Variation of parameters

  • #1
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?
 

Answers and Replies

  • #2
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5,764
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?
I don't see how you could get the matrix equation you show, with ##
\begin{bmatrix}
x_1(t)\\
x_2(t)\\
\end{bmatrix}## on the right. The zero term in the
##\begin{bmatrix}
0\\
x(t)\\
\end{bmatrix}##
vector comes from the homogeneous equation, which in this context is y'' -(a + b)y' + (ab)y = 0. The x(t) term in that vector comes from the related nonhomogeneous equation, which is y'' -(a + b)y' + (ab)y = x(t).
Bruno Tolentino said:
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)
Bruno Tolentino said:
The form above doesn't make any sense to me. the expression on the left side can't be equal to two different expressions.
Or other form?
 

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