Question

In Problems $\mathbf{13}\mathbf{-}\mathbf{32}$use variation of parameters to solve the given nonhomogeneous system.

$\mathbf{32}\mathbf{.}\mathbf{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{ccc}3& -1& -1\\ 1& 1& -1\\ 1& -1& 1\end{array}\right)\mathbf{X}\mathbf{+}\left(\begin{array}{c}0\\ t\\ 2e^t\end{array}\right)$

Short answer

The general solution of $$\begin{gathered} \mathrm{X}\text{'}=\left(\begin{array}{ccc}3& -1& -1\\ 1& 1& -1\\ 1& -1& 1\end{array}\right)\mathrm{X}+\left(\begin{array}{c}0\\ \mathrm{t}\\ 2{\mathrm{e}}^{\mathrm{t}}\end{array}\right)\mathrm{is} \\ \mathrm{X}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right){\mathrm{e}}^{\mathrm{t}}+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right){\mathrm{e}}^{2\mathrm{t}}+\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right){\mathrm{e}}^{2\mathrm{t}}+\left(\begin{array}{c}-\frac{1}{2}\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}+2{\mathrm{e}}^{\mathrm{t}}-\frac{3}{4}\\ -\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}+2{\mathrm{e}}^{\mathrm{t}}-1\\ -\frac{1}{2}\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}-\frac{3}{4}\end{array}\right) \end{gathered}$$

Step-by-step solution

Variation of parameters for nonhomogeneous linear systems:

Parameter variation is a generic approach for identifying a specific solution of a differential equation by substituting the constants in the solution of a related (homogeneous) equation with functions and defining these functions so that the original differential equation is fulfilled.

Find the characteristic equation of the coefficient matrix:

We have

$\mathrm{X}\text{'}=\left(\begin{array}{ccc}3& -1& -1\\ 1& 1& -1\\ 1& -1& 1\end{array}\right)\mathrm{X}+\left(\begin{array}{c}0\\ \mathrm{t}\\ 2{\mathrm{e}}^{\mathrm{t}}\end{array}\right)$

Where

$\mathrm{A}=\left(\begin{array}{ccc}3& -1& -1\\ 1& 1& -1\\ 1& -1& 1\end{array}\right)$

and

$\mathrm{F}\left(\mathrm{t}\right)=\left(\begin{array}{c}0\\ \mathrm{t}\\ 2{\mathrm{e}}^{\mathrm{t}}\end{array}\right)$

Obtaining the eigenvalues of the coefficient matrix,

$$\begin{gathered} \det (\mathrm{A}-\mathrm{\lambda I})=0\to \left|\begin{array}{ccc}3-\mathrm{\lambda }& -1& -1\\ 1& 1-\mathrm{\lambda }& -1\\ 1& -1& 1-\mathrm{\lambda }\end{array}\right|=0 \\ (3-\mathrm{\lambda })\left|\begin{array}{cc}1-\mathrm{\lambda }& -1\\ -1& 1-\mathrm{\lambda }\end{array}\right|+\left|\begin{array}{cc}1& -1\\ 1& 1-\mathrm{\lambda }\end{array}\right|-\left|\begin{array}{cc}1& 1-\mathrm{\lambda }\\ 1& -1\end{array}\right|=0 \\ (3-\mathrm{\lambda })[{(1-\mathrm{\lambda })}^2-1]+2-\mathrm{\lambda }+2-\mathrm{\lambda }=0 \\ (3-\mathrm{\lambda })[1-2\mathrm{\lambda }+{\mathrm{\lambda }}^2-1]+4-2\mathrm{\lambda }=0 \\ (3-\mathrm{\lambda })[{\mathrm{\lambda }}^2-2\mathrm{\lambda }]+4-2\mathrm{\lambda }=0 \\ 3{\mathrm{\lambda }}^2-6\mathrm{\lambda }-{\mathrm{\lambda }}^3+2{\mathrm{\lambda }}^2+4-2\mathrm{\lambda }=0 \\ -{\mathrm{\lambda }}^3+5{\mathrm{\lambda }}^2-8\mathrm{\lambda }+4=0 \\ -(\mathrm{\lambda }-1)〖(\mathrm{\lambda }-2){〗}^2=0 \\ \mathrm{So}\mathrm{our}\mathrm{eigenvalues}\mathrm{are}{\mathrm{\lambda }}_1=1,{\mathrm{\lambda }}_2=2.\mathrm{is}\mathrm{an}\mathrm{eigenvalue}\mathrm{of}\mathrm{multiplicity}\mathrm{two}. \end{gathered}$$

