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Homework Help: 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


    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-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


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

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

    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:
    Find out k from here.
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