System of Equations for Second-Order IVP

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Discussion Overview

The discussion revolves around converting a second-order initial value problem (IVP) into a system of first-order equations. Participants explore the definitions of variables and initial conditions associated with the system.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant proposes a transformation of the second-order IVP into a system of first-order equations using substitutions: $x_1 = y$ and $x_2 = y'$, leading to $x_1' = x_2$ and $x_2' = -x_2 + 2x_1$.
  • Another participant confirms the correctness of the transformation and inquires about the initial conditions for $x_1(0)$ and $x_2(0)$.
  • A different participant attempts to derive the initial conditions but seems to misinterpret the values, suggesting $x_1' = x_2 = 0$ and calculating $x_2' = 4$ based on incorrect assumptions.
  • One participant challenges the previous calculations, clarifying that $x_1(0) = 2$ and $x_2(0) = 0$ based on the given initial conditions of the original IVP.
  • Another participant echoes the clarification and expresses confusion regarding the multiple examples and substitutions encountered.

Areas of Agreement / Disagreement

There is disagreement regarding the interpretation of initial conditions, with some participants asserting correct values while others propose different interpretations. The discussion remains unresolved as participants have conflicting views on the calculations.

Contextual Notes

Participants express confusion stemming from multiple examples and varying substitutions, which may affect their understanding of the initial conditions and transformations.

karush
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Change the second-order IVP into a system of equations
$y''+y'-2y=0\quad y(0)= 2\quad y'(0)=0$
let $x_1=y$ and $x_2=y'$ then $x_1'= x_2$ and $y''=x_2'$
then by substitution
$x_2'+x_2-2x_1=0$
then the system of first order of equations
$x_1'=x_2$
$x_2'=-x_2+2x_1$

hopefully so far..
 
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Yes, that is correct.

Now, what are $x_1(0)$ and $x_2(0)$?
 
Country Boy said:
Yes, that is correct.
Now, what are $x_1(0)$ and $x_2(0)$?
$x_1'=x_2=y(0)=0$
and
$x_2'=-x_2+2x_1=0+2(2)=4$

its like chasing a rabbit in the briers
 
Frankly, I am not sure what you are doing, You were told that y(0)= 2 and y'(0)= 0.

Since you defined $x_1(t)$ to be y(t) and $x_2(t)$ to be y'(t),
$x_1(0)= y(0)= 2$ and $x_2(0)= y'(0)= 0$
 
Country Boy said:
Frankly, I am not sure what you are doing, You were told that y(0)= 2 and y'(0)= 0.

Since you defined $x_1(t)$ to be y(t) and $x_2(t)$ to be y'(t),
$x_1(0)= y(0)= 2$ and $x_2(0)= y'(0)= 0$

i think I get confused looking at multiple examples with all these different substitutions
 

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