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$y′ = Ay$ 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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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
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1 Answer 1

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?

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