System of differential equations using substitution Exact problem statement
Solve the system $\left\{\begin{matrix}
x_{1}'(t)=3x_{1}(t)-2x_{2}(t)+e^{2t},x_{1}(0)=a & \\
x_{2}'(t)=4x_{1}(t)-3x_{2}(t),x_{2}(0)=b
& 
\end{matrix}\right.$
by using the method of diagonalization. Hint: Substitution $x=Tz$
Progress
I have no problem solving the system 
$\left\{\begin{matrix}
x_{1}'(t)=3x_{1}(t)-2x_{2}(t),x_{1}(0)=a & \\
x_{2}'(t)=4x_{1}(t)-3x_{2}(t),x_{2}(0)=b
& 
\end{matrix}\right.$
I want to rewrite as $\boldsymbol{x'=Ax}$But I don't know how to proceed when $e^{2t}$ is added to the first equation. $T$ is probably the matrix with eigenvectors as columns but how does one set up these(and finding the eigenvectors) when $\boldsymbol{A=\begin{Bmatrix}
3 &-2  &? \\ 
 4&-3  &0? 
\end{Bmatrix}}$?
 A: We have:
$$x'(t) = \begin{bmatrix}x'_1(t)\\x'_2(t)\end{bmatrix} = Ax(t) + f(t) = \begin{bmatrix}3 & -2\\4 & -3\end{bmatrix}x(t) + \begin{bmatrix}e^{2t}\\0\end{bmatrix}, x(0) = \begin{bmatrix}a\\b\end{bmatrix}$$
Diagonalize (not always possible) the matrix $A$ and arrive at:
$$A = TDT^{-1} = \begin{bmatrix}1 & 1\\2 & 1\end{bmatrix}\begin{bmatrix}-1 & 0\\0 & 1\end{bmatrix}\begin{bmatrix}-1 & 1\\2 & -1\end{bmatrix}$$
We want to use $x = Tz \implies z=T^{-1}x \implies z'=T^{-1}x'$, which will allow us to decouple the equations. We have:


*

*$x' = A x + f(t)$

*$T^{-1}x' = T^{-1}(TDT^{-1})x + T^{-1}f(t)$

*$z' = Dz + \hat{f}(t),~ \mbox{where}~ \hat{f}(t) = T^{-1}f(t), ~z(0) = T^{-1}x(0)$


These equations are now decoupled and more easily solved. We have:
$$z'_1(t) = -z_1(t)-e^{2t}, z_1(0) = -a + b \\ z'_2(t) = z_2(t) + 2 e^{2t}, z_2(0) = 2a-b$$
You can now find $z(t)$ and then write:
$$x(t) = Tz(t)$$
Can you continue?
The solution will be:
$$x(t) = \begin{bmatrix}x_1(t)\\x_2(t)\end{bmatrix} = \begin{bmatrix}e^{-t} \left(a \left(2 e^{2 t}-1\right)+b \left(-e^{2 t}\right)+b\right)\\e^{-t} \left(2 a \left(e^{2 t}-1\right)-b \left(e^{2 t}-2\right)\right)\end{bmatrix}$$ 
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\vec{\rm r}\pars{t} \equiv { {\rm x}_{1}\pars{t} \choose {\rm x}_{2}\pars{t}}\,,\quad
\vec{\rm r}\pars{0} \equiv {a \choose b}\,,\quad
\vec{\fermi}\pars{t} \equiv { \expo{2t} \choose 0}\,,\quad
A \equiv \pars{\begin{array}{cc}3 & -2 \\ 4 & - 3\end{array}}
\end{align}

$$
\dot{\vec{\rm r}}\pars{t} = A\,\vec{\rm r}\pars{t} + \vec{\fermi}\pars{t}
\quad\imp\quad
\expo{-At}\dot{\vec{\rm r}}\pars{t} - \expo{-At}A\,\vec{\rm r}\pars{t}
= \expo{-At}\vec{\fermi}\pars{t}
$$

$$
\totald{\bracks{\expo{-At}\vec{\rm r}\pars{t}}}{t} = \expo{-At}\vec{\fermi}\pars{t}
\quad\imp\quad
\expo{-At}\vec{\rm r}\pars{t} - \vec{\rm r}\pars{0}
=\int_{0}^{t}\expo{-At'}\vec{\fermi}\pars{t'}\,\dd t'
$$

$$
\vec{\rm r}\pars{t} =\expo{At} \vec{\rm r}\pars{0}
+ \int_{0}^{t}\expo{A\pars{t - t'}}\vec{\fermi}\pars{t'}\,\dd t'
$$

However, $\ds{A^{2} = {\bf 1}}$ where $\ds{\bf 1}$ is the identity matrix.
That leads to $\ds{\totald[2]{\expo{At}}{t} = \expo{At}\quad\imp}$
$$
\expo{At} = \cosh\pars{t} + \sinh\pars{t}A
$$

$$\color{#00f}{%
\vec{\rm r}\pars{t} =\bracks{\cosh\pars{t} + \sinh\pars{t}A}\vec{\rm r}\pars{0}
+ \int_{0}^{t}\bracks{%
\cosh\pars{t - t'} + \sinh\pars{t - t'}A}\vec{\fermi}\pars{t'}\,\dd t'}
$$

It remains to complete the elementary integrals which we left to the OP.
