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Second order equation problem

  1. Aug 18, 2008 #1
    ive been doing Xdot and Ydot questions using the formula lamda2-(a+d)lamda +ad-bc=0 where Xdot=ax+bc and ydot=cx+dy no problems so far until this question where i have the second order equation xdot dot+x-1/4x3=0

    so i set Xdot dot =Ydot and therefore Xdot =y as usual but this gives me Ydot=1/4x3 -x which obviously doesnt fit the formula above, how do i calculate the equlibrium points in this case
     
  2. jcsd
  3. Aug 18, 2008 #2

    Defennder

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    This is a special type of 2nd order ODE. If you have x'' = f(x), you can use the chain rule to express it in the form:

    [tex]\frac{1}{2} \frac{d}{dx} \left( \frac{dx}{dt} \right)^2 = f(x)[/tex] which is separable.
     
  4. Aug 18, 2008 #3

    Dick

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    The first type of problems are second order linear ODE's with constant coefficients in two variables. The other one is not that type. To find equilibrium points just set x''(t)=x'(t)=0 and solve for x(t).
     
  5. Aug 19, 2008 #4
    what is x''t and x't in terms of Xdot?
     
  6. Aug 19, 2008 #5

    Dick

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    x'(t)=Xdot=dx/dt and x''(t)=Xdot dot=d^2x/dt^2.
     
  7. Aug 19, 2008 #6
    giving me 4x-xcubed=0 giving me x=0 2 or -2 ?
     
  8. Aug 19, 2008 #7

    Dick

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    Sure.
     
  9. Aug 19, 2008 #8
    cool chears
     
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