A Property of an Autonomous ODE

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SUMMARY

The discussion centers on proving that if \( y(t) = \sin(t) \) is a solution to an autonomous ordinary differential equation (ODE) of the form \( f(y, y', \ldots, y^{(n)}) = 0 \), then \( x(t) = \cos(t) \) is also a solution. The participants highlight the importance of the relationship between the derivatives of \( y \) and the nature of the function \( f \). The connection between sine and cosine through the phase shift \( \cos(t) = \sin(t + \frac{\pi}{2}) \) is suggested as a potential method for the proof.

PREREQUISITES
  • Understanding of autonomous ordinary differential equations (ODEs)
  • Familiarity with trigonometric functions and their properties
  • Knowledge of derivatives and their role in ODEs
  • Basic concepts of functional dependence in mathematical equations
NEXT STEPS
  • Study the properties of autonomous ODEs and their solutions
  • Explore the relationship between sine and cosine functions in the context of differential equations
  • Learn about phase shifts in trigonometric functions and their implications
  • Investigate specific examples of autonomous ODEs, such as the pendulum equation
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Mathematicians, physics students, and anyone interested in the properties of autonomous ordinary differential equations and their solutions.

littleHilbert
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Hi!

I wonder how to prove that if y(t)=sin(t) solves an autonomous ODE f(y,y',...,y^(n))=0, then x(t)=cos(t) is also a solution.

I mean I'm a bit distracted by the fact that all derivatives of y are present here. For example in the equation for a pendulum there are just y and y'' and a special functional dependence between them and that's why it works, isn't it?

In the above case I know almost nothing about the function f. So how to proceed?
 
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Maybe you can use that cos(x) = sin(x + pi / 2)?
 

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