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The system I am trying to solve is of this form

$x' =ax+5y$

$y' =ay+2z$

$z' =az$

so finding eigen values we solve

$ \begin{bmatrix} a-\lambda &5 &0 \\ 0 &a-\lambda &2 \\ 0 &0 &a-\lambda \end{bmatrix} $

and we will get $\lambda_{1,2,3}=a$

How do I continue on from here? Will the general solution be of the form

$Y = c_1K_1e^{at}+tc_2K_2e^{at}+t^2c_1K_3e^{at}$, where $K_1=K_2=K_3$ are the same eigen vectors since same eigen values

Note then $K_1=K_2=K_3 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ , which isn't what the answer in the book is :(

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  • $\begingroup$ You'll need to use generalized eigenvectors for repeated eigenvalues $\endgroup$ – Dylan Jun 7 '18 at 3:49
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    $\begingroup$ See that the eigenvalues are identical because the equation for $z$ is independent of the others. You can solve only for $z$ as $z=K\exp at$ and then solve for $y$ the non-homogeneous equation $y'-ay=2K\exp at$ and do the same for $x$. You will not need those fancy matrices to solve your problem :) $\endgroup$ – rafa11111 Jun 7 '18 at 3:49
  • $\begingroup$ @rafa11111 , I only learnt to solve it with matrices :( $\endgroup$ – glockm15 Jun 7 '18 at 3:53
  • $\begingroup$ You should have known something was going wrong even before checking the book’s answer: by definition, the zero vector is never an eigenvector. $\endgroup$ – amd Jun 7 '18 at 7:56
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Here's a guide on how to deal with repeated eigenvalues. The idea is to use generalized eigenvectors such that

$$ \begin{aligned} (\textbf{A}-\lambda\textbf{I})\vec{v}_1 &= \vec{0} \\ (\textbf{A}-\lambda\textbf{I})\vec{v}_2 &= \vec{v}_1 \\ (\textbf{A}-\lambda\textbf{I})\vec{v}_3 &= \vec{v}_2 \end{aligned} \tag{1} $$

EDIT: I made a mistake in my initial answer. The third linearly independent solution isn't a straightforward combination of the powers of $t^ne^{\lambda t}$. However, we can still apply the same technique by generalizing the constants

$$ \vec{r}_3 = \big(\vec{v}_1t^2 + \vec{v}_2 Bt + \vec{v}_3 C\big) e^{\lambda t} $$

Plugging this in, we obtain

\begin{align} \vec{r}_3' &= \big(2\vec{v}_1t + \lambda\vec{v}_1 t^2 + B\vec{v}_2 + \lambda B \vec{v}_2t + \lambda C \vec{v}_3\big)e^{\lambda t} \\ &= \big(\lambda \vec{v_1}\big) t^2 e^{\lambda t} + \big(2 \vec{v}_1 + \lambda B \vec{v}_2 \big)t e^{\lambda t} + \big(B\vec{v}_2 + \lambda C\vec{v}_3 \big) e^{\lambda t} \end{align}

$$ \implies \left\{ \begin{aligned} \textbf{A}\vec{v}_1 &= \lambda \vec{v}_1 \\ B\textbf{A}\vec{v}_2 &= 2 \vec{v}_1 + \lambda B \vec{v}_2 \\ C\textbf{A}\vec{v}_3 &= B\vec{v}_2 + \lambda C\vec{v}_3 \end{aligned} \right. \implies \left\{ \begin{aligned} \big(\textbf{A} - \lambda \textbf{I}\big)\vec{v}_1 &= \vec{0} \\ \big(\textbf{A} - \lambda \textbf{I}\big)\vec{v}_2 &= \frac{2}{B} \vec{v}_1 \\ \big(\textbf{A} - \lambda \textbf{I}\big)\vec{v}_3 &= \frac{B}{C}\vec{v}_2 \end{aligned} \right. $$

To get to the same form as $(1)$, we set $B=C=2$, arriving at the general solution in $(2)$.

$$ \begin{bmatrix}x \\ y \\ z \end{bmatrix} = c_1\vec{v}_1e^{\lambda t} + c_2\big(\vec{v}_1te^{\lambda t} + \vec{v}_2e^{\lambda t}\big) + c_3 \big(\vec{v}_1t^2e^{\lambda t} + \vec{v}_2 2te^{\lambda t} + \vec{v}_3 2 e^{\lambda t}\big) \tag{2} $$


Let's go on to find these vectors. Note the product first $$ (\textbf{A}-\lambda\textbf{I})\vec{v} = \begin{bmatrix} 0 & 5 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5y \\ 2z \\ 0 \end{bmatrix} $$

The first eigenvector satisfies

$$ (\textbf{A}-\lambda\textbf{I})\vec{v}_1 = \begin{bmatrix} 5y_1 \\ 2z_1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

which gives $y_1 = z_1 = 0$. There is no restriction on $x_1$, so we can pick any value we want. Note that we need $x_1 \ne 0$ to avoid a trivial solution, so let's say $x_1 = 1$

Then, the first generalized eigenvector must satisfy

$$ (\textbf{A}-\lambda\textbf{I})\vec{v}_2 = \begin{bmatrix} 5y_2 \\ 2z_2 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$

This time, $y_2=1/5$, $z_2=0$, and again any value for $x_2$ will work. We pick $x_2=0$ since there's no possibility of getting a zero vector.

