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I am working on a differential equations problem using Richardson's Arms Race Model (this isn't homework - I'm just curious). I am trying to work through the examples in the lecture located here. However, I am having trouble finding $y$ in the step at the end of the slide (below): enter image description here

I was able to get to the step where $x'' + 4x -5x = 0$. Then, using an integrating factor of $x= e^{rt} $ , I was able to find that $x_1=e^{-5t}$ and $x_2={e^t}$.

This means that $x = Ae^{-5t} + Be^t$.

Now, I am trying to solve for y, and can't get the coefficients straight. So, I am not sure if my approach to solving the equation is wrong or if I am doing something silly algebraically.

Here is what I tried:
If $x = Ae^-{5t} + Be^t$, then $x'=-5Ae^{-5t}+Be^t$. To find y, I need to plug x and x' into the the equation $ y=\frac12 x'+\frac12 x$.

This gives me:

$y = \frac12 (-5Ae^{-5t}+Be^t+Ae^{-5t}+Be^t)$

$y = \frac12 (-4Ae^{-5t}+2Be^t)$

$y = -2Ae^{-5t}+Be^t$

This is different from the equation found for y in the slides. From a differential equations, should I be taking a different approach to find y? I thought I should just be able to plug in x and x'. How did they get $y = -4Ae^{-5t}+Be^t$?

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  • $\begingroup$ Your result is correct, on the slide they obviously forgot to divide by 2 at one point. $\endgroup$ – Lutz Lehmann Nov 27 '16 at 16:28
  • $\begingroup$ @LutzL Thanks! Math has never been my strongest subject, so I didn't want to assume. $\endgroup$ – JustBlossom Nov 27 '16 at 16:33
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Semi-independent verification: The matrix $$ A=\begin{bmatrix}-1&2\\4&-3\end{bmatrix} $$ has the characteristic polynomial $$ z^2-tr(A)z+\det(A)=z^2+4z-5=(z+5)(z-1) $$ with the roots $-5$ and $1$. The eigenvector for $z=-5$ has to solve $4v_x+2v_y=0$ which has as one non-trivial solution $v_x=1,\;v_y=-2$. The eigenvector for $z=1$ has to satisfy $-w_x+w_y=0$ which has as a solution $w_x=1=w_y$.

Thus the solution can be written as $$ \begin{bmatrix}x(t)\\y(t)\end{bmatrix} = A·\begin{bmatrix}1\\-2\end{bmatrix}·e^{-5t} + B·\begin{bmatrix}1\\1\end{bmatrix}·e^t $$ which coincides with your solution.

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