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Okay, so I've been independently trying to study basic systems of differential equations as they relate to dropping a particle into a vector field. I have had two previous posts on the matter trying to intuitively understand what is happening in these cases. The posts can be found here "dropping a particle into a vector field" and here "dropping a particle into a vector field, part 2". So without looking up methods to solving them, I've been able to solve some very basic problem. For example, if I have $$\mathbf{F}(x,y)=x\mathbf{i}+y\mathbf{j}$$ I calculated that at any point $(a,b)$ where I drop my particle at $t=0$, then the path the particle traces over $t$ is $\mathbf{r}(t)=(e^t-1+a)\mathbf{i}+(e^t-1+b)\mathbf{j}$. Similarly, if my vector field is defined by $$\mathbf{F}(x,y)=x\mathbf{i}-y\mathbf{j}$$ Then $\mathbf{r}(t)=(e^t-1+a)\mathbf{i}+(e^{-t}-1+b)\mathbf{j}$ for the same conditions at $\mathbf{r}(0)$ at point $(a,b)$.

Now, I tried a different field defined by $$\mathbf{F}(x,y)=(x+y)\mathbf{i}+(x-y)\mathbf{j}$$ However, here I get the system $$\mathbf{x}'(t)=\mathbf{x}(t)+\mathbf{y}(t)$$ $$\mathbf{y}'(t)=\mathbf{x}(t)-\mathbf{y}(t)$$ If I differentiate the first equation i get $$\mathbf{x}''(t)=\mathbf{x}'(t)+\mathbf{y}'(t)$$ $$=\mathbf{x}(t)+\mathbf{y}(t)+\mathbf{x}(t)-\mathbf{y}(t)$$ $$=2\mathbf{x}(t)$$ Now with a little intuition I figured that a possible solution for $\mathbf{x}(t)$ is $e^{t\sqrt{2}}$ This implies by substitution that $$\mathbf{y}(t)=\sqrt{2}e^{t\sqrt{2}}-e^{t\sqrt{2}}$$ so $$\mathbf{y}'(t)=2e^{t\sqrt{2}}-\sqrt{2}e^{t\sqrt{2}}$$ $$\mathbf{y}'(t)=e^{t\sqrt{2}}+e^{t\sqrt{2}}-\sqrt{2}e^{t\sqrt{2}}$$ $$\mathbf{y}'(t)=e^{t\sqrt{2}}-(\sqrt{2}e^{t\sqrt{2}}-e^{t\sqrt{2}})$$ $$\mathbf{y}'(t)=\mathbf{x}(t)-\mathbf{y}(t)$$ So it works? So is this method of differentiating members of the system work for "simple" fields? basic fields with linear terms?

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Eleven-Eleven
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  • You have ended up with a second-order equation for $x$, so you should write down a solution with two unknown constants. However, since your force field is linear in $x,y$, linear algebra gives a simpler and more general approach to these types of systems... – Rhys Sep 20 '13 at 14:16
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    Okay. I was trying to do a couple of easy ones without the linear algebra and what was described In the second post a matrix exponential. But I guess it's onto linear algebra! – Eleven-Eleven Sep 20 '13 at 14:20
  • You're doing great w/ your self-study. Note that $x''=2x$ is solved by $x(t)=C{\rm e}^{\sqrt{2}t}$ - you have picked $C=1$. You're also missing a constant upon integrating the ODE for $y$, do you see where? Btw, please do post another problem where you try to work w/ linear algebra for more general linear ODEs, I'd love to see how you deal w/ those. – automaton 3 Sep 20 '13 at 23:44

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