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Homework Help: Several Differential Equation problems

  1. Nov 8, 2008 #1

    1. 1. The problem statement, all variables and given/known data
      Solve the IVP and find the interval in which the solution exists
      y'=2ty2, y(0)=y0>0

      2. Relevant equations



      3. The attempt at a solution

      y'=2ty2
      y'/y2=2t
      [tex]\int[/tex]y-2y'=[tex]\int[/tex]2t
      -1/y=t2+c[tex]\Rightarrow[/tex] c = -1/y0
      and therefore [tex]y= -1/(t^2-1/y_0)[/tex]

      So it appears that the interval is all real excluding when t2=1/y0. Is this correct?

    2. 1. The problem statement, all variables and given/known data
      A tank contains 100 gallons of water and 50 oz of salt. Water containing a salt concentration of .25(1+.5sint) oz/gal flows into the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate. Find the amount of salt in the tank at any given time.

      2. Relevant equations
      I set up the eq. like so:
      [tex]dQ/dt=r_1a+Qr_2/100[/tex] where r1 is rate in, r2 is rate out, and a is the concentration of salt coming in.

      3. The attempt at a solution
      Q'-2/100Q = .5(1+.5sin(t)) ; u = e-t/50
      (uQ)' = u(.5+.25sin(t))
      [tex]uQ=\int.5+.25sin(t)[/tex]
      uQ = .5t+1/8sin(t2)+c
      Q = et/50(.5t+1/8sin(t2)+c) ; c = 50 (from IC)
      and therefore
      Q = et/50(.5t+1/8sin(t2)+50)


    3. Sorry guys, one more if you will.

      1. The problem statement, all variables and given/known data

      Find the solution of the IVP:
      y''+y'+y=0, y(o)=1, y'(0)=-1

      2. Relevant equations



      3. The attempt at a solution
      [tex]-1/2 \pm (\sqrt{1^2-4(1)(1)})/2[/tex]
      [tex]-.5 \pm \sqrt{3/2}i[/tex] ; let [tex]\mu = \sqrt{3/2}[/tex]
      [tex]y=c_1 e^{(-t/2)}cos(\mu t)+c_2 e^{(-t/2)}sin(\mu t)[/tex]
      [tex]y(0)=1=c_1cos(0)+c_2sin(0)=1\Rightarrow c_1=1[/tex]
      [tex]y'=-\mu sin(\mu t)e^{(-t/2)}-.5e^{(-t/2)}cos(\mu t)+c_2 \mu cos(\mut)e^{(-t/2)}-.4e^{(-t/2)}sin(\mu t)[/tex]
      [tex]y'(0)=-1=-1/2+\mu c_2 \Rightarrow c_2=-.5/\mu[/tex]
      [tex]y=e^{(-t/2)}cos(\mu t)+c_2e^{(-t/2)}sin(\mu t)[/tex] where c2 is defined above


    Thanks a lot for your time. I would really appreciate any replies as to the correctness of the solution, or of I made some mistakes where they might be and how I might go about fixing them. Hopefully I didn't butcher the formatting :).
     
  2. jcsd
  3. Nov 8, 2008 #2

    HallsofIvy

    User Avatar
    Science Advisor


    1. Looks good!

      What happened to the "u" on the right hand side? Shouldn't this be
      [tex]uQ= \int e^{-t/50}(.5+ .25 sin(t))dt[/tex]

      Yes but I don't see anything good accomplished by using [itex]\mu[/itex] rather than [itex]\sqrt{3}/2[/itex]. In particular, using [itex]\sqrt{3}/2[/itex] makes it clearer that [itex]C_2= 1/\sqrt{3}= \sqrt{3}/3[/itex].

     
  4. Nov 8, 2008 #3
    Thanks for the fast reply.

    In question 2 though.. I have no idea how to integrate that though :(. It's been a long while since I've taken my calc classes. Any ideas?

    In question 3, I just used mu because it was easier to write. Quick question though.. [tex]c_2=-1/ \sqrt{3}[/tex] should be [tex]c_2=- \sqrt { 3 } /3[/tex] right?
     
    Last edited: Nov 8, 2008
  5. Nov 8, 2008 #4

    HallsofIvy

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    Science Advisor

    Do you mean
    [tex]uQ= \int e^{-t/50}(.5+ .25 sin(t))dt[/tex]?

    separate it into
    [tex].5\int e^(t/50)dt[tex]
    which is easy and
    [tex].25\int e^{-t/50}sin(t)dt[/tex]
    Use integration by parts for that- you will have use integration by parts twice.

    Yes, I accidently dropped the "-".
     
  6. Nov 8, 2008 #5
    Alright, so doing the integration here's what I get:
    1. [tex]1/2\int e^{-t/50}sin(t)dt[/tex]

    2. [tex]u=e^{-t/50},du=-1/50e^{-t/50},dv=sin(t)dt,v=-cos(t)[/tex]
      [tex]1/2\int e^{-t/50}sin(t)dt=1/2(-e^{-t/50}cos(t)-\int 1/50e^{-t/50}cos(t)dt[/tex]

    3. [tex]u=1/50e^{-t/50},du=-1/2500e^{-t/50}dt,dv=cos(t)dt, v=sin(t)[/tex]
      [tex]1/2e^{-t/50}sin(t)dt=1/2[-e^{-t/50}cos(t)-(1/50e^{-t/50}sin(t)+ \int 1/2500e^{-t/50}sin(t)dt)][/tex]
    4. [tex]\int e^{-t/50}sin(t)dt=-e^{-t/50}cos(t)-1/50e^{-t/50}sin(t)-1/2500\int e^{-t/50}sin(t)dt[/tex]
    5. [tex]0=-e^{-t/50}cos(t)-1/50e^{-t/50}sin(t)-1/2500\int e^{-t/50}sin(t)dt-\int e^{-t/50}sin(t)dt[/tex]
    6. [tex]e^{-t/50}cos(t)+1/50e^{-t/50}sin(t)=\int e^{-t/50}sin(t)dt [-1/2500-1][/tex]

    7. [tex]\frac{e^{-t/50}cos(t)+1/50e^{-t/50}sin(t)+C}{-1/2500-1}=\int e^{-t/50}sin(t)dt[/tex]

    Before I go back and plug this back in, hopefully someone can verify that I did this correctly. Thanks again for everyone's time.
     
  7. Nov 9, 2008 #6
    Bump.

    Still awaiting further guidance, thank you.
     
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