A question on second order linear equations

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Discussion Overview

The discussion revolves around the reasoning behind the use of two linearly independent solutions when solving second order linear differential equations, specifically focusing on the equation y'' - y = 0 and its initial conditions. Participants explore the implications of linear combinations of solutions and the derivation of the general solution.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that any linear combination of solutions to the differential equation is also a solution, which leads to the necessity of two linearly independent solutions to form a general solution.
  • One participant expresses confusion regarding the requirement for two solutions and seeks clarification on why the general solution is derived from any two solutions.
  • Another participant explains the process of finding the complementary solution by using the characteristic polynomial derived from the differential equation.
  • There is a question about the method of guessing the complementary solution, with reference to a standard technique for solving linear ODEs with constant coefficients.

Areas of Agreement / Disagreement

Participants generally agree on the importance of linear combinations of solutions in forming the general solution, but there is some confusion and lack of clarity regarding the reasoning for needing two solutions and the method of deriving them.

Contextual Notes

Some assumptions about the nature of the solutions and the methods used to derive them are not fully explored, and the discussion does not resolve the confusion expressed by some participants regarding the derivation of the complementary solution.

mech-eng
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Hi, all. While solving a second order linear differential equation why do we have to use linear independent but two solutions. For example, when solving y''- y = 0 , y(0) = 5 and y'(0) = 3 , we use ex and e-x and then we write y = c1*ex+ c2*e-x-
 
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That's because any linear combination of a solution of the DE is a solution of the DE.
Initial conditions restrict the set of possible solutions.
 
That's because any linear combination of a solution of the DE is a solution of the DE.
Initial conditions restrict the set of possible solutions.
 
maajdl said:
That's because any linear combination of a solution of the DE is a solution of the DE.
Initial conditions restrict the set of possible solutions.

I am especially interested in the number "two", "two solutions". Why we derive general solution
from any two?
 
You seem to be confused about the solutions to ODEs.

For the equation y" - y = 0, with y(0) = 5 and y'(0) = 3, we have a homogeneous ODE for which we find the complementary solution by assuming y = e^(rx). Differentiating and substituting into the original ODE gives:

r^2 * e^(rx) - e^rx = 0

Dividing both sides by e^rx (which is always > 0, for any x), we obtain the characteristic polynomial

r^2 - 1 = 0 or r^2 = 1, which has solutions r = 1 and r = -1.

This means that e^x and e^-x are both solutions to the original ODE. Since the original ODE is linear, the sum of the individual solutions obtained from solving the characteristic polynomial must also be a solution, thus

y = C1*e^-x + C2 * e^x

By applying the initial conditions y(0) = 5 and y'(0) = 3, the coefficients of the general solution C1 and C2 can be determined.
 
How do we guess that complementary solution? In my example you did it as erx
 
mech-eng said:
How do we guess that complementary solution? In my example you did it as erx

That's the standard technique for solving linear ODEs with constant coefficients. See:

http://en.wikipedia.org/wiki/Linear_differential_equation

Section: Homogeneous equations with constant coefficients
 
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