Is There a Theorem Relating Exponential and Trigonometric Functions?

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Discussion Overview

The discussion revolves around the relationship between exponential functions and trigonometric functions in the context of ordinary differential equations (ODEs). Participants explore whether a specific form of a function can be a solution to a second-order linear homogeneous ODE.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions if the function \( y = e^{px}(A\cos(nx) + B\sin(mx)) \) can be a solution to the ODE \( ay'' + by' + cy = 0 \).
  • Another participant asserts that the proposed function does not generally satisfy the ODE and suggests solving the ODE to find the conditions on the coefficients.
  • A different participant clarifies that the function can solve the ODE if the auxiliary equation has complex roots, relating it to the form of the solutions involving exponential and trigonometric functions.
  • One participant acknowledges the connection to the complementary function of the ODE when the auxiliary equation has complex roots, but expresses uncertainty about the generality of the statement they read.
  • Another participant notes that the indices \( n \) and \( m \) should be equal for the proposed function to fit the context correctly.

Areas of Agreement / Disagreement

Participants express differing views on the generality of the relationship between the exponential and trigonometric functions in the context of ODEs. There is no consensus on whether the initial claim holds true in a general sense.

Contextual Notes

The discussion highlights the dependence on the nature of the roots of the auxiliary equation and the specific conditions required for the proposed function to be a solution, which remain unresolved.

Who May Find This Useful

This discussion may be of interest to those studying differential equations, particularly in understanding the relationships between different types of solutions and their forms.

daster
I was just wondering if there's a known result or theorem that states if:

[tex]y=e^{px}(A\cos (nx)+B\sin (mx))[/tex]

Then:

[tex]ay'' + by' + cy = 0[/tex]

I don't see why this should be the case, but I read this somewhere and was intrigued.
 
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No,in general,that is not true...
Try to do it viceversa...Solve the ODE...And then find the conditions the coefficients need to fulfil,as the solution would have the form u've given.

Daniel.
 
Thanks Daniel.
 
perhaps you read that e^(px)[Acos(qx) + Bsin(qx)] solves an equation of form

ay'' + by' + cy = 0, where p+qi and p-qi are imaginary roots of the equation

aX^2 + bX + c = 0.

the reason for this is the solutions of this equation are all linear combinations of

e^(p+qi)x and e^(p-qi)x, i.e. of e^(px)[cos(qx) +isin(qx)]

and e^(px)[cos(qx) -isin(qx)], equivalently, of

e^(px)[cos(qx)] and e^(px)[sin(qx)].

I.e. of form Ae^(px)[cos(qx) + B e^(px)[sin(qx)] =

e^(px)[Acos(qx) + Bsin(qx)].
 
Yeah, I know that the form I posted was the complementary function of the ODE if the auxilary equation produced complex roots, but what I read was stating something generally, which I wasn't too sure about.
 
i'm just saying you were very close, except you need to take n =m.
 

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