- #1
Jhenrique
- 685
- 4
Given system like ##\begin{bmatrix}
x'(t)\\
y'(t)\\
\end{bmatrix}
=
\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
\begin{bmatrix}
x(t)\\
y(t)\\
\end{bmatrix}## can be rewritten as ##(-1)u''(t) + (bd)u'(t) + (ad+c)u(t) = 0## with ##\begin{bmatrix}
x(t)\\
y(t)\\
\end{bmatrix}
=
\begin{bmatrix}
u(t)\\
u'(t)\\
\end{bmatrix}##
In other words, if a diff equation of 2nd order can be disassembled in a system of 1nd order, thus a system of 1nd order can be assembled in a diff equation of 2nd order, correct? This way, I don't need to find the eigenvalues and eigenvectors...
x'(t)\\
y'(t)\\
\end{bmatrix}
=
\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
\begin{bmatrix}
x(t)\\
y(t)\\
\end{bmatrix}## can be rewritten as ##(-1)u''(t) + (bd)u'(t) + (ad+c)u(t) = 0## with ##\begin{bmatrix}
x(t)\\
y(t)\\
\end{bmatrix}
=
\begin{bmatrix}
u(t)\\
u'(t)\\
\end{bmatrix}##
In other words, if a diff equation of 2nd order can be disassembled in a system of 1nd order, thus a system of 1nd order can be assembled in a diff equation of 2nd order, correct? This way, I don't need to find the eigenvalues and eigenvectors...