- #1
Bruno Tolentino
- 97
- 0
Given a ODE like this:
y''(t) - (a + b) y'(t) + (a b) y(t) = x(t)
The general solution is: y(t) = A exp(a t) + B exp(b t) + u(t) exp(a t) + v(t) exp(b t)
So, for determine u(t) and v(t), is used the method of variation of parameters:
[tex]
\begin{bmatrix}
u'(t)\\
v'(t)\\
\end{bmatrix}
=
\begin{bmatrix}
y_1(t) & y_2(t) \\
y_1'(t) & y_2'(t) \\
\end{bmatrix}^{-1}
\begin{bmatrix}
0\\
x(t)\\
\end{bmatrix}[/tex] Where:
y1(t) = exp(a t)
y2(t) = exp(b t)
So, my question is: AND IF the matrix equation above woud be like this:
[tex]\begin{bmatrix}
u'(t)\\
v'(t)\\
\end{bmatrix}
=
\begin{bmatrix}
y_1(t) & y_2(t) \\
y_1'(t) & y_2'(t) \\
\end{bmatrix}^{-1}
\begin{bmatrix}
x_1(t)\\
x_2(t)\\
\end{bmatrix}[/tex]
How would be the right side of the ODE for matrix equation above?
Would be like this:
y''(t) - (a + b) y'(t) + (a b) y(t) = x1(t) + x2(t)
Or like this:
y''(t) - (a + b) y'(t) + (a b) y(t) = x1(t)
y''(t) - (a + b) y'(t) + (a b) y(t) = x2(t)
Or other form?
y''(t) - (a + b) y'(t) + (a b) y(t) = x(t)
The general solution is: y(t) = A exp(a t) + B exp(b t) + u(t) exp(a t) + v(t) exp(b t)
So, for determine u(t) and v(t), is used the method of variation of parameters:
[tex]
\begin{bmatrix}
u'(t)\\
v'(t)\\
\end{bmatrix}
=
\begin{bmatrix}
y_1(t) & y_2(t) \\
y_1'(t) & y_2'(t) \\
\end{bmatrix}^{-1}
\begin{bmatrix}
0\\
x(t)\\
\end{bmatrix}[/tex] Where:
y1(t) = exp(a t)
y2(t) = exp(b t)
So, my question is: AND IF the matrix equation above woud be like this:
[tex]\begin{bmatrix}
u'(t)\\
v'(t)\\
\end{bmatrix}
=
\begin{bmatrix}
y_1(t) & y_2(t) \\
y_1'(t) & y_2'(t) \\
\end{bmatrix}^{-1}
\begin{bmatrix}
x_1(t)\\
x_2(t)\\
\end{bmatrix}[/tex]
How would be the right side of the ODE for matrix equation above?
Would be like this:
y''(t) - (a + b) y'(t) + (a b) y(t) = x1(t) + x2(t)
Or like this:
y''(t) - (a + b) y'(t) + (a b) y(t) = x1(t)
y''(t) - (a + b) y'(t) + (a b) y(t) = x2(t)
Or other form?