# Let A be a 2 × 2 matrix and consider the differential equation [closed]

$y′ = Ay$ where $y = y(t)$ is a vector function of $t$ in the form $y(t) = [y_1(t), y_2(t)]$. Suppose that the matrix $A$ has characteristic polynomial with a root $\lambda$ of multiplicity two. Suppose also that there are linearly independent vectors $\vec{v}_1$ and $\vec{v}_2$ in $\mathbb{R}^2$ such that $$y = e^{λt} \vec{v}_2 + te^{λt} \vec{v}_1$$ is a solution to the differential equation.

Prove that $\vec{v}_1$ and $\vec{v}_2$ must satisfy the equations \begin{eqnarray} A\vec{v}_1 &=& λ\vec{v}_1\\ A\vec{v}_2 &=& λ\vec{v}_2 + \vec{v}_1. \end{eqnarray}

Part B

$$A= \begin{bmatrix}6 & 9\\-1 & 0\end{bmatrix}$$

Show that $CA(λ) = (λ−3)^2$. Find the eigenspace of $A$ for $λ = 3$, and show that it is one dimensional. Find a specific non-zero eigenvector $\vec{v}_1$. Find a vector $\vec{v}_2$ such that $A\vec{v}_2 = 3\vec{v}_2 + \vec{v}_1$.

(c) Use the method of part (a) so give one solution to the differential equation $y′ = Ay$ for $A$ as in part (b). Check your result.

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## closed as off-topic by 900 sit-ups a day, Ivo Terek, Jack D'Aurizio, Pedro Tamaroff, Tunk-FeyAug 11 at 4:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 900 sit-ups a day, Ivo Terek, Jack D'Aurizio, Pedro Tamaroff, Tunk-Fey
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So, have you done anything yourself? –  Pedro Tamaroff Nov 25 '13 at 22:26
I've been working on it for quite a bit now and I'm pretty sure I'm on the wrong track, any nudge in the right direction would be greatly appreciated –  Tony Nov 25 '13 at 22:41
I have written something. Let me know, and try the other two parts. People like to see some effort from your side before they put some effort themselves. –  Pedro Tamaroff Nov 25 '13 at 23:00

We have that $$y'=\lambda e^{λt} v_2+e^{λt} v_1 +\lambda t e^{λt} v_1=e^{\lambda t}(\lambda v_2+v_1)+te^{\lambda t}\lambda v_1$$
One the other hand, $Ay=e^{\lambda t}Av_2+te^{\lambda t}Av_1$. Using $y$ is a solution to $y'=Ay$, what does this give?