Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$y′ = Ay$ 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.

share|improve this question
6  
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
add comment

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?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.