- #1

Bruno Tolentino

- 97

- 0

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:

y

_{1}(t) = exp(a t)

y

_{2}(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) = x

_{1}(t) + x

_{2}(t)

Or like this:

y''(t) - (a + b) y'(t) + (a b) y(t) = x

_{1}(t)

y''(t) - (a + b) y'(t) + (a b) y(t) = x

_{2}(t)

Or other form?