( 1\right ) =1,x_{2}\left ( 1\right ) =0\). The two equations to solve are\begin{align} x_{1}^{\prime } & =-\frac{4}{t}x_{1}-\frac{2}{t^{2}}x_{2}\tag{1}\\ x_{2}^{\prime } & =x_{1}\tag{2} \end{align}
From the second equation\[ \frac{d}{dt}x_{2}=x_{1}\] Integrate both sides\begin{align*} \int _{1}^{t}dx_{2} & =\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \\ x_{2}\left ( t\right ) -x_{2}\left ( 1\right ) & =\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \end{align*}
But \(x_{2}\left ( 1\right ) =0\), hence \(x_{2}\left ( t\right ) =\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \). Substituting this in (1) gives\[ x_{1}^{\prime }=-\frac{4}{t}x_{1}-\frac{2}{t^{2}}\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \] Multiply both sides by \(\frac{t^{2}}{2}\)\[ \frac{t^{2}}{2}x^{\prime }=-2tx_{1}-\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \] Taking derivative of both sides with respect to \(t\) gives\begin{align} tx_{1}^{\prime }+\frac{t^{2}}{2}x_{1}^{\prime \prime } & =-2x_{1}-2tx_{1}^{\prime }-x_{1}\left ( t\right ) \nonumber \\ \frac{t^{2}}{2}x_{1}^{\prime \prime }+3tx_{1}^{\prime }+3x_{1} & =0\nonumber \\ t^{2}x_{1}^{\prime \prime }+6tx_{1}^{\prime }+3x_{1} & =0\tag{3} \end{align}
This second order differential is now solved for \(x_{1}\left ( t\right ) \). The initial conditions is \(x_{1}\left ( 1\right ) =1\) and \(x_{1}^{\prime }\left ( 1\right ) \). However, we do not know \(x_{1}^{\prime }\left ( 1\right ) \), as not given, but we can obtain it from the first equation (1) by noting that at \(t=1\) we find \(x^{\prime }\left ( 1\right ) =-\frac{4}{1}x_{1}\left ( 1\right ) -\frac{2}{1^{2}}x_{2}(1)=-4\). Therefore (3) can now be solved for \(x_{1}\) since we have two initial conditions. Hence the problem to solve is\begin{align*} t^{2}x_{1}^{\prime \prime }+6tx_{1}^{\prime }+6x_{1} & =0\\ x_{1}\left ( 1\right ) & =1\\ x_{1}^{\prime }\left ( 1\right ) & =-4 \end{align*}
Equation (3) is in the form of Euler equation. Euler ODE has solution of the form \(x_{1}\left ( t\right ) =t^{\alpha }\). Substituting this trial solution in (3) gives\begin{align*} t^{2}\left ( \alpha \left ( \alpha -1\right ) t^{\alpha -2}\right ) +6t\alpha t^{\alpha -1}+6t^{\alpha } & =0\\ \alpha \left ( \alpha -1\right ) t^{\alpha }+6\alpha t^{\alpha }+6t^{\alpha } & =0 \end{align*}
For non-trivial solution, and assuming \(t>0\) which is the case here, dividing the above by \(t^{\alpha }\) gives\begin{align*} \alpha \left ( \alpha -1\right ) +6\alpha +6 & =0\\ \alpha ^{2}+5\alpha +6 & =0 \end{align*}
Hence \[ \alpha =\left \{ -2,-3\right \} \] Therefore the solution is a combination of solutions using these, which is\begin{equation} x_{1}\left ( t\right ) =\frac{c_{1}}{t^{2}}+\frac{c_{2}}{t^{3}}\tag{4} \end{equation} Now we apply the initial conditions. At \(t=1\,,x_{1}\left ( 1\right ) =1,\) hence\begin{equation} 1=c_{1}+c_{2}\tag{5} \end{equation} And \[ x_{1}^{\prime }\left ( t\right ) =-2\frac{c_{1}}{t^{3}}-3\frac{c_{2}}{t^{4}}\] And we have \(x_{1}^{\prime }\left ( 1\right ) =-4\) hence\begin{equation} -4=-2c_{1}-3c_{2}\tag{6} \end{equation} We now have (5),(6), which is two equations in two unknowns. The solution is\begin{align*} 1 & =c_{1}+c_{2}\\ -4 & =-2c_{1}-3c_{2} \end{align*}
The solution is: \(c_{1}=-1,c_{2}=2\). Hence the the solution is now found, using (4), it is\[ \fbox{$x_1\left ( t\right ) =\frac{-1}{t^2}+\frac{2}{t^3}$}\] Now that we know \(x_{1}\left ( t\right ) \,\), we can find \(x_{2}\left ( t\right ) \) from \(x_{2}\left ( t\right ) =\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \), therefore\[ x_{2}\left ( t\right ) =\int _{1}^{t}\frac{-1}{\tau ^{2}}+\frac{2}{\tau ^{3}}d\tau \] Hence\[ \fbox{$x_2\left ( t\right ) =\frac{t-1}{t^2}$}\] This gives us the first column of \[ \Psi ^{1}=\begin{pmatrix} \frac{-1}{t^{2}}+\frac{2}{t^{3}}\\ \frac{t-1}{t^{2}}\end{pmatrix} \] Now we need to do the same the \(X^{02}\).
