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Nonlinear differential equation

  • Thread starter mbadin
  • Start date
  • #1
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Homework Statement



In one problem I had got to this equations, but I was not able to solve it, because I'm actually
on high school.

The equation :
d^2/dt^2(x) = -h*g/(h+x)

I tried use separation of variables but I was not able to use the chain rule.
Can anybody show me the steps with explanation?
 
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Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi mbadin! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
d2x/dt2 = -h*g/(h+x)
either multiply both sides by dx/dt

or use d2x/dt2 = vdv/dx

(where v = dx/dt … you can prove it using the chain rule :wink:)
 
  • #3
DryRun
Gold Member
838
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OK, here is how your equation appears to me:
[tex]\frac{d^2x}{dt^2} = -\frac{hg}{h+x}[/tex]
If i understood your question correctly, then you need to integrate twice to get rid of the differentiation.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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OK, here is how your equation appears to me:
[tex]\frac{d^2x}{dt^2} = -\frac{hg}{h+x}[/tex]
If i understood your question correctly, then you need to integrate twice to get rid of the differentiation.
It's not that easy because the right side, that you want to integrate with respect to t depends upon the unknown function x.

I would do what tiny-tim suggested: Let v= dx/dt so that [itex]d^2x/dt^2= dv/dt[/itex] and then, by the chain rule,
[tex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}[/tex]

Your equation becomes
[tex]v\frac{dv}{dx}= -\frac{hg}{h+ x}[/tex]
which can be integrated as
[tex]\int v dv= hg\int \frac{dx}{h+ x}[/tex]

It might well give you a function v= dx/dt that is difficult to integrate but that is the most direct method to solve this equation.
 

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