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Find the equations of the trajectories of y"+y^3=0

  1. Feb 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the equations of the trajectories of y"+y^3=0.

    2. Relevant equations
    None.

    3. The attempt at a solution
    y"+p(y)=0
    v(dv/dy)+p(y)=0
    integrate
    v^2/2+P(y)=C
    so I got v^2/2+y^4/4=C. Is v^2/2+y^4/4=C the correct answer?
     
  2. jcsd
  3. Feb 13, 2015 #2

    Ray Vickson

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    Substitute it into the DE and check to see if it works. You can do it just as easily as we can, and (as I have said before), doing your own checking first is a good habit to develop. You won't be able to ask for help when you write an exam!
     
  4. Feb 13, 2015 #3

    vela

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    Nope. You're looking for an expression where y is expressed as a function of x (or whatever your independent variable is). Right now, what you have is
    $$\frac 12 (y')^2 = C - \frac 14 y^4.$$ You need to solve that differential equation now.
     
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