Using the second initial conditions \(x_{1}\left ( 1\right ) =0,x_{2}\left ( 1\right ) =1\). The two equations to solve are\begin{align} x_{1}^{\prime } & =-\frac{4}{t}x_{1}-\frac{2}{t^{2}}x_{2}\tag{1A}\\ x_{2}^{\prime } & =x_{1} \tag{2A} \end{align}
From the second equation\[ \frac{d}{dt}x_{2}=x_{1}\] Integrate both sides\begin{align*} \int _{1}^{t}dx_{2} & =\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \\ x_{2}\left ( t\right ) -x_{2}\left ( 1\right ) & =\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \end{align*}
But \(x_{2}\left ( 1\right ) =1\), hence \(x_{2}\left ( t\right ) =1+\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \). Substituting this in (1A) gives\[ x_{1}^{\prime }=-\frac{4}{t}x_{1}-\frac{2}{t^{2}}\left ( 1+\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \right ) \] Multiply both sides by \(\frac{t^{2}}{2}\)\[ \frac{t^{2}}{2}x^{\prime }=-2tx_{1}-1-\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \] Taking derivative of both sides with respect to \(t\) gives\begin{align*} tx_{1}^{\prime }+\frac{t^{2}}{2}x_{1}^{\prime \prime } & =-2x_{1}-2tx_{1}^{\prime }-x_{1}\left ( t\right ) \\ \frac{t^{2}}{2}x_{1}^{\prime \prime }+3tx_{1}^{\prime }+3x_{1} & =0\\ t^{2}x_{1}^{\prime \prime }+6tx_{1}^{\prime }+3x_{1} & =0 \end{align*}
This is the same second order differential as was found for \(X^{01}\) but the initial conditions are now different. The initial conditions are \(x_{1}\left ( 1\right ) =0\) and \(x_{1}^{\prime }\left ( 1\right ) \). However, we do not know \(x_{1}^{\prime }\left ( 1\right ) \), as not given, but we can obtain it from the first equation (1) by noting that at \(t=1\) we find \(x^{\prime }\left ( 1\right ) =-\frac{4}{1}x_{1}\left ( 1\right ) -\frac{2}{1^{2}}x_{2}(1)=-2\). Therefore (3A) can now be solved for \(x_{1}\) since we have two initial conditions. Hence the problem to solve is\begin{align} t^{2}x_{1}^{\prime \prime }+6tx_{1}^{\prime }+6x_{1} & =0\tag{3A}\\ x_{1}\left ( 1\right ) & =0\nonumber \\ x_{1}^{\prime }\left ( 1\right ) & =-2\nonumber \end{align}
Equation (3A) is in the form of Euler equation. Euler ODE has solution of the form \(x_{1}\left ( t\right ) =t^{\alpha }\). Substituting this trial solution in (3A) gives\begin{align*} t^{2}\left ( \alpha \left ( \alpha -1\right ) t^{\alpha -2}\right ) +6t\alpha t^{\alpha -1}+6t^{\alpha } & =0\\ \alpha \left ( \alpha -1\right ) t^{\alpha }+6\alpha t^{\alpha }+6t^{\alpha } & =0 \end{align*}
For non-trivial solution, and assuming \(t>0\) which is the case here, diving the above by \(t^{\alpha }\) gives\begin{align*} \alpha \left ( \alpha -1\right ) +6\alpha +6 & =0\\ \alpha ^{2}+5\alpha +6 & =0 \end{align*}
