Solve x'' = cox(x)

  • Thread starter peacex
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  • #1
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im new to 2nd order differential eqns and im getting myself all confused.
ive tried so hard but i just cant solve it.
can someone please help me with tips to solve/solutions?

x'' = cos(x)
 

Answers and Replies

  • #2
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Try using [tex]\ddot{x} = \dot{x}\frac{d\dot{x}}{dx}[/tex]

That will allow you to turn it into a first order equation at the very least.
 
  • #3
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thankyou!! i didnt think of that!
 
  • #4
dextercioby
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Try using [tex]\ddot{x} = \dot{x}\frac{d\dot{x}}{dx}[/tex]

That will allow you to turn it into a first order equation at the very least.
I don't think that's correct.
You should have written

[tex] \ddot{x}=\frac{1}{\dot{x}}\frac{d}{dt}\left(\frac{\dot{x}^{2}}{2}\right) [/tex]
 
  • #5
HallsofIvy
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No, Dextercioby, the first response was correct. If you let u= dx/dt, then du/dt= d2x/dt2. But we can apply the chain rule to the left side: du/dt= (du/dx)(dx/dt)= u du/dx since u= dx/dt. d2u/dt2= u du/dx.

This method is called "quadrature" (because the integral involves u2) and can be used to reduce the order of a differential equation when the independent variable, t, does not appear explicitely in the equation.
 
  • #6
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x" = cosx

its x-missing

v = x'

v' = x" = v dv/dx

v dv/dx = cosx

vdv = cosx dx

v^2 = 2 sinx

(x')^2 = 2 sinx

x' = ( 2 sinx )^1/2

dx / (sinx )^1/2 = 2^1/2 dy

Q = 2^1/2 y + C


such that Q = http://integrals.wolfram.com/index.jsp
 
Last edited:
  • #7
You cannot neglect the constant of integration.

if

[tex]y^{\prime \prime} = \cos{y}[/tex]

multiply by [tex]y^{\prime}[/tex]

gives you

[tex]y^{\prime} y^{\prime \prime} = y^{\prime} \cos{y}[/tex]

integrate, and you get

[tex] \frac{1}{2}y^{\prime 2} = \sin{y} + C[/tex]

Note the constant of integration which must not be ignored.

Therefore, you have

[tex]y^{\prime} = \sqrt{2 \sin{y} + C} [/tex]

which isn't easily soluble -- the only way to get an answer is to use elliptic integrals -- have a look at this Wolfram page.
 
  • #9
HallsofIvy
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Where in the world did you get that idea?

If x(t)= et, then x'= et= x. In other words,
[tex]\frac{dx'}{dx}= 1[/tex]
not 0.
 

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