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Homework Help: Differential equation second order;]

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    I have differential equation;] this is equation from central force thing in Lagrange mechanics, you know, you know, its second order hahaha:D
    [tex]y^3y^{\prime\prime}=ay+b[/tex]
    2. Relevant equations
    I will using method of making a first degree of this equation
    3. The attempt at a solution
    I have
    [tex]\frac{\mbox{d}y}{\mbox{d}x}=u(y)[/tex]
    [tex]\frac{\mbox{d}^2y}{\mbox{d}x^2}=\frac{\mbox{d}u}{\mbox{d}x}=\frac{\mbox{d}u}{\mbox{d}y}\frac{\mbox{d}y}{\mbox{d}x}=u\frac{\mbox{d}u}{\mbox{d}y}[/tex]
    and now I enter this new equation to before:
    [tex]uy^3\frac{\mbox{d}u}{\mbox{d}y}=ay+b[/tex]
    [tex]u\frac{\mbox{d}u}{\mbox{d}y}=\frac{ay+b}{y^3}[/tex]
    [tex]u^2=-2\frac{a}{y}-4\frac{b}{y^2}+C[/tex]
    [tex]u=\frac{\sqrt{Cy^2-2ay-4b}}{y}[/tex]
    [tex]\frac{y}{\sqrt{Cy^2-2ay-4b}}\frac{\mbox{d}y}{\mbox{d}x}=1[/tex]
    someone who is good in mathematic, please tell me if this calculations was good and what can I do now, I was thinking about do this with area functions, but not sure, please tell me simplest way to solve it;] thank you!
     
  2. jcsd
  3. Jan 18, 2010 #2
    What is that -4 in [tex]u^2=-2\frac{a}{y}-4\frac{b}{y^2}+C[/tex]? It must be -1 since if you take the derivative of the second term wrt y, a 2y would appear in the numerator which cancels out a factor 1/2 coming from another side so the term b/y^3 is then retrieved. Now that dx should be brought to the other side of equation and then there you are left with a simple integration. Just note that since you are taking the root of [tex]u^2=-2\frac{a}{y}-4\frac{b}{y^2}+C[/tex], don't forget to have a sign [tex]\pm[/tex] accompanied with the radical.

    AB
     
    Last edited: Jan 18, 2010
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