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Differential equation y(x)''=f(y(x))

  1. May 21, 2012 #1
    Hi,
    simple quetion, as you can see in the title.
    How can I solve differential equation y(x)''=f(y(x))
    I know I can write first derivative like dy/dx. But how can I write second derivative in such form?
    If it would be y(x)'=y, then it can be writen dy/dx=y
    => (1/y)dy=(1)dx
    => I can integrate ln(y)+C=x
    => its clear from now on

    I need some similar solution for situation when I have second derivative on the left side and some function consisting just y(x) (not x itself) on the right.
    Thanks for helping
     
  2. jcsd
  3. May 21, 2012 #2

    tiny-tim

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    Hi Holali! :smile:

    Multiply both sides by y' :wink:

    (or use the chain rule … y'' = dy'/dx = y' dy'/dy)
     
  4. May 22, 2012 #3
    Hmm, chain rule.I found it at wikipedia,but cant understand it.
    I understand equation y''=dy'/dx, but not y''=y' dy'/dy.
    Could you show me some easy example, like y''=2y^2, or y''=y^2 -y ?
     
  5. May 22, 2012 #4

    HallsofIvy

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    What tiny-time is referring to is often called "quadrature"- you'll see why in a moment. The crucial point is that "x", the independent variable does not appear explicitely in the equation y''= f(7).

    If you let u= y', you can write
    [tex]\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{du}{dx}[/tex]
    We can then say , using the chain rule,
    [tex]\frac{du}{dx}= \frac{du}{dy}\frac{dy}{dx}= u\frac{du}{dy}= f(y)[/tex]
    which is a separable equation:
    [tex]u du= f(y)dy[/tex]
    The left side is, of course, [itex](1/2)u^2[/itex], the reason for the name "quadrature". we have
    [tex](1/2)u^2= \int f(x)dx[/tex]
    so
    [tex]\frac{dy}{dx}= u= \sqrt{2\int f(x)dx}[/tex]
    [tex]y= \int^x \sqrt{2\int^u f(t)dt} du[/tex]

    Of course, if f(y) was not a "nice" function to begin with, those integrals may be difficult to do!
     
  6. Jun 20, 2012 #5
    ""
    udu=f(y)dy

    The left side is, of course, (1/2)u2, the reason for the name "quadrature". we have

    (1/2)u2=∫f(x)dx

    ""
    DIdn't you change y for x in this part?
     
  7. Jun 23, 2012 #6
    Simple manipulation of left hand side of the equation will solve this

    d(dy/dx)/dx= dy'/dy * dy/dx
    Moving the 1/dy factor to the right hand side, we get

    y' dy' = f(y) dy

    Integrating this will give us

    (1/2)y'^2=F(y) (F'(y)=f(y))

    Rearranging, we get

    dy/dx=[2F(y)]^1/2

    And of course, the second order diff eq. is reduced to a simple ODE. Using the method of separation of variables, we obtain

    x=integral{dy/[2F(y)]^1/2}
     
  8. Nov 24, 2012 #7
    For the same general equation, how can i solve

    y''(x)=a*sin(y(x)) ; a is a constant
    I think it would be a numerical method.
     
  9. Nov 24, 2012 #8

    tiny-tim

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    welcome to pf!

    hi etpatati1! welcome to pf! :smile:
    Multiply both sides by y' :wink:

    (or use the chain rule … y'' = dy'/dx = y' dy'/dy)
     
  10. Nov 25, 2012 #9
    Thanks to the Jacobi am function, a closed form exists to express y(x).
     
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