Change of variable for leibniz notation

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SUMMARY

The discussion focuses on the change of variable in Leibniz notation for the differential equation xy'' + y' + λxy = 0, relevant to free vibrations of circular elastic membranes. The user seeks clarification on deriving the second derivative transformation d²y/dx² = λ d²y/dt² from the first derivative transformation dy/dx = √λ dy/dt. The solution involves applying the chain rule, specifically d/dx = d/dt (dt/dx), leading to the conclusion that d/dx = d/dt (1/√λ). Additionally, a resource for practicing MathJax syntax is provided.

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  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with Leibniz notation for derivatives.
  • Knowledge of the chain rule in calculus.
  • Basic concepts of free vibrations in physics.
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progenitor
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Hi

I have a question for change of notation.

Quote from textbook:

As an example of a singular problem on a finite interval, consider the equation
xy'' + y' + λxy = 0, (6)
or
−(xy')' = λxy, (7)
on the interval 0 < x < 1, and suppose that λ > 0. This equation arises in the study of
free vibrations of a circular elastic membrane, and is discussed further in Section 11.5.
If we introduce the new independent variable t defined by t =sqrt(λ)x, then
dy/dx=sqrt(λ)*dy/dt,
d2y/dx2 = λ*d2y/dt2.
...

my question is the steps involved in deriving d2y/dx2 = λ d2y/dt2.

My understanding in deriving dy/dx=sqrt(λ)*dy/dt is as follows:

since dy/dt = dy/dx*dx/dt
and dx/dt ->(using differentials) dt = sqrt(λ)dx -> dx/dt = 1/sqrt(λ) -> dy/dx = sqrt(λ)*dy/dt.

However, how do you derive d2y/dx2 = λ d2y/dt2?
I get 1=1 which is no help at all?

Sorry if this is noobie, I'm kind of new.

Also, is there a online notepad for practicing mathjax synatax? I like pretty equations.

Thank you.
 
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welcome to pf!

hi progenitor! welcome to pf! :smile:

(have a square-root: √ and try using the X2 icon just above the Reply box :wink:)
progenitor said:
If we introduce the new independent variable t defined by t =sqrt(λ)x, then

… how do you derive d2y/dx2 = λ d2y/dt2?

use the chain rule, d/dx = d/dt dt/dx = d/dt 1/√λ :wink:

(and you can practise your mathjax at https://www.physicsforums.com/mathjax/test/preview.html" :smile:)
 
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