You can do the last case. The 3 vectors are

$$ \vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} 0 \\ 1/5 \\ 0 \end{bmatrix}, \quad \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1/10 \end{bmatrix} $$

Plugging in $(2)$, we get the full solution $$ \begin{bmatrix}x \\ y \\ z \end{bmatrix} = \begin{bmatrix} c_1 e^{at} + c_2 te^{at} + c_3t^2e^{at} \\ \frac{1}{5}\big(c_2e^{at} + 2c_3te^{at}\big) \\ \frac{1}{5}c_3e^{at} \end{bmatrix} $$

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  • $\begingroup$ where did you find the formula for the generalized solutions. I tried this technique with another problem and checked the answer with wolfram alpha and the answer always differs by what I got for x(t). y(t) and z(t) matches wolfram alpha. $\endgroup$ – glockm15 Jun 7 '18 at 5:48
  • $\begingroup$ Follow the link at the beginning of my answer for a basic explanation. WolframAlpha doesn't always give the answer in a form that you want, i.e. the constants may be redefined $\endgroup$ – Dylan Jun 7 '18 at 6:00
  • $\begingroup$ omm, I guess I might have done something wrong then. In my other problem instead of 5 and 2 I have 1 so I would get $\begin{bmatrix} y \\ z\\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 0 \end{bmatrix}$ then my eigen vectors would be $\begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 1\\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 0\\ 1 \end{bmatrix}$ $\endgroup$ – glockm15 Jun 7 '18 at 6:08
  • $\begingroup$ I've updated my answer. The generalized eigenvectors are still what they are, but some constants in the general solution were inaccurate. You can also look into the Jordan decomposition and its function form for a more straightforward application of the matrix exponential $\endgroup$ – Dylan Jun 7 '18 at 6:46
  • $\begingroup$ My answer only differs from wolfram alpha's by the term $c_3e^at^2$ as it should be $\frac{1}{2}c_3e^at^2$. The new equation (2) still doesn't get me the same answer but it changed what I should be getting for y(t) and z(t). $\endgroup$ – glockm15 Jun 7 '18 at 6:53
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Dylan’s answer takes you through the general method of dealing with eigenvalues for which the geometric multiplicity is less than the algebraic multiplicity, but in this case there’s a much more direct way to find a solution, one that doesn’t require computing any eigenvectors whatsoever.

Note first that since the coefficient matrix $A$ is triangular, its eigenvalues are just its diagonal elements, so they can be read directly from $A$ without going through the usual process of finding the roots of the characteristic polynomial. The only eigenvalue is $a$, so you can decompose $A$ into the sum of the diagonal matrix $aI$ and $N=A-aI$. These two matrices commute, which means that $$e^{tA} = e^{t(aI+N)} = e^{taI}e^{tN}.$$ Now, $N^2\ne0$ and $N^3=0$, so the power series for $e^{tN}$ will have only three terms: $$e^{tN} = I+tN+\frac12 t^2 N^2.$$ On the other hand, $e^{taI}=e^{at}I$, therefore the general solution to the system is $$e^{at}\left(I+tN+\frac12 t^2 N^2\right)\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix} = e^{at} \begin{bmatrix}1 & 5t & 5t^2 \\ 0 & 1 & 2t \\ 0&0&1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix},$$ with the constants determined by the initial conditions, as usual. This likely doesn’t look exactly like the book answer, but keep in mind that the three constants are arbitrary, so you can rename combinations of them as needed.

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  • $\begingroup$ Thanks amd, I think I might stick with Mohammad and rafa's method for now. It appears I don't understand the theory enough right now to solve it any other way. $\endgroup$ – glockm15 Jun 7 '18 at 8:02
  • $\begingroup$ This should be the best answer. $\endgroup$ – Dylan Jun 7 '18 at 11:16
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Another way to solve this system is to solve $$ z'=az$$ and plug the result in $$ y' =ay+2z$$

Then solve $ y' =ay+2z$ for $y$ and plug the results in $$x' =ax+5y$$ and solve for $x$

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