Thread: Nonlinear System of two ODEs

1. Nonlinear System of two ODEs

Can the system

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

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

be solved explicitly? Where $\displaystyle m$, $\displaystyle g$ and $\displaystyle k$ are constants and the functions $\displaystyle x$ and $\displaystyle y$ depend only on $\displaystyle t$

2. 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.

3. Are you allowed to use a small-angle approximation?

4. 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.

5. 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

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

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

And then I get this

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

Do I treat $\displaystyle v$ as a constant. I'm not sure because $\displaystyle v$ and $\displaystyle u$ are not independent

6. 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?

7. For the first equation we have

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

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

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

8. 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?

9. It's

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

and the first one is

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

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

11. 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)

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

12. I can get it down to

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

13. 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?

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

15. Right. Now what can you say about that equation?

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