Find the eigenvector and corresponding solution vector:

$$\begin{gathered} \mathrm{For}\mathrm{\lambda \_}1=1: \\ (\mathrm{A}-\mathrm{I}\mid 0)=\left(\begin{array}{cccc}3-1& -1& -1& 0\\ 1& 1-1& -1& 0\\ 1& -1& 1-1& 0\end{array}\right) \\ =\left(\begin{array}{cccc}2& -1& -1& 0\\ 1& 0& -1& 0\\ 1& -1& 0& 0\end{array}\right) \end{gathered}$$

From the second and third-row we see that

${\mathrm{k}}_1={\mathrm{k}}_3,"\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{and}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}"{\mathrm{k}}_1={\mathrm{k}}_2$

Choosing ${\mathrm{k}}_1=1$yields ${\mathrm{k}}_2=1$and ${\mathrm{k}}_3=1$. This gives an eigenvector and a corresponding solution vector

${\mathrm{K}}_1=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),{\mathrm{X}}_1=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right){\mathrm{e}}^{\mathrm{t}}$

$\mathrm{For}{\mathrm{\lambda }}_{-}2=2:$

$$\begin{gathered} (\mathrm{A}-2\mathrm{I}\mid 0)=\left(\begin{array}{cccc}3-2& -1& -1& 0\\ 1& 1-2& -1& 0\\ 1& -1& 1-2& 0\end{array}\right) \\ =\left(\begin{array}{cccc}1& -1& -1& 0\\ 1& -1& -1& 0\\ 1& -1& -1& 0\end{array}\right) \end{gathered}$$

Here we have

$k_1-k_2-k_3=0\to k_1=k_2+k_3$

Choosing ${\mathrm{k}}_2=1$and ${\mathrm{k}}_3=0$yields ${\mathrm{k}}_1=1$. This gives an eigenvector and a corresponding solution vector

${\mathrm{K}}_2=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right),{\mathrm{X}}_2=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right){\mathrm{e}}^{2\mathrm{t}}$

Choosing ${\mathrm{k}}_2=1$and ${\mathrm{k}}_3=1$yields . This gives an eigenvector and a corresponding solution vector

${\mathrm{K}}_2=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right),{\mathrm{X}}_2=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right){\mathrm{e}}^{2\mathrm{t}}$

Choosing ${\mathrm{k}}_2=0$and ${\mathrm{k}}_3=1$yields ${\mathrm{k}}_1=1$. This gives an eigenvector and a corresponding solution vector

${\mathrm{K}}_3=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right),{\mathrm{X}}_2=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right){\mathrm{e}}^{2\mathrm{t}}$

Hence, the complementary solution is

$\mathrm{X}={\mathrm{X}}_{\mathrm{c}}+{\mathrm{X}}_{\mathrm{p}}$

localid="1668012318747" $=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right){\mathrm{e}}^{\mathrm{t}}+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right){\mathrm{e}}^{2\mathrm{t}}+\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right){\mathrm{e}}^{2\mathrm{t}}$

Find the matrix of cofactors:

