Question

Consider the initial value problem

dx/dt = [-9 -3 30 9]x, x(0) = [6 7]

Solve the initial value problem. Give your solution in real form.

x(t) = [ _____ _____]

Answer 1

(a) The eigenvalues are
`$\lambda_1 = 3$` and
`$\lambda_2 = -6$` and the corresponding eigenvectors are
`$\vec{r}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$` and
`$\vec{r}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$`.

(b) The real form is:

`$$x(t) = \begin{bmatrix} 5e^(3t) + 3e^(-6t) \\ 10e^(3t) - 3e^(-6t) \end{bmatrix}$$`

The initial value problem is:

`$(dx)/(dt) = \begin{bmatrix} -9 & -3 \\ 30 & 9 \end{bmatrix} x, \quad x(0) = \begin{bmatrix} 8 \\ 2 \end{bmatrix}$$`

(a) We need to find the eigenvalues and eigenvectors for the coefficient matrix. The eigenvalues are
`$\lambda_1 = 3$` and
`$\lambda_2 = -6$`, and the corresponding eigenvectors are
`$\vec{r}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$` and
`$\vec{r}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$`.

(b) To solve the initial value problem, we can use the general solution of the homogeneous equation:

`$$x(t) = c_1 e^(\lambda_1 t) \vec{r}_1 + c_2 e^(\lambda_2 t) \vec{r}_2$$`

Then we can use the initial condition to find the values of
`$c_1$` and
`$c_2$`. Substituting t = 0
`$x(t) = \begin{bmatrix} 8 \\ 2 \end{bmatrix}$` into the general solution, we get the system of equations:

`\begin{align*}c_1 + c_2 &= 8 \\2c_1 - c_2 &= 2\end{align*}`

Solving this system, we find that $c_1 = 5$ and $c_2 = 3$. Therefore, the solution to the initial value problem is:

`$$x(t) = 5e^(3t) \begin{bmatrix} 1 \\ 2 \end{bmatrix} + 3e^(-6t) \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$`

In real form, this is:

`$$x(t) = \begin{bmatrix} 5e^(3t) + 3e^(-6t) \\ 10e^(3t) - 3e^(-6t) \end{bmatrix}$$`