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2nd order diff Eq help!

  1. Mar 14, 2008 #1
    2nd order diff Eq with t missing

    I am trying to find y as a function of t
    and y'' - y = 0
    The two IV given are y(0) = 7, and y(1) = 5 .. Remark: the initial condition involves values at two points.

    Well since y = {y,y''} and the independent variable t does not appear, I went about it by setting z = y' and trying to reduce it to a first order DE.

    z = dy/dt = y'
    z * dz/dy - y = 0
    z(dz/dy) = y
    Separating the variables and integrating...
    zdz = ydy
    (z^2)/2 = (y^2)/2 + c
    z^2 = (y^2) + C (where C = 2c)
    z = sqrt (y^2 + C)
    Normally here I would use the IVs to determine C before..however both there is no IV involving y'(t)...
    Anyways...
    z = sqrt (y^2 + C)
    y' = sqrt (y^2 + C)
    dy/dt = sqrt (y^2 + C)
    dy/sqrt(y^2 + C) = dt
    So do I just integrate from here? Isn't this a composite function on the left hand side...so the variable y would appear as well as y^2 ?

    How do I solve this problem and get an answer in the form of y(t) = _________________
    Any help would be appreciated.
     
  2. jcsd
  3. Mar 14, 2008 #2
    I am trying to find y as a function of t (Determine y(t) = _____________)
    and y'' - y = 0
    The two IV given are y(0) = 7, and y(1) = 5 .. Remark: the initial condition involves values at two points.

    Well since y = {y,y''} and the independent variable t does not appear, I went about it by setting z = y' and trying to reduce it to a first order DE.

    z = dy/dt = y'
    z * dz/dy - y = 0
    z(dz/dy) = y
    Separating the variables and integrating...
    zdz = ydy
    (z^2)/2 = (y^2)/2 + c
    z^2 = (y^2) + C (where C = 2c)
    z = sqrt (y^2 + C)
    Normally here I would use the IVs to determine C before..however both there is no IV involving y'(t)...
    Anyways...
    z = sqrt (y^2 + C)
    y' = sqrt (y^2 + C)
    dy/dt = sqrt (y^2 + C)
    dy/sqrt(y^2 + C) = dt
    So do I just integrate from here? Isn't this a composite function on the left hand side...so the variable y would appear as well as y^2 ?

    How do I solve this problem and get an answer in the form of y(t) = _________________
    Any help would be appreciated.
     
  4. Mar 14, 2008 #3

    Vid

    User Avatar

    The solution here is going to be an exponential.
    [tex]
    L(y) = y'' - y = 0[/tex]
    [tex]
    L(e^{rt}) = (r^{2} - 1)(e^{rt})
    [/tex]

    That expression is zero when r = [itex]\pm 1[/itex]
    This gives us two different solutions

    [tex]\phi _{1} = e^{t} [/tex]
    [tex]\phi _{2} = e^{-t}
    [/tex]
    It turns out every solution to L(y) = 0 is in the span of these two solutions.
    [tex]
    y = c_{1}\phi _{1} + c_{2}\phi_{2}
    [/tex]
    Use the two initial value points to get two equations in two unknowns then solve the system
     
    Last edited: Mar 14, 2008
  5. Mar 14, 2008 #4
    Can you please explain to me how you got to this step:

    This gives us two different solutions
    phi1 = e^t*phi2 = e^(-t)

    Thanks
     
  6. Mar 14, 2008 #5

    Vid

    User Avatar

    I plugged e^(rt) into L(y). This gave us (r^2 - 1)e^(rt) = 0
    e^(rt) can't be zero so this is only zero when r^2 - 1 = 0 which gives r = 1 or r = -1.
    Taking r = 1 gives us e^t, r = -1 gives us e^-t. The two solutions aren't being multiplied; they're two different equations. I'll edit my post to make that more clear.
     
  7. Mar 14, 2008 #6
    oh, okay thanks, thats what I found confusing it makes sense now.
     
  8. Mar 14, 2008 #7
    Hello cheeee,

    You are on the right track. Indeed you need to integrate this function, but keep in mind that there are two roots, a positive and a negative one, so you have to check two integrals. Now take p.e. the negative one, which was:

    [tex]z=-\sqrt{y^2+A^2}=\frac{dy}{dt}[/tex]

    This gives as an integral:

    [tex]-\int dt= \int \frac{dy}{\sqrt{y^2+A^2}}[/tex]

    Or:

    [tex]-t= ln(y+\sqrt{y^2+A^2})+B[/tex]

    Now, if you rewrite this as:

    [tex]e^{(-t-B)}=y+\sqrt{y^2+A^2}[/tex]

    Take the square, and use this equation as a substitution, you will find y=f(t). There will be two constants involved as it should be. Calculate the same for the positive root, it will be the same and then you can apply the initial values. Note that it is a standard equation you are solving, to be honest I had to think a bit before I could follow your reasoning. To me it is so common that I know the solution by heart. There is a different way of writing the solution if you consider the definition of the hyperbolic sin and cosine function.

    If anything is unclear, please post.
     
  9. Mar 14, 2008 #8
    I have other problems of similar form
    such as 4y'' - 81y = 0
    and 2500y'' + 9y = 0

    Should I go about solving these using the same method?
     
  10. Mar 14, 2008 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's ok for the first one. But the second one will have solutions of the form cos(rt) and sin(rt). Haven't you learned a more general method for handling these types of equations?
     
  11. Mar 14, 2008 #10
    No unfortunately the professor has mentioned anything about it and gave us this web assignment anyways... also we don't use a textbook just his lectures...which why I am utterly confused.. if someone can point me in the direction of a good link that might explain the general methods that would be great.
     
  12. Mar 14, 2008 #11

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    By the way, you don't have an initial condition involving "values at 2 different points". An "initial condition", by defition has all values given at a single point. What you have is a "boundary value condition".

    It's not just a "technical point". "Existance and Uniqueness" for an initial value problem depend entirely upon the equation. "Existance and Uniquenexx" for boundary value problems also depends upon the precise boundary values.
     
  13. Mar 14, 2008 #12

    HallsofIvy

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    Staff Emeritus
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    This same problem appeared in "homework" so I am merging the threads
     
  14. Mar 14, 2008 #13

    CompuChip

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    Yep, that will definitely work and in fact it will work with any equation of the form
    [tex]a y + b y' + c y'' + d y''' + e y'''' + .... = 0[/tex],
    as long as you can solve the polynomial you get for the r in the trial solution [itex]e^{r t}[/itex] (which might be very hard, and impossible to do non-numerically for fifth-order equations)

    But, as Dick said, the second one might give you problems if you are not used to working with complex numbers. If you are, however, you can just use the same method and use Euler's identity
    [tex]e^{i\theta} = \cos\theta + \mathrm i \sin\theta[/tex]
    to write it in terms sines and cosines (which is completely equivalent, but looks better).

    If you're not used to working with complex numbers: just try the first method. If you find there are no solutions for r (usually I call it [itex]\lambda[/itex]) then try plugging in something like
    [tex]y(t) = A \cos(\lambda t) + B \sin(\lambda t)[/tex]
    which seems like a reasonable thing to do, since y'' already looks very much like y (just need to determine the constants)
     
    Last edited: Mar 14, 2008
  15. Mar 14, 2008 #14

    Dick

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    Science Advisor
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    Your examples so far are linear equations with constant coefficients. You might check out http://en.wikipedia.org/wiki/Linear_differential_equation
     
  16. Mar 14, 2008 #15

    Vid

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