# Repeated Eigenvalues Initial Value Problem

If someone could help me step by step in solving this initial value problem. There's a lot I'm confused about since the solution is supposed to be expressed in the form $x_1(t) =$ and $x_2(t) =$: $$x' = \left[\begin{matrix} \frac{22}{3} & \frac{1}{3} \\ -\frac{25}{3} & \frac{32}{3} \\ \end{matrix}\right]$$

$$x(0) = \left[\begin{matrix} 3 \\ 24 \end{matrix}\right]\\$$

The characteristic polynomial is given by $|A - \lambda I| = 0$, yielding:

$$(\lambda-9)^2 = 0 \rightarrow \lambda_{1,2} = 9$$

Next we need to find two linearly independent eigenvectors. We setup and solve $[A - \lambda I]v_1 = 0$, yielding:

$$\begin{bmatrix}-\dfrac{5}{3} & \dfrac{1}{3}\\ -\dfrac{25}{3} & \dfrac{5}{3}\end{bmatrix}v_1 = 0$$

The RREF gives:

$$\begin{bmatrix}1 & -\dfrac{1}{5}\\ 0 & 0 \end{bmatrix}v_1 = 0$$

This gives us an eigenvector of:

$$v_1 = \left(\dfrac{1}{5},1\right)$$

Unfortunately, this yields a single eigenvector, so to find a second one we try:

$$[A - \lambda I]v_2 = v_1$$

This yields (work the details) a second, generalized eigenvector of:

$$v_2 = \left( -\dfrac{3}{25},0 \right)$$

Now, we can write the solution as:

$$X(t) = e^{\lambda t}\left(c_1 v_1 + c_2(v_2 + tv_1)\right)$$

We substitute the initial condition:

$$X(0) = \begin{bmatrix}3 \\ 24 \end{bmatrix}$$

The final solution is (spoiler):

$$X(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = 3e^{9t}\begin{bmatrix} t + 1 \\ 5t+8 \end{bmatrix}$$

• Thank you! A lot simpler than I originally thought! Commented Dec 6, 2013 at 3:43
• Make sure you work each step and understand the details! You are welcome. Regards Commented Dec 6, 2013 at 3:44
• This deserves a TU! +1 And Happy POETS day! Commented Dec 7, 2013 at 0:06
• That's wonderful! Tuesday's just around the corner, you'll be a smashing hit, and then be sure to breathe and enjoy some holiday spirit. Commented Dec 7, 2013 at 0:10