The entries in ${\mathrm{X}}_1$form the first column of $\mathrm{\phi }\left(\mathrm{t}\right)$, and the entries in ${\mathrm{X}}_2$form the second column of $\mathrm{\phi }\left(\mathrm{t}\right)$, and the entries in ${\mathrm{X}}_3$form the third column of $\mathrm{\phi }\left(\mathrm{t}\right)$. Therefore,

$\mathrm{\phi }\left(\mathrm{t}\right)=\left(\begin{array}{ccc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& 0\\ {\mathrm{e}}^{\mathrm{t}}& 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right)$

We want to make sure that $\mathrm{\phi }\left(\mathrm{t}\right)$is an invertible matrix by checking the determinant, where

$$\begin{gathered} \left|\mathrm{\Phi }\right(\mathrm{t}\left)\right|=\left|\begin{array}{ccc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& 0\\ {\mathrm{e}}^{\mathrm{t}}& 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right| \\ \left|\begin{array}{cc}{\mathrm{e}}^{2\mathrm{t}}& 0\\ 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right|-{\mathrm{e}}^{2\mathrm{t}}\left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{t}}& 0\\ {\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right|+{\mathrm{e}}^{2\mathrm{t}}\left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{\mathrm{t}}& 0\end{array}\right| \\ {\mathrm{e}}^{\mathrm{t}}={\mathrm{e}}^{5\mathrm{t}}-{\mathrm{e}}^{5\mathrm{t}}-{\mathrm{e}}^{5\mathrm{t}} \\ =-{\mathrm{e}}^{5\mathrm{t}}\ne 0 \end{gathered}$$

Since the determinant does not equal zero, the matrix is in fact, an invertible matrix.

So now we find the inverse of the matrix $\mathrm{\phi }\left(\mathrm{t}\right)$, since this is a $3\times 3$matrix, finding the inverse will be a bit lengthy, first we transpose the original matrix

localid="1668013490477" ${\mathrm{\Phi }}^{\mathrm{T}}\left(\mathrm{t}\right)=\left(\begin{array}{ccc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& 0\\ {\mathrm{e}}^{2\mathrm{t}}& 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right)$

Then we find the determinant of each of the $2 \times 2$ minor matrices.

$$\begin{gathered} \left|\begin{array}{cc}{\mathrm{e}}^{2\mathrm{t}}& 0\\ 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right|={\mathrm{e}}^{4\mathrm{t}}\left|\begin{array}{cc}{\mathrm{e}}^{2\mathrm{t}}& 0\\ {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right|={\mathrm{e}}^{4\mathrm{t}}\left|\begin{array}{cc}{\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& 0\end{array}\right|=-{\mathrm{e}}^{4\mathrm{t}} \\ \left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{\mathrm{t}}\\ 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right|={\mathrm{e}}^{3\mathrm{t}}\left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right|=0\left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& 0\end{array}\right|=-{\mathrm{e}}^{3\mathrm{t}} \\ \left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& 0\end{array}\right|=-{\mathrm{e}}^{3\mathrm{t}}\left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& 0\end{array}\right|=-{\mathrm{e}}^{3\mathrm{t}}\left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right|=0 \end{gathered}$$

Now we create the matrix of cofactors, we must then reverse the sign of alternating terms of this new matrix,

$\mathrm{Adj}\left(\mathrm{\phi }\right(\mathrm{t}\left)\right)=\left(\begin{array}{ccc}{\mathrm{e}}^{4\mathrm{t}}& {\mathrm{e}}^{4\mathrm{t}}& -{\mathrm{e}}^{4\mathrm{t}}\\ {\mathrm{e}}^{3\mathrm{t}}& 0& -{\mathrm{e}}^{3\mathrm{t}}\\ -{\mathrm{e}}^{3\mathrm{t}}& -{\mathrm{e}}^{3\mathrm{t}}& 0\end{array}\right)\left(\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right)=\left(\begin{array}{ccc}{\mathrm{e}}^{4\mathrm{t}}& -{\mathrm{e}}^{4\mathrm{t}}& -{\mathrm{e}}^{4\mathrm{t}}\\ -{\mathrm{e}}^{3\mathrm{t}}& 0& {\mathrm{e}}^{3\mathrm{t}}\\ -{\mathrm{e}}^{3\mathrm{t}}& {\mathrm{e}}^{3\mathrm{t}}& 0\end{array}\right)$

