What is the second derivative of a given function with a constant value?

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Discussion Overview

The discussion revolves around solving a mathematical problem involving the second derivative of a function, specifically in the context of a constant coefficient ordinary differential equation (ODE). Participants explore the implications of different values of the constant k on the nature of the solutions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the equation z" / z = k and seeks assistance in solving for z, where z is a function of t.
  • Another participant suggests rewriting the equation as z'' - kz = 0, identifying it as a constant coefficient ODE that can be solved using a characteristic polynomial.
  • It is proposed that if k > 0, the solutions will be exponential, with one solution tending to zero and the other growing to infinity.
  • For k < 0, oscillatory motion is suggested, with solutions involving sine and cosine functions, and the angular frequency being \sqrt{|k|}.
  • A further clarification states that if k = 0, the solution is linear, adding completeness to the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical approach to rewriting the equation and the implications of different values of k on the solutions. However, there is no explicit consensus on the overall solution or its application.

Contextual Notes

The discussion does not address potential limitations or assumptions underlying the mathematical steps, nor does it explore the implications of the solutions in a broader context.

bokche
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Can someone, please, help me solve this math problem?

z" / z = k

z=f(t)

z=?
z" - second derivative of function z=f(t)

k=const.

Thanks
 
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Re write this as z''-kz=0
This is a fairly simple constant coeff. ode, solved with a characteristic polynomial.
If k>0 then you have an exponential solution (two basic solution, one tending to zero and the second grows to infinity)
If k<0 you have an oscillatory motion, with two basic solution being sine and cosine, and the angular frequency being [tex]\sqrt{|k|}[/tex]
 
elibj123 said:
Re write this as z''-kz=0
This is a fairly simple constant coeff. ode, solved with a characteristic polynomial.
If k>0 then you have an exponential solution (two basic solution, one tending to zero and the second grows to infinity)
If k<0 you have an oscillatory motion, with two basic solution being sine and cosine, and the angular frequency being [tex]\sqrt{|k|}[/tex]
And if k= 0, of course, the solution is linear.

(Just wanted to be complete!:smile:)
 
Thanks for replies
Cheers
 

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