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I would appreciate any assistance working out the first problem:

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Here is what I know so far:

Given the general solution for our system of differential equations, we have eigenvalues of $\lambda_1 = -2$ and $\lambda_2 = -1$. Since we have 2 given eigenvectors, I believe that the two straight line solutions would be $y = -4x$ and $y = \frac{x}{2}$. A simple substitution should assure that these are indeed the straight line solutions. My trouble here is how to come up with an appropriate graph, assuming that all that I said is correct up to this point.

Kyogre
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  • The substitution $s=e^{-t}$ might make these parametric curves easier to recognize, and, for that matter, eliminate the parameter. – amd Dec 06 '17 at 22:49
  • I would turn to software to do that, but I can't visualize it if I write it down by hand. – Kyogre Dec 06 '17 at 22:49
  • The resulting parameterization is a pair of quadratics in $s$, which represents a parabola. – amd Dec 06 '17 at 23:13
  • With that substitution, I get $x(t) = -k_1s^2+2k_2s$ and $y(t) = 4k_1s^2+k_2s$ – Kyogre Dec 06 '17 at 23:26
  • Actually, you’ve now got $x(s)$ and $y(s)$. At any rate, now eliminate $s$ from this system of equations and you’ll get the equation of a conic. You can then use standard methods (such as described here) to determine its nature. – amd Dec 06 '17 at 23:44
  • BTW, in the future please take the time to enter key parts of your question as text. Images are neither searchable nor accessible to people using screen readers. – amd Dec 06 '17 at 23:48
  • Sounds good amd. Yes I will refrain from posting snapshots in the future and will instead type out the problem. – Kyogre Dec 06 '17 at 23:54

1 Answers1

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I would start by making the substitution $s=e^{-t}$ in the solution to get the parametric equations $$\begin{align} x &= -k_1s^2+2k_2s \\ y &= 4k_1s^2+k_2s. \end{align}$$ This is a parameterization of a parabola, and, as described here, we can read several of its properties directly from this parameterization: its axis is parallel to $[-1,4]^T$ and it passes through the origin with tangent parallel to $[2,1]^T$ there. Not surprisingly, these correspond to the two linear trajectories that you’ve already computed. The “openness” of the parabola and the direction in which it opens is determined by the ratio ${k_2^2\over k_1}$. You can also find the parabola’s vertex by using the condition that its tangent there is orthogonal to its axis.

If you didn’t recognize the parametric equations as that of a parabola, you could produce a Cartesian equation for the curve by eliminating $s$. You can do this by proceeding through a series of rearrangements and back-substitutions, or use the method of resultants illustrated here: a Cartesian equation of the curve is $$\det\begin{bmatrix} -k_1 & 2k_2 &-x & 0 \\ 0 & -k_1 & 2k_2 & -x \\ 4k_1 & k_2 & -y & 0 \\ 0 & 4k_1 & k_2 & -y \end{bmatrix} = k_1(16k_1x^2+8k_1xy+k_1y^2+9k_2^2x-18k_2^2y) = 0,$$ which is some sort of conic section. For $k_1\ne0$ the discriminant of this conic vanishes, so it represents a (possibly degenerate) parabola. (You’ve already found that for $k_1=0$ the trajectory is a line.) The direction of its axis can be found by computing an eigenvector of $0$ of the quadratic part, but in this case the quadratic part is easily factored into $k_1(4x+y)^2$, from which you can see that the axis is perpendicular to $[4,1]^T$, i.e., is parallel to $[-1,4]^T$, just as determined previously. Other attributes of the parabola can also be extracted from the coefficients of the Cartesian equation, but frankly, it’s easier to work directly with the parametric form.

Last, but not least, the exponents in the solution are negative, so the flow along any of the solution curves with increasing $t$ is toward the origin.

amd
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  • Hi amd. The explanation is clear as can be. I will try this general approach to other similar problems so I can get some practice. – Kyogre Dec 07 '17 at 03:59