How Does Changing the Right-Hand Side of the Matrix Affect the ODE Solution?

In summary: I am unable to provide a summary for the conversation as the question and the responses do not seem to be coherent or complete. It is unclear what the conversation is trying to convey or what information is being requested.
  • #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?
 
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  • #2
Bruno Tolentino said:
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?
 

Related to How Does Changing the Right-Hand Side of the Matrix Affect the ODE Solution?

1. What is the "Variation of Parameters" method?

The Variation of Parameters method is a technique used in solving ordinary differential equations. It is an alternative to the more commonly used method of undetermined coefficients, and is especially useful for solving non-homogeneous equations.

2. How does the "Variation of Parameters" method work?

The method involves finding a particular solution to a non-homogeneous equation by using a set of functions, called variation parameters, instead of constant coefficients. These parameters are then determined by substituting them into the original equation and solving for their values.

3. When is the "Variation of Parameters" method most useful?

This method is most useful when solving non-homogeneous equations with variable coefficients, where the method of undetermined coefficients would not work. It is also useful when the non-homogeneous term has a particularly complicated form.

4. What are the advantages of using the "Variation of Parameters" method?

One advantage of this method is that it can be used to find a particular solution to a non-homogeneous equation without needing to know the general solution beforehand. This allows for a more efficient and direct approach to solving the equation. Additionally, the variation parameters used in this method can often be chosen in a way that simplifies the calculations.

5. Are there any limitations to the "Variation of Parameters" method?

One limitation of this method is that it can only be used to find a particular solution, not the general solution, to a non-homogeneous equation. Therefore, it may not be applicable in certain situations where the general solution is needed. Additionally, the method can become more complicated when dealing with higher order differential equations or more complex non-homogeneous terms.

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