# Nonlinear System of two ODEs

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• Jul 8th 2010, 12:10 PM
fobos3
Nonlinear System of two ODEs
Can the system

$m\dfrac{d^2 x}{dt^2}=-k \dfrac{dx}{dt} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{d y}{dt}\right)^2}$

$m \dfrac{d^2 y}{dt^2}+mg=-k \dfrac{dy}{dt} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{d y}{dt}\right)^2}$

be solved explicitly? Where $m$, $g$ and $k$ are constants and the functions $x$ and $y$ depend only on $t$
• Jul 8th 2010, 12:32 PM
Ackbeet
You can integrate once immediately by reducing the order. You have no x(t)'s or y(t)'s in there. Once you do that, you might be able to use a polar coordinate transformation to simplify.
• Jul 8th 2010, 12:56 PM
Ackbeet
Are you allowed to use a small-angle approximation?
• Jul 8th 2010, 01:14 PM
Ackbeet
Actually, you don't have to use small-angle approximation. The equations separate out, and you get two separated first-order linear equations. Just solve for the square roots and equate.
• Jul 8th 2010, 02:40 PM
fobos3
Quote:

Originally Posted by Ackbeet
You can integrate once immediately by reducing the order. You have no x(t)'s or y(t)'s in there. Once you do that, you might be able to use a polar coordinate transformation to simplify.

Let

$u=\dfrac{dx}{dt}$

$v=\dfrac{dy}{dt}$

And then I get this

$\int{\dfrac{1}{u\sqrt{u^2+v^2}}\,du}$

Do I treat $v$ as a constant. I'm not sure because $v$ and $u$ are not independent
• Jul 8th 2010, 03:34 PM
Ackbeet
I'm not sure where that integral came from. Your substitution is fine. I would write out your DE's again with that substitution. What do you get?
• Jul 8th 2010, 04:12 PM
fobos3
For the first equation we have

$m\dfrac{du}{dt}=-ku\sqrt{u^2+v^2}$

$\dfrac{1}{u\sqrt{u^2+v^2}}\dfrac{du}{dt}=-\dfrac{k}{m}$

$\int{\dfrac{1}{u\sqrt{u^2+v^2}}\,du}=-\dfrac{k}{m}t+A$
• Jul 8th 2010, 04:13 PM
Ackbeet
I'm not sure I'd be in such a rush to integrate just yet. I like your first equation. What's the analog of that for the second equation?
• Jul 8th 2010, 04:15 PM
fobos3
It's

$m\dfrac{dv}{dt}+mg=-kv\sqrt{u^2+v^2}$

and the first one is

$m\dfrac{du}{dt}=-ku\sqrt{u^2+v^2}$
• Jul 8th 2010, 04:18 PM
Ackbeet
Now, what if you were to solve both equations for the square roots: what would you get?
• Jul 8th 2010, 04:23 PM
fobos3
Quote:

Originally Posted by Ackbeet
Now, what if you were to solve both equations for the square roots: what would you get?

I actually solve for -k (square root)

$u\dfrac{dv}{dt}+gu=\dfrac{du}{dt}v$
• Jul 8th 2010, 04:26 PM
fobos3
I can get it down to

$\dfrac{1}{v}\dfrac{dv}{dt}+\dfrac{g}{v}=\dfrac{1}{ u}\dfrac{du}{dt}$
• Jul 8th 2010, 04:27 PM
Ackbeet
I can see what you did there. Multiplying both equations by what you need in order to get the same thing on the RHS's. That's fine. Now what if you were to divide the entire equation by uv?

: I see you anticipated me. Excellent. What can you say now?
• Jul 8th 2010, 04:34 PM
fobos3
$u\dfrac{dv}{dt}+gu=\dfrac{du}{dt}v$ becomes $\dfrac{1}{v}\dfrac{dv}{dt}+\dfrac{g}{v}=\dfrac{1}{ u}\dfrac{du}{dt}$
• Jul 8th 2010, 04:37 PM
Ackbeet
Right. Now what can you say about that equation?
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