1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential equation

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data
    differential equation
    (dr/dy)+r=8 ; r(1)=0.3
    3. The attempt at a solution
    (dr/dy)=(8-r)

    ∫[dr/(8-r)]=∫dy

    ln|8-r|+c1=y+c2 ; k=c2-c1

    ln|8-r|=y+k ; r(1)=0.3
    then take of e^ both sides which gives
    8-r=e^(y+k)
    8-e^(y+k)=r and we know r(1)=0.3 which means 0.3=8-e^(1+k); then solve for k
    k=ln|7.7|-1 then when plug this in to r(y)= 8-e^(y+k) for some reason webassign says
    r(y)= 8-7.7*e^(1-y)

    but i get r(y)=8-7.7*e^(y-1) are they equivalent
     
    Last edited by a moderator: Apr 13, 2013
  2. jcsd
  3. Apr 12, 2013 #2
    what am i doing wrong?
     
  4. Apr 13, 2013 #3
    You have done almost right but check the integral of dr/(8-r).
     
  5. Apr 13, 2013 #4
    i think when i was taking the intgral of dr/(8-r) i was forgetting the absolute value sign for the natural log which is -ln|8-r|+c1=y+c2 ;c2-c1=k

    then giving me: -ln|8-r|=y+k then becomes ln|8-r|=-y-k ; then take e^ of both sides giving me

    e^(ln|8-r|)=e^(-y-k)

    r-8=e^(-y-k) this then becomes

    r=e^(-y-k)+1 we know r(1)=0.3

    0.3=e^(-1-k)+8
    -7.7=e^(-1-k)
    im stuck here because now i cant take the log of both sides becuase of the negative symbol on left hand side.
     
    Last edited: Apr 13, 2013
  6. Apr 13, 2013 #5
    What's the integral of dr/(8+ax) ?
     
  7. Apr 13, 2013 #6
    i did a quick edit check it out i made the -8 in the ln|8-R| into ones now its right but im still stuck
     
  8. Apr 13, 2013 #7

    im stuck at -7.7=e^(-1-k)
     
  9. Apr 13, 2013 #8
    You solved the differential equation correctly.

    Why did you change |8-r| to r-8? :confused:
     
  10. Apr 13, 2013 #9
    becuase when you take the the e of the absolute value of e^(ln|8-x|) i get |x-8|. dont the absolute values do this?
     
  11. Apr 13, 2013 #10
    No, I don't think it changes to |x-8|. Instead of this, you can simply put y=1 and r=0.3 in the equation:
    [tex]-ln|8-x|=y+k[/tex]
    Find out k from here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Differential equation
  1. Differential equations (Replies: 1)

  2. Differential equation (Replies: 4)

  3. Differential Equation (Replies: 12)

  4. Differential Equations (Replies: 1)

Loading...