simple autonomous ODE Watch

Chaoslord
Badges: 4
Rep:
?
#1
Report Thread starter 8 years ago
#1
\frac{dx}{dt} = 1

am i right in thinking this is not homogeneous, since the 1 is there

am i right in thinking this is autonomous since the dependent variable does not appear??*

*since this can be written \frac{dx}{dt} = t^0 making it non autonomous but i dont think it matters otherwise \frac{dx}{dt} = 0 could be written \frac{dx}{dt} = 0t i just want to clarify
0
reply
TheEd
Badges: 0
Rep:
?
#2
Report 8 years ago
#2
(Original post by Chaoslord)
\frac{dx}{dt} = 1

am i right in thinking this is not homogeneous, since the 1 is there
Yes it is not homogeneous as the RHS would = 0 if it was.

(Original post by Chaoslord)
am i right in thinking this is autonomous since the dependent variable does not appear??
You are right in saying it is autonomous but your reason is wrong. It is autonomous because the independent variable does not appear.
0
reply
Chaoslord
Badges: 4
Rep:
?
#3
Report Thread starter 8 years ago
#3
(Original post by TheEd)
Yes it is not homogeneous as the RHS would = 0 if it was.


You are right in saying it is autonomous but your reason is wrong. It is autonomous because the independent variable does not appear.

sweet, that was just a typo xD my ts in my example should have made that clear =P but yea thanks
0
reply
RBarack
Badges: 2
Rep:
?
#4
Report 8 years ago
#4
(Original post by Chaoslord)
\frac{dx}{dt} = 1

am i right in thinking this is not homogeneous, since the 1 is there

am i right in thinking this is autonomous since the dependent variable does not appear??*

*since this can be written \frac{dx}{dt} = t^0 making it non autonomous but i dont think it matters otherwise \frac{dx}{dt} = 0 could be written \frac{dx}{dt} = 0t i just want to clarify
dx = dt
integration of dx = integration of dt
Therefore x = t + c.

EDIT: And uh yes it is autonomous because t does not appear on the RHS.
0
reply
DFranklin
Badges: 18
Rep:
?
#5
Report 8 years ago
#5
(Original post by Chaoslord)
\frac{dx}{dt} = 1

am i right in thinking this is not homogeneous, since the 1 is there

am i right in thinking this is autonomous since the dependent variable does not appear??*
Note that there is more than one meaning for "homogeneous differential equation".
0
reply
Chaoslord
Badges: 4
Rep:
?
#6
Report Thread starter 8 years ago
#6
(Original post by DFranklin)
Note that there is more than one meaning for "homogeneous differential equation".
is that the linearity link?

edit: scratch that, just googled - cheers (though i dont think i'll need to know all that for the course i'm doing)
0
reply
Chaoslord
Badges: 4
Rep:
?
#7
Report Thread starter 8 years ago
#7
(Original post by RBarack)
dx = dt
integration of dx = integration of dt
Therefore x = t + c.

EDIT: And uh yes it is autonomous because t does not appear on the RHS.
um, cheers but integrating that wasn't the issue... i made that equation up just to clarify my understanding of autonomous and homogeneous, sorry
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of the West of England, Bristol
    Undergraduate Open Afternoon - Frenchay Campus Undergraduate
    Wed, 23 Jan '19
  • University of East London
    Postgraduate Open Evening Postgraduate
    Wed, 23 Jan '19
  • University of Gloucestershire
    School of Education Open Day Postgraduate
    Wed, 23 Jan '19

Brexit: Given the chance now, would you vote leave or remain?

Remain (1619)
79.13%
Leave (427)
20.87%

Watched Threads

View All
Latest
My Feed