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Some higher order differential equations

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data

    a) s^2*t''+st*t'=s
    b) y(dx/dy)^2=x^2+1
    c) 17y''''-t^6*y"-4.2y^5=3cost(t)

    2. Relevant equations



    3. The attempt at a solution
    For part a I thought about doing a reduction of order but I can't because I have the variable s present. Not sure what my other options are.

    Part b I solved for dx/dy but now I'm stuck. Can't think clearly

    Part c I'm not even remotely sure on how to do this. I was thinking integrating factor but I don't know.
     
  2. jcsd
  3. Jan 30, 2008 #2

    HallsofIvy

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    You can solve (2) algebraically for dx/dy:
    [tex]\frac{dx}{dy}= \pm \frac{\sqrt{x^2+ 1}}{\sqrt{y}}[/tex]
    and have two separable differential equations.

    As for the other two, they are badly non-linear and I have no clue!
     
  4. Feb 5, 2008 #3
    If the 1st one is

    [tex]s^2\,t''+s\,t\,t'=s\Rightarrow t\,x''+x\,x'=1[/tex]

    then as HallsofIvy said it is non-linear. (I changed the notation a bit).
    The general strategy for these ODE's are to search for Lie point symmetries. Unfortunately this one does not have any :frown:
    A second approach is to search for a first integral of the equation, which is based on a search and a try method.

    For this ODE if you try a first integral of the type

    [tex]I=f(x,t)\,x'+g(x,t)=C[/tex]

    you can find that

    [tex]I=t\,x'+\frac{1}{2}x^2-x-t=C[/tex]

    which is a Riccati type 1st order ODE. With the standard transformation you can convert it in to a 2nd order homogeneous linear ODE which can be solved the help of Bessel functions.
     
  5. Feb 9, 2008 #4
    By carefully following Rainbow Child's instructions I think I was able to solve the first equation. The first part was to find the first integral, which was indeed as stated:

    [tex]f(x,t)\cdot x' +g(x,t)=C[/tex]

    The functions f and g can be found by taking the derivative with respect to t giving:

    [tex]fx''+f'x'+g'=0[/tex]

    Setting this equal to the original equation gives then:

    [tex]f(x,t)=t[/tex]

    and thus f'=0, and also

    [tex]g(x,t)=\frac{x^2}{2}-x-t[/tex]

    We have now:

    [tex]t\cdot x'+ \frac{x^2}{2}-x-t=C[/tex]

    or:

    [tex]\frac{dx}{dt}-\frac{1}{t}\cdot x + \frac{1}{2t}x^2=\frac{C}{t}+1[/tex]

    The Riccati equation:

    [tex]\frac{dx}{dt}+Q(t)x+R(t)x^2=P(t)[/tex]

    can be transformed by using:

    [tex]x=\frac{u'}{Ru}[/tex]

    Doing this transforms the equation into:

    [tex]t^2\frac{d^2u}{dt^2}-\frac{1}{2}(t+C)u=0[/tex]

    Which is a special case of the following extended Bessel equation:

    [tex]t^2\frac{d^2u}{dt^2}+(2k+1)t\frac{du}{dt}- \left(\alpha^2t^{2r}+\beta^2 \right)u=0[/tex]

    Which has the solution:

    [tex]u(t)=t^{-k}\left[AI_{\frac{\gamma}{r}}\left(\frac{\alpha}{r}t^r\right) + BK_{\frac{\gamma}{r}}\left(\frac{\alpha}{r}t^r\right) \right][/tex]

    with:

    [tex]\gamma=\sqrt{k^2-\beta^2}[/tex]

    The functions are the modified Bessel functions of the first and second kind of order [tex]\frac{\gamma}{r}[/tex].

    In our case we have:

    [tex]k=-\frac{1}{2} \qquad r=\frac{1}{2} \qquad \beta^2=\frac{C}{2} \qquad \alpha^2=\frac{1}{2}[/tex]

    and:

    [tex]\frac{\gamma}{r}=\sqrt{1-2C}=n \qquad \frac{\alpha}{r}=\sqrt{2}[/tex]

    The solution is thus:

    [tex]u(t)=\sqrt{t} \left[ A I_{n}\left(\sqrt{2t}\right) + B K_{n}\left(\sqrt{2t}\right) \right][/tex]

    After transforming back to x(t) we get:

    [tex]x(t) = 1+\sqrt{2t} \left[ \frac{\displaystyle \frac{A}{2} \left[I_{n-1}\left(\sqrt{2t}\right) + I_{n+1}\left(\sqrt{2t}\right) \right]- \frac{B}{2} \left[K_{n-1}\left(\sqrt{2t}\right) +K_{n+1}\left(\sqrt{2t}\right) \right]}{\displaystyle A I_{n}\left(\sqrt{2t}\right) + B K_{n}\left(\sqrt{2t}\right)} \right][/tex]

    Rewriting a little bit gives now:

    [tex]x(t) = 1+\sqrt{\frac{t}{2}} \left[ \frac{\displaystyle \left[I_{n-1}\left(\sqrt{2t}\right) + I_{n+1}\left(\sqrt{2t}\right) \right]- D \left[K_{n-1}\left(\sqrt{2t}\right) +K_{n+1}\left(\sqrt{2t}\right) \right]}{\displaystyle I_{n}\left(\sqrt{2t}\right) + D K_{n}\left(\sqrt{2t}\right)} \right][/tex]

    This solution has indeed two parameters C (in n) and D. To see if it is the correct one, it must be validated using the original equation. I haven't done this yet, the algebra is somewhat involved.

    Non-linear differential equations are indeed not easy. Thanks to Rainbow Child for the instructions, it was a pleasure doing this calculation.

    @ roldy: Is this the result you were looking for? Where exactly do these equations come from?
     
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