Hence \[ \alpha =\left \{ -2,-3\right \} \] Therefore the solution is a combination of solutions using these, which is\begin{equation} x_{1}\left ( t\right ) =\frac{c_{1}}{t^{2}}+\frac{c_{2}}{t^{3}} \tag{4A} \end{equation} Now we apply the initial conditions. At \(t=1\,,x_{1}\left ( 1\right ) =0,\) hence\begin{equation} 0=c_{1}+c_{2} \tag{5A} \end{equation} And \[ x_{1}^{\prime }\left ( t\right ) =-2\frac{c_{1}}{t^{3}}-3\frac{c_{2}}{t^{4}}\] And we have \(x_{1}^{\prime }\left ( 1\right ) =-2\) hence\begin{equation} -2=-2c_{1}-3c_{2} \tag{6A} \end{equation} We now have (5A),(6A), which is two equations in two unknowns. The solution is\begin{align*} 0 & =c_{1}+c_{2}\\ -2 & =-2c_{1}-3c_{2} \end{align*}
The solution is: \(c_{1}=-2,c_{2}=2\). Hence the the solution is now found, using (4A), it is\[ x_{1}\left ( t\right ) =\frac{-2}{t^{2}}+\frac{2}{t^{3}}\] Now that we know \(x_{1}\left ( t\right ) \,\), we can find \(x_{2}\left ( t\right ) \) from \(x_{2}\left ( t\right ) =1+\int _{1}^{t}x_{1}\left ( \tau \right ) d\tau \), therefore\[ x_{2}\left ( t\right ) =1+\int _{1}^{t}\frac{-2}{\tau ^{2}}+\frac{2}{\tau ^{3}}d\tau \] Hence\[ \fbox{$x_2\left ( t\right ) =\frac{2t-1}{t^2}$}\] This gives us the second column of \[ \Psi ^{2}=\begin{pmatrix} \frac{-2}{t^{2}}+\frac{2}{t^{3}}\\ \frac{2t-1}{t^{2}}\end{pmatrix} \] Hence the fundamental matrix is\[ \Psi =\begin{pmatrix} \frac{-1}{t^{2}}+\frac{2}{t^{3}} & \frac{-2}{t^{2}}+\frac{2}{t^{3}}\\ \frac{t-1}{t^{2}} & \frac{2t-1}{t^{2}}\end{pmatrix} \] The inverse is now found. \[ \Psi ^{-1}=\frac{\begin{pmatrix} \frac{2t-1}{t^{2}} & \frac{2}{t^{2}}-\frac{2}{t^{3}}\\ -\frac{t-1}{t^{2}} & \frac{-1}{t^{2}}+\frac{2}{t^{3}}\end{pmatrix} }{1/t^{4}}=\begin{pmatrix} 2t^{3}-t^{2} & 2t^{2}-2t\\ t^{2}-t^{3} & -\frac{t^{2}-t^{3}}{t-t^{2}}\left ( t-2\right ) \end{pmatrix} \] Therefore the state transition function is\begin{align*} \Phi \left ( t,\tau \right ) & =\Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) \\ & =\begin{pmatrix} \frac{-1}{t^{2}}+\frac{2}{t^{3}} & \frac{-2}{t^{2}}+\frac{2}{t^{3}}\\ \frac{t-1}{t^{2}} & \frac{2t-1}{t^{2}}\end{pmatrix}\begin{pmatrix} 2\tau ^{3}-\tau ^{2} & 2\tau ^{2}-2\tau \\ \tau ^{2}-\tau ^{3} & -\frac{\tau ^{2}-\tau ^{3}}{\tau -\tau ^{2}}\left ( \tau -2\right ) \end{pmatrix} \\ & =\begin{pmatrix} -\frac{1}{t^{3}}\tau ^{2}\left ( t-2\tau \right ) & -\frac{2}{t^{3}}\tau \left ( t-\tau \right ) \\ \frac{1}{t^{2}}\tau ^{2}\left ( t-\tau \right ) & -\frac{1}{t^{2}}\tau \left ( \tau -2t\right ) \end{pmatrix} \\ & =\frac{\tau }{t^{2}}\begin{pmatrix} -\frac{\tau ^{2}}{t}\left ( t-2\tau \right ) & -\frac{2}{t}\left ( t-\tau \right ) \\ \tau \left ( t-\tau \right ) & -\left ( \tau -2t\right ) \end{pmatrix} \end{align*}