Finally,

localid="1668016733251" ${\mathrm{\phi }}^{-1}\left(\mathrm{t}\right)=-\frac{1}{{\mathrm{e}}^{5\mathrm{t}}}\left(\begin{array}{ccc}{\mathrm{e}}^{4\mathrm{t}}& -{\mathrm{e}}^{4\mathrm{t}}& -{\mathrm{e}}^{4\mathrm{t}}\\ -{\mathrm{e}}^{3\mathrm{t}}& 0& {\mathrm{e}}^{3\mathrm{t}}\\ -{\mathrm{e}}^{3\mathrm{t}}& {\mathrm{e}}^{3\mathrm{t}}& 0\end{array}\right)=\left(\begin{array}{ccc}-{\mathrm{e}}^{-\mathrm{t}}& {\mathrm{e}}^{-\mathrm{t}}& {\mathrm{e}}^{-\mathrm{t}}\\ {\mathrm{e}}^{-2\mathrm{t}}& 0& -{\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& -{\mathrm{e}}^{-2\mathrm{t}}& 0\end{array}\right)$

Find the particular solution of the system:

Obtaining the particular solution,

$$\begin{gathered} {\mathrm{X}}_{\mathrm{p}}=\mathrm{\Phi }\left(\mathrm{t}\right)\int {\mathrm{\Phi }}^{-1}\left(\mathrm{t}\right)\mathrm{F}\left(\mathrm{t}\right)\mathrm{dt} \\ =\left(\begin{array}{ccc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& 0\\ {\mathrm{e}}^{\mathrm{t}}& 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right)\int \left(\begin{array}{ccc}-{\mathrm{e}}^{-\mathrm{t}}& {\mathrm{e}}^{-\mathrm{t}}& {\mathrm{e}}^{-\mathrm{t}}\\ {\mathrm{e}}^{-2\mathrm{t}}& 0& -{\mathrm{e}}^{-2\mathrm{t}}\\ {\mathrm{e}}^{2\mathrm{t}}& -{\mathrm{e}}^{-2\mathrm{t}}& 0\end{array}\right)\left(\begin{array}{c}0\\ \mathrm{t}\\ 2{\mathrm{e}}^{\mathrm{t}}\end{array}\right)\mathrm{dt} \\ =\left(\begin{array}{ccc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& 0\\ {\mathrm{e}}^{\mathrm{t}}& 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right)\int \left(\begin{array}{c}{\mathrm{te}}^{-\mathrm{t}}+2\\ -2{\mathrm{e}}^{-\mathrm{t}}\\ -{\mathrm{te}}^{-2\mathrm{t}}\end{array}\right)\mathrm{dt} \end{gathered}$$

To integrate

$\int {\mathrm{te}}^{(}-\mathrm{t})\mathrm{dt}$

we will use integration by parts:

$$\begin{gathered} \mathrm{u}=\mathrm{t}\mathrm{dv}={\mathrm{e}}^{-\mathrm{t}} \\ \mathrm{du}=1\mathrm{v}=-{\mathrm{e}}^{-\mathrm{t}} \end{gathered}$$

Where,

$$\begin{gathered} \int {\mathrm{te}}^{-\mathrm{t}}\mathrm{dt}=\mathrm{uv}-\int \mathrm{vdu} \\ =-{\mathrm{te}}^{-\mathrm{t}}+\int {\mathrm{e}}^{-\mathrm{t}}\mathrm{dt} \\ =-{\mathrm{te}}^{-\mathrm{t}}-\mathrm{e}-\mathrm{t} \end{gathered}$$

