# Lineary DE in Kinematics

1. Jan 14, 2005

### P3X-018

Linear DE in Kinematics

Hey
I would like to know, how it is possible to solve the following
differentialequations

$$\ddot{x}(t)+\alpha \dot{x}(t)^2=0$$

and the one that really gives troubles

$$\ddot{y}(t)+\alpha\dot{y}(t)^2=-g$$

when given that $$x(0)=0$$, $$y(0)=0$$, $$\dot{y}(0)=v_0\sin \theta$$ and $$\dot{x}(0)=v_0\cos \theta$$. Where $$\alpha=k/m$$. The problem in equation 2 is, that i can't even solve the integral

$$\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}$$

Doing the substitution here doesn't get get me anywhere. Maybe someone have made this problem before, it's motion in 2-dimensions, with
airresistance. The 2 equations comes from

$$m\vec{a}=\vec{F}_g - \vec{F}_{airr}$$

Last edited by a moderator: Jan 14, 2005
2. Jan 14, 2005

### dextercioby

HINT:not [\tex],but [/tex]

Daniel.

3. Jan 14, 2005

### dextercioby

2.Write them using the velocity.E.g.
$$m\frac{dv(t)}{dt}-kv^{2}(t)=-mg$$

Can u solve it,now??

Daniel.

Last edited: Jan 14, 2005
4. Jan 14, 2005

### P3X-018

That doesn't help, I started by writing it in that way, but then rewrited it here. The problem is the integral

$$\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}$$

How can you solve this. By using the substitution, you end up with an even more complicated integral, and partial integration doesn't help. Can you that problem? I hope you can help me. Thanks in advance.

5. Jan 14, 2005

### dextercioby

Can't u solve this integral??

$$\int \frac{dv}{-mg+kv^{2}}$$

Daniel.

PS.If u can't,what are u doing solving ODE-s??

6. Jan 14, 2005

### arildno

Check up an arcus tangens and artanh (inverse of hyperbolic tangens)

7. Jan 14, 2005

### dextercioby

Let's not go there,Arildno... :tongue2: Maybe separation into simple fractions...??

Daniel.

8. Jan 14, 2005

### P3X-018

No I can't solve this integral:

$$\int \frac{dv}{-mg+kv^{2}}$$

Because when I substitude lets say

$$u=-mg+kv^{2}$$

then

$$v=\sqrt{\frac{u+mg}{k}}$$

and

$$\frac{du}{dv}=2kv \Leftrightarrow dv=\frac{du}{2kv}$$

But when I use the partial integration technique, I end up with a even more complicated integral.... Try solve it and you will see. Perhaps I should check up an arcus tangens and artanh?

9. Jan 14, 2005

### dextercioby

Do u have any experience doing integrals??In this case,u can decompose the fraction/integrand into 2 simple fractions.

Daniel.

PS.Do you know this integration technique??

10. Jan 14, 2005

### Nylex

If you want to use an inverse hyperbolic/trig integral, don't subsitute like this. What you need to do is make the denominator into the form 1 - av^2 (where a is a constant (can you see how to do that?), then you can look up what this integral is. It's not arctan, as d/dx (arctan x) = 1/(1 + x^2), but I can't remember which one it is.

11. Jan 16, 2005

### urble

looking up integrals ? just draw a right triangle... make pythagoras proud :rofl: [/hint]

if you use trig substitution, you'll get to $$\int\sec\theta\ d\theta = \ln | \sec\theta + \tan\theta |+C$$ which will be the same as if you were to go by seperation.

Last edited: Jan 16, 2005
12. Jan 17, 2005

### P3X-018

Well, I havn't been online for 2 days now, but I have solved the ODE. I just jused arctan Nylex says.

13. Jan 17, 2005

### dextercioby

WHAT??U mean "arctanh" as from "arcus tangens hyperbolicus",right???
As i said,partial fraction would have dunnit much easier.

Daniel.

14. Jan 17, 2005

### P3X-018

No, I used arctan;
$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$$
And yes I sepretated into simple fractions.

Last edited by a moderator: Jan 17, 2005
15. Jan 17, 2005

### dextercioby

Then your result was WRONG,WRONG!!!!!!You should have used "arctanh".

Daniel.

PS.Are u sure it was the integral discussed above,the one with one minus at the denominator??

16. Jan 17, 2005

### dextercioby

$$\frac{1}{1+x^{2}}$$

Daniel.

17. Jan 17, 2005

### P3X-018

I have the calculations in Mathcad, how can I upload a file in here, so that you can see them..

18. Jan 17, 2005

### dextercioby

If it's not more than 50KB,you can post them as an attachement.OOPS,can u transfer them into another format??Attachment manager doesn't support "mathcad/matlab" files,sorry.

Try to post the general idea,at least.

Valid extensions:
"Valid file extensions: bmp doc gif jpe jpeg jpg pdf png psd txt zip"

Daniel.

19. Jan 17, 2005

### P3X-018

Here is how i solve it;

$$m\frac{d}{dt}v_y(t)+kv_y(t)^2=-mg$$

$$\frac{d}{dt}v=-g-\alpha v^2$$

where \alpha = k/m, then

$$\frac{dv}{-g-\alpha v ^2}=dt$$

$$\int\frac{1}{-g-\alpha v^2}dv=t + c$$

$$\frac{1}{-g}\int\frac{1}{1+\frac{\alpha}{g}v^2}dv=t + c$$

$$\frac{1}{-g}\int\frac{1}{1+\left(\sqrt{\frac{\alpha}{g}}v\right)^2}dv=t + c$$

Then i substitude

$$u=\sqrt{\frac{\alpha}{g}}v$$

$$\frac{1}{-g\sqrt{\frac{\alpha}{g}}}\int\frac{1}{1+u^2}du$$

therefor

$$\frac{-1}{\sqrt{g\alpha}}\arctan\left(\sqrt{\frac{\alpha}{g}}v\right)=t+c$$

Where does it go wrong :(?

Last edited by a moderator: Jan 17, 2005
20. Jan 17, 2005

### dextercioby

I thought so.The initial equation is wrong.Gravity and aerodynamic force have opposite signs...

Daniel.

P.S.Redo calculations.