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Re: Units

  • To: mathgroup at
  • Subject: [mg23810] Re: [mg23718] Units
  • From: "Richard Finley" <rfinley at>
  • Date: Sat, 10 Jun 2000 03:00:19 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

Notice that you have your units wrong...since t is time in seconds, your acceleration should be 
{3 t Meter/Second^3, t^2 Meter/Second^4}  
when you enter the ODE you need to be careful of the units:

DSolve[{x''[t]==3 t Meter/Second^3,y''[t]==t^2 Meter/Second^4, 
x[0 Second]==2 Meter,y[0 Second]==5 Meter,x'[1 Second]==2 
Meter/Second,y'[1 Second]==3 Meter/Second },{x[t],y[t]},t]

This will give the solution:
{{ x[t] -> Meter (4 Second^3 + Second^2 t + t^3)/(2 Second^3), 
y[t]-> Meter (60 Second^4 + 32 Second^3 t + t^4)/{12 Second^4) }}

 Which gives the result in Meter as it should.  When you plug in a value for t...say
t -> 3 Second you will get the correctly dimensioned answer.  
Of course, the easier (and usual) way to do it is to just leave the units out until the end.

regards, RF

>>> "Enrique Cao" <cao at> 06/04/00 11:09PM >>>


I need help in order  to solve this problem.

I have the vector aceleration {3 t Meter/Second^2,t^2 Meter/Second^2}. 
This is a vector dependent of time (t).

I want to get the  position vector with initial conditions {x[0]==2 
Meter,y[0]==5 Meter,x'[1]==2 Meter/Second,y'[1]==3 

I do this with the follow  expression:

<< Miscellaneous`Units`

<< Miscellaneous`SIUnits`

DSolve[{x''[t]==3 t Meter/Second^2,y''[t]==t^2 Meter/Second^2, 
x[0]==2 Meter,y[0]==5 Meter,x'[1]==2 =
Meter/Second,y'[1]==3 Meter/Second },{x[t],y[t]},t]

The solution is

{x[t]->(Meter(4 Second^2+ 8 Second t +t^3 ))/(2 Second^2),

y[t]->Mter(3+(6 t)/Second+t^4/(12 Second^2))}

The unit of position is Meter but the  solution is not correct.

How can I get the correct solution ??

Thank you very much


cao at

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