Solving 2nd Order Linear DE with Constant Coefficients

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SUMMARY

The discussion focuses on solving second-order linear differential equations (DE) with constant coefficients, specifically the form ay'' + by' + cy = 0. The general solution is expressed as y(t) = c1 e^(r1 t) + c2 e^(r2 t) when there are two distinct roots of the characteristic equation. The participants explore the reasoning behind the uniqueness of this solution, referencing methods used in first-order cases to demonstrate the absence of additional solutions.

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with characteristic equations and their roots
  • Knowledge of exponential functions and their properties
  • Basic concepts of differential calculus
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  • Study the derivation of the characteristic equation for second-order linear DEs
  • Explore the method of undetermined coefficients for solving linear DEs
  • Learn about the Wronskian and its role in determining the linear independence of solutions
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Students and professionals in mathematics, particularly those studying differential equations, as well as educators looking to deepen their understanding of solution methods for linear DEs.

Apteronotus
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Hi,

When solving a 2nd order Linear DE with constant coefficients (ay''+by'+cy=0) we are told to look for solutions of the form y=e^{rt} and then the solution (if we have 2 distinct roots of the characteristic) is given by
y(t)=c_1 e^{r_1 t}+c_2 e^{r_2 t}

This is clearly a solution, but how do we know there are no other solutions?
That is, how do we know this is the general solution?
 
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Hi Apteronotus! :smile:
Apteronotus said:
… how do we know there are no other solutions?

It's easy to prove for the first-order case …

if y' - ry = 0, put y = zert, then (z' + rz)ert = rzert

so ert = 0 (which is impossible),

or z' + rz = rz, ie z' = 0, ie z is constant :wink:

and now try (y' - ry)(y' - sy) = 0, using the same trick twice :smile:
 

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