System of Equations for Second-Order IVP

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SUMMARY

The discussion focuses on transforming the second-order initial value problem (IVP) represented by the equation $y'' + y' - 2y = 0$ with initial conditions $y(0) = 2$ and $y'(0) = 0$ into a system of first-order equations. The variables are defined as $x_1 = y$ and $x_2 = y'$, leading to the system $x_1' = x_2$ and $x_2' = -x_2 + 2x_1$. The initial conditions are confirmed as $x_1(0) = 2$ and $x_2(0) = 0$, clarifying the confusion around the substitutions used in the transformation.

PREREQUISITES
  • Understanding of second-order differential equations
  • Familiarity with initial value problems (IVP)
  • Knowledge of first-order systems of equations
  • Basic concepts of substitution in differential equations
NEXT STEPS
  • Study the method of converting higher-order differential equations to first-order systems
  • Explore numerical methods for solving initial value problems
  • Learn about stability analysis of first-order systems
  • Investigate applications of systems of equations in engineering problems
USEFUL FOR

Mathematics students, engineers, and anyone involved in solving differential equations or studying dynamic systems will benefit from this discussion.

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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