To integrate

$-\int {\mathrm{te}}^{-2\mathrm{t}}\mathrm{dt}$

We will also use integration by parts:

$$\begin{gathered} \mathrm{u}=-\mathrm{t}\mathrm{dv}={\mathrm{e}}^{-2\mathrm{t}} \\ \mathrm{du}=-1\mathrm{v}=-\frac{1}{2}{\mathrm{e}}^{-2\mathrm{t}} \end{gathered}$$

Where,

$$\begin{gathered} -\int {\mathrm{te}}^{-2\mathrm{t}}\mathrm{dt}=\mathrm{uv}-\int \mathrm{vdu} \\ =\frac{1}{2}{\mathrm{te}}^{-2\mathrm{t}}-\frac{1}{2}\int {\mathrm{e}}^{-2\mathrm{t}}\mathrm{dt} \\ =\frac{1}{2}{\mathrm{te}}^{-2\mathrm{t}}+\frac{1}{4}{\mathrm{e}}^{-2\mathrm{t}} \end{gathered}$$

Find the general solution of a Nonhomogeneous system:

The other terms in the matrix are easy to integrate. Back to obtaining the particular solution,

$$\begin{gathered} {\mathrm{X}}_{\mathrm{p}}=\left(\begin{array}{ccc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& 0\\ {\mathrm{e}}^{\mathrm{t}}& 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right)\int \left(\begin{array}{c}{\mathrm{te}}^{-\mathrm{t}}+2\\ -2{\mathrm{e}}^{-\mathrm{t}}\\ -{\mathrm{te}}^{-2\mathrm{t}}\end{array}\right)\mathrm{dt} \\ =\left(\begin{array}{ccc}{\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{e}}^{\mathrm{t}}& {\mathrm{e}}^{2\mathrm{t}}& 0\\ {\mathrm{e}}^{\mathrm{t}}& 0& {\mathrm{e}}^{2\mathrm{t}}\end{array}\right)\left(\begin{array}{c}-{\mathrm{te}}^{-\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}+2\mathrm{t}\\ 2{\mathrm{e}}^{-\mathrm{t}}\\ \frac{1}{2}{\mathrm{te}}^{-2\mathrm{t}}+\frac{1}{4}{\mathrm{e}}^{-2\mathrm{t}}\end{array}\right) \\ =\left(\begin{array}{c}-\frac{1}{2}\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}+2{\mathrm{e}}^{\mathrm{t}}-\frac{3}{4}\\ -\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}+2{\mathrm{e}}^{\mathrm{t}}-1\\ -\frac{1}{2}\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}-\frac{3}{4}\end{array}\right) \\ =\left(\begin{array}{c}-\frac{1}{2}\\ -1\\ -\frac{1}{2}\end{array}\right)\mathrm{t}+\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right){\mathrm{te}}^{\mathrm{t}}+\left(\begin{array}{c}2\\ 2\\ 0\end{array}\right){\mathrm{e}}^{\mathrm{t}}+\left(\begin{array}{c}-\frac{3}{4}\\ -1\\ -\frac{3}{4}\end{array}\right) \end{gathered}$$

Therefore, the general solution of the system is

$\mathrm{X}={\mathrm{X}}_{\mathrm{c}}+{\mathrm{X}}_{\mathrm{p}}$

$=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right){\mathrm{e}}^{\mathrm{t}}+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right){\mathrm{e}}^{2\mathrm{t}}+\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right){\mathrm{e}}^{2\mathrm{t}}+\left(\begin{array}{c}-\frac{1}{2}\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}+2{\mathrm{e}}^{\mathrm{t}}-\frac{3}{4}\\ -\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}+2{\mathrm{e}}^{\mathrm{t}}-1\\ -\frac{1}{2}\mathrm{t}+2{\mathrm{te}}^{\mathrm{t}}-\frac{3}{4}\end{array}\right)$