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

[Bruno Tolentino]

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:
<br /> \begin{bmatrix}<br /> u&#039;(t)\\ <br /> v&#039;(t)\\<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> y_1(t) &amp; y_2(t) \\<br /> y_1&#039;(t) &amp; y_2&#039;(t) \\<br /> \end{bmatrix}^{-1}<br /> \begin{bmatrix}<br /> 0\\ <br /> x(t)\\<br /> \end{bmatrix} 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:
\begin{bmatrix}<br /> u&#039;(t)\\ <br /> v&#039;(t)\\<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> y_1(t) &amp; y_2(t) \\<br /> y_1&#039;(t) &amp; y_2&#039;(t) \\<br /> \end{bmatrix}^{-1}<br /> \begin{bmatrix}<br /> x_1(t)\\ <br /> x_2(t)\\<br /> \end{bmatrix}

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?

 

[Mark44]

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:
<br /> \begin{bmatrix}<br /> u&#039;(t)\\<br /> v&#039;(t)\\<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> y_1(t) &amp; y_2(t) \\<br /> y_1&#039;(t) &amp; y_2&#039;(t) \\<br /> \end{bmatrix}^{-1}<br /> \begin{bmatrix}<br /> 0\\<br /> x(t)\\<br /> \end{bmatrix} 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:
\begin{bmatrix}<br /> u&#039;(t)\\<br /> v&#039;(t)\\<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> y_1(t) &amp; y_2(t) \\<br /> y_1&#039;(t) &amp; y_2&#039;(t) \\<br /> \end{bmatrix}^{-1}<br /> \begin{bmatrix}<br /> x_1(t)\\<br /> x_2(t)\\<br /> \end{bmatrix}

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).

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)

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?