For \(A\left ( t\right ) =\begin{pmatrix} 2 & -e^{t}\\ e^{t} & 1 \end{pmatrix} \,\)we first need to find the fundamental matrix \(\Psi \left ( t\right ) \) and then \(\Phi \left ( t,\tau \right ) =\Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) \). Let the two linearly independent initial conditions be \[ X^{01}=\begin{pmatrix} 1\\ 0 \end{pmatrix} ,X^{02}=\begin{pmatrix} 1\\ 0 \end{pmatrix} \] We know solve \(x^{\prime }=A\left ( t\right ) x\) using both of these initial conditions and obtain two linearly independent solutions to use to construct \(\Psi \left ( t\right ) \) with. Using the first initial conditions \(x_{1}\left ( 1\right ) =1,x_{2}\left ( 1\right ) =0\). The two equations to solve are\begin{align} x_{1}^{\prime } & =2x_{1}-e^{t}x_{2}\tag{1}\\ x_{2}^{\prime } & =e^{-t}x_{1}+x_{2} \tag{2} \end{align}
Starting with (2), \(x_{2}^{\prime }-x_{2}=e^{-t}x_{1},\) this is in the form \(x^{\prime }+p\left ( t\right ) x=f\left ( t\right ) \) \(,\) hence the integrating factor is \(e^{\int p\left ( t\right ) dt}=e^{^{-\int dt}}=e^{-t}\) and the solution is \[ \frac{d}{dt}\left ( e^{-t}x_{2}\right ) =e^{-t}\left ( e^{-t}x_{1}\right ) \] Integrating both sides\begin{align*} \left [ e^{-\tau }x_{2}\left ( \tau \right ) \right ] _{1}^{t} & =\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \\ e^{-t}x_{2}\left ( t\right ) -\overset{\text{zero}}{\overbrace{e^{-1}x_{2}\left ( 1\right ) }} & =\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \\ e^{-t}x_{2}\left ( t\right ) & =\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \end{align*}
Hence\begin{equation} x_{2}=e^{t}\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \tag{3} \end{equation} Substituting this solution in (1) gives\begin{align*} x_{1}^{\prime } & =2x_{1}-e^{2t}\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \\ e^{-2t}x_{1}^{\prime }-2x_{1}e^{-2t} & =-\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \end{align*}
Differentiating\begin{align*} -2e^{-2t}x_{1}^{\prime }+e^{-2t}x_{1}^{\prime \prime }-2x_{1}^{\prime }e^{-2t}+4x_{1}e^{-2t} & =-e^{-2t}x_{1}\left ( t\right ) \\ e^{-2t}x_{1}^{\prime \prime }-4e^{-2t}x_{1}^{\prime }+5e^{-2t}x_{1} & =0\\ x_{1}^{\prime \prime }-4x_{1}^{\prime }+5x_{1} & =0 \end{align*}
This is a constant coefficient ODE. Its solution can be found from the characteristic polynomial. \(\lambda ^{2}-4\lambda +5=0,\) the solution is \(\left \{ 2+i,2-i=0\right \} \), hence \[ x_{1}=c_{1}e^{\left ( 2+i\right ) t}+c_{2}e^{\left ( 2-i\right ) t}\] Since the roots are complex, this can be written as \(\sin /\cos \), \begin{align*} x_{1} & =c_{1}e^{2t}e^{it}+c_{2}e^{2t}e^{-it}\\ & =e^{2t}\left ( c_{1}e^{it}+c_{2}e^{-it}\right ) \\ & =e^{2t}\left ( c_{1}\left ( \cos t+i\sin t\right ) +c_{2}\left ( \cos t-i\sin t\right ) \right ) \\ & =e^{2t}\left ( \cos t\left ( c_{1}+c_{2}\right ) +\sin t\left ( ic_{1}-ic_{2}\right ) \right ) \end{align*}
