# Variation of parameters

1. Oct 12, 2015

### 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:
$$\begin{bmatrix} u'(t)\\ v'(t)\\ \end{bmatrix} = \begin{bmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \\ \end{bmatrix}^{-1} \begin{bmatrix} 0\\ x(t)\\ \end{bmatrix}$$ 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:
$$\begin{bmatrix} u'(t)\\ v'(t)\\ \end{bmatrix} = \begin{bmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \\ \end{bmatrix}^{-1} \begin{bmatrix} x_1(t)\\ x_2(t)\\ \end{bmatrix}$$

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?

2. Oct 13, 2015

### Staff: Mentor

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).