A differential equation question

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Discussion Overview

The discussion revolves around verifying whether a given function is a solution to the differential equation y'' - 2y' + 2y = 0. Participants explore various methods for addressing the problem, including differentiation and substitution techniques, while also considering the context of the question and its potential origins.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant outlines a method involving differentiation of the proposed solution and substitution into the differential equation to verify it.
  • Another participant suggests solving the differential equation directly by assuming a solution of the form y = e^(mx) and deriving conditions for m.
  • A different approach is proposed using a change of variable y = e^x u, leading to a simpler second-order equation for u.
  • Some participants emphasize that the original question is about verification rather than solving the equation, suggesting it may not be from a differential equations course.
  • There is a discussion about the general applicability of the method involving the change of variable and whether one could arrive at it without prior knowledge of the solution.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to the problem, with some focusing on verification and others on solving the equation. No consensus is reached on a single preferred method.

Contextual Notes

Participants note that the equation is linear and homogeneous with constant coefficients, which influences the methods discussed. There is also mention of the ease of solving the equation for those familiar with the techniques.

Who May Find This Useful

This discussion may be useful for students or practitioners interested in differential equations, particularly those exploring verification methods and solution techniques.

mech-eng
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Verify that given function is a solution.
y'' - 2y' + 2y = 0 , y=e^x(Acos x + Bsin x)

First I take derivative of y which is y+e^x(-Asin x + B cos x) then I asign e^x(-Asin x + B cos x) to y'-y. Then I take derivative of y' which is y'+(y'-y)-y which equals 2y'-2y=y'' then I use y'' as 2y'-2y which satifies. Do you have a faster or more exciting way?

Have a nice day.
 
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Just solve the DE. Let y = emx. Plug into the DE and get the condition m2 -2m + 2 = 0, which has solutions m = 1 ± i.

Thus the DE has solutions y = a e(1 + i)x + b e(1 - i)x = ex(a eix + b e-ix), which by Euler's formula can be reduced to the form you give.
 
That equation begs for the change of variable y=e^x u
y'=e^x(u'+u)
y''=e^x(u''+2u'+u)
so
y'' - 2y' + 2y = 0 , y=e^x(Acos x + Bsin x)
becomes
u''+u=0 , u=Acos x + Bsin x
 
The method I gave works for any linear homogeneous DE with constant coefficients, reducing it to an algebraic equation. And in this case all that remains to do is solve a quadratic.

lurflurf said:
That equation begs for the change of variable y=e^x u
Would you have been able to guess this change of variable if you did not already have the solution to look at?
 
The question doesn't ask the OP to solve the equation. It asks to verify the given function is a solution.

Of course this is a very easy ODE to solve if you know how to solve it, but the question might not even be from a differential equations course. It might just be an exercise in differentiation.
 
Bill_K said:
The method I gave works for any linear homogeneous DE with constant coefficients, reducing it to an algebraic equation. And in this case all that remains to do is solve a quadratic.
Not sure what you are getting at. Solving polynomials (other than quadratics) is not easy.
Bill_K said:
Would you have been able to guess this change of variable if you did not already have the solution to look at?
Sure
m2 -2m + 2=(m-1)2+1
 
Last edited:

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