Angelina Lai
Posted on Saturday, 26 April, 2003 - 07:30 pm:
I've got a 2nd order d.e. to solve in the form

x'' + w2 *(x')2 = k

Can this be solved using the auxiliary method?

Kerwin Hui
Posted on Saturday, 26 April, 2003 - 07:40 pm:
If w and k do not depend on t, then the substitution u=x'^2 gives 2x''=du/dx, and you get a 1st order ODE.

Kerwin


Angelina Lai
Posted on Saturday, 26 April, 2003 - 10:01 pm:
Oh yes of course! Thank you very much!
Michael Doré
Posted on Saturday, 26 April, 2003 - 10:14 pm:
In fact the substitution v = x' gives an 1st order ODE for v.
Kerwin Hui
Posted on Saturday, 26 April, 2003 - 10:21 pm:
Yes. The substitution will work as long as k and w have no x dependencies. In general,

f(x'',x',x)=0 - use u=x'^2, du/dx=2x''
f(x'',x',t)=0 - use u=x', u'=x''

Kerwin
Angelina Lai
Posted on Sunday, 27 April, 2003 - 04:58 pm:
No k and w are constants. This is for my M4 coursework which is on the effect of air resistance on falling objects and we are using paper cake cups. In the revised model assuming R = kv2 the new d.e. I obtained came to the form above.

I've substituted v = x' and the d.e. then becomes
v' + w 2 *v2 = g

Using separation of variables I somehow got the following solution
v = w tanh [Aw SQRT(g)]
which is not dependent on time...

Did I make a mistake somewhere in the integration? Please help!!!
Angelina Lai
Posted on Sunday, 27 April, 2003 - 06:05 pm:
Actually it's ok, I've made a stupid mistake but everything is sorted now!

Anyhow, can someone just verify the particular solution with the initial conditions v = 0, x = 0 when t = 0
v=(k/mg)tanh((kg/m)t)


x=log|cosh((kg/m)*t)|/g

It looks rather complicated and I feel a littler nervous about them...

Kerwin Hui
Posted on Sunday, 27 April, 2003 - 09:49 pm:
Errr... what is your w ? Presumably it is k/m.

The form of the solution looks right, just need to check some constants:-

The terminal velocity is sqrt(g/w 2 )=sqrt(mg/k), but your solution gives sqrt(k/mg). (alternatively, use dimension analysis)

Kerwin