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Show that it satisfies the differential equation

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that

    [tex]y=tanh(t)[/tex]

    satisfies the differential equation

    [itex]\frac{dy}{dt}=1-y^{2}[/itex]

    with initial conditions y(0) = 0

    2. Relevant equations



    3. The attempt at a solution

    I'm not sure how to start, but we have only dealt with one D.E., and we had moved all the variables to a respective side then integrated.

    [itex]dy=(1-y^{2})dt[/itex]

    [itex]\frac{dy}{1-y^{2}}=dt[/itex]

    Integrate both sides to get

    [itex]ln|1-y^{2}|=t[/itex]

    Not sure where to go from here
     
  2. jcsd
  3. Jan 25, 2012 #2

    tiny-tim

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    hi crybllrd! :smile:
    nooo :redface:

    d/dy (ln(1 - y2)) = 2y/(1 - y2)

    try again, splitting 1/(1 - y2) into partial fractions first :wink:
     
  4. Jan 26, 2012 #3
    Oh ok , I got it. I am in Calc 2 and when I read Diff. Eq. it just scared me off.
    That was easy enough. It is a seven part question, and the third part looks very similar:

    Show that for the arbitrary constants A and B, the function y=Atanh(Bt) satisfies:

    [itex]\frac{dy}{dt}=AB-\frac{B}{A}y^{2}[/itex]

    I will try to get it to match the form:

    [itex]\int\frac{1}{A^{2}+x^{2}}=\frac{1}{A}tan^{-1}(\frac{x}{A})+c[/itex]
    _______________________________________________________

    [itex]dy=B(A-\frac{1}{A}y^{2})(dt)[/itex]<--- factored out the B and multiplied by dt

    [itex]\frac{dy}{A-\frac{1}{A}y^{2}}=Bdt[/itex]

    It looks like I need to do a little more manipulation/simplification to get the left side where I want it (namely getting rid of the 1/A in the denominator) but I can't get it to the form I want.
     
    Last edited: Jan 26, 2012
  5. Jan 26, 2012 #4

    HallsofIvy

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    No, no, no! Did you not understand what sheriff89 said? You are not asked to solve the differential equation. In Calculus II you wouldn't be expected to do that.
    Differentiate y= Atanh(Bt) and put it and [itex]y^2= A^2tanh^2(Bt)[/itex] into the equation and determine if you get a true equation or not.
     
  6. Jan 26, 2012 #5
    Why would I put the y^2 into the equation and square the other side? And what is a "true equation"?
    I have a feeling these will be on our exam next week, so I want to understand it well.
    Thanks
     
    Last edited: Jan 26, 2012
  7. Jan 26, 2012 #6

    Dick

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    You still aren't getting the point of just substituting into the differential equation to show you have a solution. Go back to the first one. If y=tanh(x) then dy/dx=sech(x)^2. The other side is 1-y^2=1-tanh(x)^2. Aren't they equal? You don't have to solve any ODE's, you just have to show you have a solution.
     
  8. Jan 28, 2012 #7
    Oh ok I get it.
    On part A I algebraically rearranged to get dy/(1-y^2 )=dt, then I integrated both sides to get (after simplification) y= tanh⁡t, done. Just did it the long way, not even thinking because all the other problems were integration.

    then, when I started integrating part b, people starting getting upset :P

    So this is what I have for part b:

    y(t)=Atanh(Bt)
    y(t)'=ABsech2(Bt)

    y(t)'=AB-(B/A)y2

    ABsech2(Bt)=AB-(B/A)Atanh2(Bt)

    B's cancel out


    Asech2(Bt)=A-tanh2(Bt)

    I can see this being true if A=1
    Is that all I need?
    And, where does the y(0)=0 come in to play? I see where that is true, but is that information actually needed to solve it?
     
  9. Jan 28, 2012 #8

    Dick

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    y(t)^2=A^2*tanh(Bt)^2. If you work it out correctly the A factor will cancel as well. And you don't need y(0)=0 to show that. But given the solution, you can say y(0)=0.
     
  10. Jan 28, 2012 #9
    Cool, thanks a lot.
    I think I'm definitely ready for an exam on Monday.

    EDIT: I had a math error on my last post when I substituted for y2. When done properly, all the A's cancelled out.
     
    Last edited: Jan 28, 2012
  11. Jan 29, 2012 #10
    Ugh, this problem is killing me. On a later part, it asks:
    Let v(t) be the velocity of a falling object of mass m that started from rest. For large velocities, air resistance is proportional to the square of velocity v(t)2.
    If we choose coordinates so that v(t)>0 for a falling object, then there is a constant k>0 such that
    [itex]\frac{dv}{dt}=g-\frac{k}{m}v^{2}[/itex]
    Solve for v(t) by applying the rsults of part (b) (what I just worked out) with
    [itex]A=\sqrt{gm/k}, B=\sqrt{gk/m}[/itex]

    I am not sure what "result" he is wanting from part B, I just had to show that a function satisfied a differential, and in doing so I cancelled out all of the constants except for the B within the hyperbolic.
     
  12. Jan 29, 2012 #11

    Dick

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    You know that v(t)=A*tanh(Bt) satisfies [itex]\frac{dv}{dt}=A B - \frac{B}{A}v^{2}[/itex]. So to find a solution to [itex]\frac{dv}{dt}=g-\frac{k}{m}v^{2}[/itex] you just set [itex]A B=g[/itex] and [itex]\frac{B}{A}=\frac{k}{m}[/itex] and solve for A and B. But it looks like they already did that for you as well. So maybe they just want to to substitute those values for A and B into v(t)=A*tanh(Bt)? Or maybe they want you to show explicitly that it works?
     
  13. Jan 29, 2012 #12
    Alright I just did both. Thanks again.
     
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