-b.3.1.1 find the general solution of the second order y''+2y'-3y=0

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Discussion Overview

The discussion revolves around finding the general solution of the second-order differential equation \(y'' + 2y' - 3y = 0\). Participants explore the method of characteristic equations and the implications of linear independence in the context of solutions.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants begin by assuming a solution of the form \(y = e^{rt}\) and derive the characteristic equation \(r^2 + 2r - 3 = 0\).
  • It is noted that the roots of the characteristic equation are \(r = -3\) and \(r = 1\), leading to two linearly independent solutions: \(y = e^{-3t}\) and \(y = e^{t}\).
  • One participant asks how to write down the general solution given the two linearly independent solutions.
  • The general solution is proposed as \(y = c_1 e^{-3t} + c_2 e^{t}\), where \(c_1\) and \(c_2\) are constants.
  • There is a question regarding the concept of linear independence of solutions, which is subsequently explained in terms of the conditions for linear dependence.

Areas of Agreement / Disagreement

Participants generally agree on the roots of the characteristic equation and the form of the general solution. However, there is an ongoing exploration of the concept of linear independence, indicating that some aspects of the discussion remain open for further clarification.

Contextual Notes

The discussion does not resolve all potential nuances regarding the implications of linear independence and the conditions under which solutions are considered independent.

karush
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$\tiny{3.1.1}$
find the general solution of the second order differential equation.
$$y''+2y'-3y=0$$
assume that $y = e^{rt}$ then,
$$r^2+2r-3=0\implies (r+3)(r-1)=0$$
new stuff... so far..
 
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karush said:
$\tiny{3.1.1}$
find the general solution of the second order differential equation.
$$y''+2y'-3y=0$$
assume that $y = e^{rt}$ then,
$$r^2+2r-3=0\implies (r+3)(r-1)=0$$
new stuff... so far..
Good so far. So r = -3 and r = 1. Therefore two linearly independent solutions would be [math]y = e^{-3t}[/math] and [math]y = e^{t}[/math].

One last little bit for you. We have two linearly independent solutions to a second order linear homogenous differential equation. How do you write down the general solution?

-Dan
 
topsquark said:
Good so far. So r = -3 and r = 1. Therefore two linearly independent solutions would be [math]y = e^{-3t}[/math] and [math]y = e^{t}[/math].

One last little bit for you. We have two linearly independent solutions to a second order linear homogenous differential equation. How do you write down the general solution?

-Dan
$$y=c_1e^{-3t}+c_2e^{t}$$

guess that's it?

what is linear independent solutions?
 
karush said:
$$y=c_1e^{-3t}+c_2e^{t}$$

guess that's it?

what is linear independent solutions?
Yes, that's it.

Let f(t) and g(t) be differentiable functions. Then they are called linearly dependent if there are nonzero constants c1 and c2 with [math]c_1 f(t) + c_2 g(t) = 0[/math] for all t. If they are not linearly dependent then they are linearly independent.

-Dan
 

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