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Math Help - Unfamiliar ode

  1. #1
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    Unfamiliar ode

    Here's the ode i set up,
     \frac{d^2 x}{d t^2} + \frac{k}{c -x} =0
    assuming i set it up properly, how would i deal with the x in the denominator?
    Substituting y= c-x gives this:
     \frac {-d^2 y}{dt^2} + \frac{k}{y}=0
    the y in the denominator has me mixed up
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  2. #2
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    Why not just substitute y= (1/c-x) instead, giving:

    (-d2y)/(dt2) + ky = 0
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  3. #3
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    Quote Originally Posted by crymorenoobs View Post
    Why not just substitute y= (1/c-x) instead, giving:

    (-d2y)/(dt2) + ky = 0
    Because that substitution does NOT give that equation.

    If y= \frac{1}{c-x} then c- x= \frac{1}{y}= y^{-1}.

    \frac{d(c-x)}{dt}= -\frac{dx}{dt}= -y^{-2}\frac{dy}{dt}
    -\frac{d^2x}{dt^2}= -y^{-2}\frac{d^2y}{dt^2}+ 2y^{-3}\frac{dy}{dt}.

    with that substitution the equation becomes -y^{-2}\frac{d^2y}{dt^2}+ 2y^{-3}\frac{dy}{dt}+ \frac{k}{y}= 0

    Multiplying by -y^3 makes that
    y\frac{d^2y}{dt^2}- 2\frac{dy}{dt}- ky^2= 0
    which is certainly not an improvement.
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  4. #4
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    That's a case of independent variable missing. So if x''+\frac{k}{c-x}=0 and I let x'=p, then I get the new equation:

    p\frac{dp}{dx}=\frac{k}{x-c}

    with variables separated. That's ok if we get a p^2. Just take it's square root and solve both first-order odes resulting from the root.

    Guess it's kinda' messy though and you'll have to invert the expression and get it in terms of an inverse function.
    Last edited by shawsend; March 12th 2010 at 05:16 AM.
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  5. #5
    MHF Contributor chisigma's Avatar
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    The second order DE has the form...

    x^{''} = f(x) (1)

    ... where...

    f(x)= \frac{k}{x-c} (2)

    The approach to solve (1) is the following: because is...

    x^{''} = \frac{d x^{'}}{dt} = \frac{d x^{'}}{dx}\cdot \frac{dx}{dt} = x^{'}\cdot \frac{dx^{'}}{dx} (3)

    ... the (1) becomes...

    x^{'}\cdot \frac{dx^{'}}{dx}= f(x) \rightarrow x^{'}\cdot dx^{'} = f(x)\cdot dx (4)

    ... so that we have a first order DE in x^{'} and x. Integrating (4) we obtain...

    \frac{x^{'2}}{2} = \phi (x) + c_{1} \rightarrow x^{'} = \sqrt{2\cdot \phi(x) + c_{1}} (5)

    ... where \phi(x) is a primitive of f(x). Separating variables again in (5) we obtain...

    dt = \frac{dx}{ \sqrt{2\cdot \phi(x) + c_{1}}} (6)

    ... so that is...

    t= \int \frac{dx}{ \sqrt{2\cdot \phi(x) + c_{1}}} + c_{2} (7)

    Now what remains to do is compute a primitive of f(x) given by (2), insert it in (7), and then integrating again... very easy ...

    Kind regards

    \chi \sigma
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  6. #6
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    Shawsend and ChiSigma are giving essentially the same answer.
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  7. #7
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    Quote Originally Posted by shawsend View Post
    That's a case of independent variable missing. So if x''+\frac{k}{c-x}=0 and I let x'=p, then I get the new equation:

    p\frac{dp}{dx}=\frac{k}{x-c}

    with variables separated. That's ok if we get a p^2. Just take it's square root and solve both first-order odes resulting from the root.

    Guess it's kinda' messy though and you'll have to invert the expression and get it in terms of an inverse function.
    thanks for the reply,
    here 's the try
     p \frac{dp}{dx} = \frac{k}{x-c}
    and separating variables;
     p dp =k \frac{dx}{x-c}
    integrating:
     \frac{p ^2}{2} = k ln|x-c| + c_1
    and if im still on the right track,
     \frac{dx}{dt} =2 (k ln|x-c| + c_1) ^{\frac{1}{2}}
    which looks like something i need to take a while to work on, did i get you right though? thanks again.
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  8. #8
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    Quote Originally Posted by chisigma View Post
    The second order DE has the form...

    x^{''} = f(x) (1)

    ... where...

    f(x)= \frac{k}{x-c} (2)

    The approach to solve (1) is the following: because is...

    x^{''} = \frac{d x^{'}}{dt} = \frac{d x^{'}}{dx}\cdot \frac{dx}{dt} = x^{'}\cdot \frac{dx^{'}}{dx} (3)

    ... the (1) becomes...

    x^{'}\cdot \frac{dx^{'}}{dx}= f(x) \rightarrow x^{'}\cdot dx^{'} = f(x)\cdot dx (4)

    ... so that we have a first order DE in x^{'} and x. Integrating (4) we obtain...