Let \(c_{1}+c_{2}=A\) and \(i\left ( c_{1}-c_{2}\right ) =B\), some new constants. Hence the above becomes\begin{equation} x_{1}\left ( t\right ) =e^{2t}\left ( A\cos t+B\sin t\right ) \tag{4} \end{equation} From initial conditions, \(x_{1}\left ( 1\right ) =1\). But we are not given \(x_{1}^{\prime }\left ( 1\right ) \). We can find this from (1) \(x_{1}^{\prime }=2x_{1}-e^{t}x_{2}\) by noting that at \(t=1,\)\begin{align*} x^{\prime }\left ( 1\right ) & =2x_{1}\left ( 1\right ) -e^{1}x_{2}\left ( 1\right ) \\ & =2 \end{align*}
Hence now we have the two initial conditions to find \(A,B\) from (4)\(.\) At \(t=1\), (4) becomes\begin{equation} 1=e^{2}\left ( A\cos 1+B\sin 1\right ) \tag{5} \end{equation} Taking derivative of (4)\[ x_{1}^{\prime }\left ( t\right ) =2e^{2t}\left ( A\cos t+B\sin t\right ) +e^{2t}\left ( -A\sin t+B\cos t\right ) \] And at \(t=1\) this becomes\begin{equation} 2=2e^{2}\left ( A\cos 1+B\sin 1\right ) +e^{2}\left ( -A\sin 1+B\cos 1\right ) \tag{6} \end{equation} From (5),(6) we can solve for \(A,B\), \begin{align*} 1 & =e^{2}\left ( A\cos 1+B\sin 1\right ) \\ 2 & =2e^{2}\left ( A\cos 1+B\sin 1\right ) +e^{2}\left ( -A\sin 1+B\cos 1\right ) \end{align*}
The solution is \(A=\frac{\cos 1}{e^{2}},B=\frac{\sin 1}{e^{2}}.\) Therefore from (4) we obtain\begin{align*} x_{1}\left ( t\right ) & =e^{2t}\left ( \frac{\cos 1\cos t}{e^{2}}+\frac{\sin 1\sin t}{e^{2}}\right ) \\ & =e^{2t}\left ( \frac{\cos 1\cos t+\sin 1\sin t}{e^{2}}\right ) \end{align*}
But \(\cos 1\cos t+\sin 1\sin t=\cos \left ( 1-t\right ) \), hence\[ x_{1}\left ( t\right ) =e^{2\left ( t-1\right ) }\cos \left ( 1-t\right ) \] Now that we found \(x_{1}\left ( t\right ) \) we go to (3) and find \(x_{2}\left ( t\right ) \)\begin{align*} x_{2} & =e^{t}\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \\ & =e^{t}\int _{1}^{t}e^{-2\tau }e^{2\left ( \tau -1\right ) }\cos \left ( 1-\tau \right ) d\tau \\ & =e^{t}\int _{1}^{t}e^{-2}\cos \left ( 1-\tau \right ) d\tau \end{align*}
Hence\[ x_{2}=-e^{t-2}\sin \left ( 1-t\right ) \] Therefore, the first columns of the fundamental matrix is found\[ \Psi ^{1}=\begin{pmatrix} e^{2t-2}\cos \left ( 1-t\right ) \\ -e^{t-2}\sin \left ( 1-t\right ) \end{pmatrix} \] We now find the second column \(\Psi ^{2}\). Using the second initial conditions \(x_{1}\left ( 1\right ) =0,x_{2}\left ( 1\right ) =1\). The two equations to solve are\begin{align} x_{1}^{\prime } & =2x_{1}-e^{t}x_{2}\tag{1A}\\ x_{2}^{\prime } & =e^{-t}x_{1}+x_{2} \tag{2A} \end{align}
