System of differential equations using substitution

Question

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 $\boldsymbol{A=\begin{Bmatrix} 3 &-2 &? \\ 4&-3 &0? \end{Bmatrix}}$?

Answer 1

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:

$$x(t) = \begin{bmatrix}x_1(t)\\x_2(t)\end{bmatrix} = \begin{bmatrix}e^{-t} \left(a \left(2 e^{2 t}-1\right)+b \left(-e^{2 t}\right)+b\right)\\e^{-t} \left(2 a \left(e^{2 t}-1\right)-b \left(e^{2 t}-2\right)\right)\end{bmatrix}$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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.