    \frac{x^{'2}}{2} = \phi (x) + c_{1} \rightarrow x^{'} = \sqrt{2\cdot \phi(x) + c_{1}} (5)

    ... where \phi(x) is a primitive of f(x). Separating variables again in (5) we obtain...

    dt = \frac{dx}{ \sqrt{2\cdot \phi(x) + c_{1}}} (6)

    ... so that is...

    t= \int \frac{dx}{ \sqrt{2\cdot \phi(x) + c_{1}}} + c_{2} (7)

    Now what remains to do is compute a primitive of f(x) given by (2), insert it in (7), and then integrating again... very easy ...

    Kind regards

    \chi \sigma




    thanks, iv not quite encountered primitives but il sure look that up . thanks.
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  9. #9
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    Hi. How about start at the end and work backwards. Suppose I solve a DE for x(t) and I end up with a messy implicit expression in terms of x(t) like \text{myfunction}(x)=t+c and I can't invert it to get x(t). One thing to do is just create a table of x and t values starting with the initial value x(0)=x_0. Something like:

    myInversionTable=Table[{myFunction[x]-c,x},{x,x_0,xend}]

    Now, I've just constructed an "inverse" table of (t,x(t)) values. Now, in the range of values that is one-to-one, I can then fit that table to some function like a polynomial of high degree and that polynomial will probably be a good empirical representation of the solution for some range of values of t.

    So, we have:

    pdp=\frac{k}{x-c}dx

    Let's integrate that explicitly based on the initial conditions x(0)=x_0, x'(0)=x_1:

    \int_{x'(0)}^{x'(t)} pdp=k\int_{x(0)}^{x(t)}\frac{1}{x-c}dx

    \frac{p^2}{2}\biggr|_{x'(0)}^{x'(t)}=k\log(x-c)\biggr|_{x(0)}^{x(t)}

    1/2 x'(t)^2=1/2 x'(0)^2+k\log(x-c)-k\log(x_0-c)

    \left(\frac{dx}{dt}\right)^2=2k\log(x-c)+K_1

    \frac{dx}{dt}=\pm\sqrt{2k\log(x-c)+K_1}

    Now integrate explicitly again:

    \pm\int_{x(0)}^{x(t)}\frac{dx}{\sqrt{2k\log(x-c)+K_1}}= \int_{t_0}^t dt

    \pm\text{myFunction(x)}=t+K_2
    Last edited by shawsend; March 13th 2010 at 04:22 AM. Reason: corrected syntax for table
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  10. #10
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    The second order DE has the form...

    x^{''} = f(x) (1)

    ... where...

    f(x)= \frac{k}{x-c} (2)

    The approach to solve (1) is the following: because is...

    x^{''} = \frac{d x^{'}}{dt} = \frac{d x^{'}}{dx}\cdot \frac{dx}{dt} = x^{'}\cdot \frac{dx^{'}}{dx} (3)

    ... the (1) becomes...

    x^{'}\cdot \frac{dx^{'}}{dx}= f(x) \rightarrow x^{'}\cdot dx^{'} = f(x)\cdot dx (4)

    ... so that we have a first order DE in x^{'} and x. Integrating (4) we obtain...

    \frac{x^{'2}}{2} = \phi (x) + c_{1} \rightarrow x^{'} = \sqrt{2\cdot \phi(x) + c_{1}} (5)

    ... where \phi(x) is a primitive of f(x). Separating variables again in (5) we obtain...

    dt = \frac{dx}{ \sqrt{2\cdot \phi(x) + c_{1}}} (6)

    ... so that is...

    t= \int \frac{dx}{ \sqrt{2\cdot \phi(x) + c_{1}}} + c_{2} (7)

    Now what remains to do is compute a primitive of f(x) given by (2), insert it in (7), and then integrating again... very easy ...

    Kind regards

    \chi \sigma
    Lets try to effectively perform double integration. From (2) we have...

    \phi (x) = \int \frac{k}{x-c}\cdot dx = k\cdot \ln (x+c) (8)

    ... so that (7) becomes...

    t = \frac{1}{\sqrt{2k}}\cdot \int \frac{dx}{\sqrt{\ln (x-c) + \frac{c_{1}}{2k}}}+ c_{2} (9)

    Now first we set \ln (x-c)= y \rightarrow dx= e^{y}\cdot dy so that the integral in (9) becomes...

    \int \frac{dx}{\sqrt{\ln (x-c) + \frac{c_{1}}{2k}}}= \int \frac{e^{y}}{\sqrt{y + \frac{c_{1}}{2k}}}\cdot dy (10)

    ... and then \sqrt{y + \frac{c_{1}}{2k}}= \xi \rightarrow dy = 2\cdot \xi \cdot d\xi so that the integral (10) becomes...

    \int \frac{e^{y}}{\sqrt{y + \frac{c_{1}}{2k}}}\cdot dy= e^{-\frac{c_{1}}{2k}}\cdot \int e^{\xi^{2}}\cdot d\xi = -i\cdot \frac{\sqrt{\pi}}{2}\cdot e^{-\frac{c_{1}}{2k}}\cdot erf (i\cdot \xi) (11)

    ... so that the solution of (1) is...

     t = -\frac{i}{\sqrt{2k}}\cdot \frac{\sqrt{\pi}}{2}\cdot e^{-\frac{c_{1}}{2k}}\cdot erf \{i\cdot \sqrt{\ln (x-c) + \frac{c_{1}}{2k}} \}+ c_{2} (12)

    Kind regards

    \chi \sigma
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