Starting with (2), \(x_{2}^{\prime }-x_{2}=e^{-t}x_{1},\) this is in the form \(x^{\prime }+p\left ( t\right ) x=f\left ( t\right ) \) \(,\) hence the integrating factor is \(e^{\int p\left ( t\right ) dt}=e^{^{-\int dt}}=e^{-t}\) and the solution is \[ \frac{d}{dt}\left ( e^{-t}x_{2}\right ) =e^{-t}\left ( e^{-t}x_{1}\right ) \] Integrating both sides\begin{align} \left [ e^{-\tau }x_{2}\left ( \tau \right ) \right ] _{1}^{t} & =\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \nonumber \\ e^{-t}x_{2}\left ( t\right ) -e^{-1}x_{2}\left ( 1\right ) & =\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \nonumber \\ e^{-t}x_{2}\left ( t\right ) -e^{-1} & =\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau \nonumber \\ x_{2}\left ( t\right ) & =e^{t}\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau +e^{t-1} \tag{3A} \end{align}
Substituting this solution in (1A) gives\begin{align*} x_{1}^{\prime } & =2x_{1}-e^{t}\left ( e^{t}\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau +e^{t-1}\right ) \\ x_{1}^{\prime } & =2x_{1}-e^{2t}\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau -e^{2t-1}\\ e^{-2t}x_{1}^{\prime }-2x_{1}e^{-2t} & =-\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau -e^{-1} \end{align*}
Differentiating\begin{align*} -2e^{-2t}x_{1}^{\prime }+e^{-2t}x_{1}^{\prime \prime }-2x_{1}^{\prime }e^{-2t}+4x_{1}e^{-2t} & =-e^{-2t}x_{1}\left ( t\right ) \\ e^{-2t}x_{1}^{\prime \prime }-4e^{-2t}x_{1}^{\prime }+5e^{-2t}x_{1} & =0\\ x_{1}^{\prime \prime }-4x_{1}^{\prime }+5x_{1} & =0 \end{align*}
This is a constant coefficient ODE. Its solution can be found from the characteristic polynomial. \(\lambda ^{2}-4\lambda +5=0,\) the solution is \(\left \{ 2+i,2-i=0\right \} \), hence \[ x_{1}=c_{1}e^{\left ( 2+i\right ) t}+c_{2}e^{\left ( 2-i\right ) t}\] Since the roots are complex, this can be written as \(\sin /\cos \), giving, as above\begin{equation} x_{1}\left ( t\right ) =e^{2t}\left ( A\cos t+B\sin t\right ) \tag{4A} \end{equation} From initial conditions, \(x_{1}\left ( 1\right ) =0\). But we are not given \(x_{1}^{\prime }\left ( 1\right ) \). We can find this from (1A) \(x_{1}^{\prime }=2x_{1}-e^{t}x_{2}\) by noting that at \(t=1,\)\begin{align*} x^{\prime }\left ( 1\right ) & =2x_{1}\left ( 1\right ) -e^{1}x_{2}\left ( 1\right ) \\ & =-e^{1} \end{align*}
Hence now we have the two initial conditions to find \(A,B\) from (4)\(.\) At \(t=1\), (4A) becomes\begin{align} 0 & =e^{2}\left ( A\cos 1+B\sin 1\right ) \nonumber \\ 0 & =A\cos 1+B\sin 1 \tag{5A} \end{align}
