Nonlinear System of two ODEs

**fobos3** · July 8th 2010, 12:10 PM

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 $t$

**Ackbeet** · July 8th 2010, 12:32 PM

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.

**Ackbeet** · July 8th 2010, 12:56 PM

Are you allowed to use a small-angle approximation?

**Ackbeet** · July 8th 2010, 01:14 PM

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.

**fobos3** · July 8th 2010, 02:40 PM

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

**Ackbeet** · July 8th 2010, 03:34 PM

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?

**fobos3** · July 8th 2010, 04:12 PM

For the first equation we have

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

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

$\int{\dfrac{1}{u\sqrt{u^2+v^2}}\,du}=-\dfrac{k}{m}t+A$

**Ackbeet** · July 8th 2010, 04:13 PM

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?

**fobos3** · July 8th 2010, 04:15 PM

It's

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

and the first one is

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

**Ackbeet** · July 8th 2010, 04:18 PM

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

**fobos3** · July 8th 2010, 04:23 PM

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)

$u\dfrac{dv}{dt}+gu=\dfrac{du}{dt}v$

**fobos3** · July 8th 2010, 04:26 PM

I can get it down to

$\dfrac{1}{v}\dfrac{dv}{dt}+\dfrac{g}{v}=\dfrac{1}{ u}\dfrac{du}{dt}$

**Ackbeet** · July 8th 2010, 04:27 PM

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?

[Edit]: I see you anticipated me. Excellent. What can you say now?

**fobos3** · July 8th 2010, 04:34 PM

$u\dfrac{dv}{dt}+gu=\dfrac{du}{dt}v$ becomes $\dfrac{1}{v}\dfrac{dv}{dt}+\dfrac{g}{v}=\dfrac{1}{ u}\dfrac{du}{dt}$

**Ackbeet** · July 8th 2010, 04:37 PM

Right. Now what can you say about that equation?

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