Taking derivative of (4A)\[ x_{1}^{\prime }\left ( t\right ) =2e^{2t}\left ( A\cos t+B\sin t\right ) +e^{2t}\left ( -A\sin t+B\cos t\right ) \] And at \(t=1\) this becomes\begin{equation} -e^{1}=2e^{2}\left ( A\cos 1+B\sin 1\right ) +e^{2}\left ( -A\sin 1+B\cos 1\right ) \tag{6A} \end{equation} From (5A),(6A) we can solve for \(A,B\), \begin{align*} 0 & =e^{2}\left ( A\cos 1+B\sin 1\right ) \\ -e^{1} & =2e^{2}\left ( A\cos \left ( 1\right ) +B\sin \left ( 1\right ) \right ) +e^{2}\left ( -A\sin \left ( 1\right ) +B\cos \left ( 1\right ) \right ) \end{align*}
The solution is \(A=\frac{\sin 1}{e},B=\frac{-\cos 1}{e}\). Therefore (4A) becomes\begin{align*} x_{1}\left ( t\right ) & =e^{2t}\left ( A\cos t+B\sin t\right ) \\ & =e^{2t}\left ( \frac{\sin 1\cos t}{e}-\frac{\cos 1\sin t}{e}\right ) \\ & =e^{2t}\left ( \frac{\sin 1\cos t-\cos 1\sin t}{e}\right ) \end{align*}
But \(\sin 1\cos t-\cos 1\sin t=\sin \left ( 1-t\right ) \), hence \[ x_{1}\left ( t\right ) =e^{2t-1}\sin \left ( 1-t\right ) \] Now that we found \(x_{1}\left ( t\right ) \) we go to (3A) and find \(x_{2}\left ( t\right ) \)\begin{align*} x_{2}\left ( t\right ) & =e^{t}\int _{1}^{t}e^{-2\tau }x_{1}\left ( \tau \right ) d\tau +e^{t-1}\\ & =e^{t-1}+e^{t}\int _{1}^{t}e^{-2\tau }e^{2\tau -1}\sin \left ( 1-\tau \right ) d\tau \\ & =e^{t-1}+e^{t}\int _{1}^{t}e^{-1}\sin \left ( 1-\tau \right ) d\tau \\ & =e^{t-1}+e^{t-1}\int _{1}^{t}\sin \left ( 1-\tau \right ) d\tau \\ & =e^{t-1}+e^{t-1}\left ( -1+\cos \left ( 1-t\right ) \right ) \\ & =e^{t-1}\cos \left ( 1-t\right ) \end{align*}
Therefore, the second column of the fundamental matrix is found\[ \Psi ^{2}=\begin{pmatrix} e^{2t-1}\sin \left ( 1-t\right ) \\ e^{t-1}\cos \left ( 1-t\right ) \end{pmatrix} \] Hence the fundamental matrix is\[ \Psi =\begin{pmatrix} e^{2t-2}\cos \left ( 1-t\right ) & e^{2t-1}\sin \left ( 1-t\right ) \\ -e^{t-2}\sin \left ( 1-t\right ) & e^{t-1}\cos \left ( 1-t\right ) \end{pmatrix} \] The inverse is now found. \[ \Psi ^{-1}=\begin{pmatrix} e^{2-2t}\cos \left ( 1-t\right ) & -e^{2-t}\sin \left ( 1-t\right ) \\ e^{1-2t}\sin \left ( 1-t\right ) & e^{1-t}\cos \left ( 1-t\right ) \end{pmatrix} \] Therefore the state transition function, after some simplification, is\begin{align*} \Phi \left ( t,\tau \right ) & =\Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) \\ & =\begin{pmatrix} e^{2t-2}\cos \left ( 1-t\right ) & e^{2t-1}\sin \left ( 1-t\right ) \\ -e^{t-2}\sin \left ( 1-t\right ) & e^{t-1}\cos \left ( 1-t\right ) \end{pmatrix}\begin{pmatrix} e^{2-2\tau }\cos \left ( 1-\tau \right ) & -e^{2-\tau }\sin \left ( 1-\tau \right ) \\ e^{1-2\tau }\sin \left ( 1-\tau \right ) & e^{1-\tau }\cos \left ( 1-\tau \right ) \end{pmatrix} \\ & =\begin{pmatrix} e^{2\left ( t-\tau \right ) }\cos \left ( t-\tau \right ) & -e^{2t-\tau }\sin \left ( t-\tau \right ) \\ e^{t-2\tau }\sin \left ( t-\tau \right ) & e^{t-\tau }\cos \left ( t-\tau \right ) \end{pmatrix} \end{align*}
