097 Change the second-order IVP into a system of equations

[karush]

$\tiny{2.1.5.1}$
Change the second-order initial-value problem into a system of equations
$x''+6x'-2x= 0\quad x(0)=1\quad x'(0)=1$
ok my first step was to do this
$e^{rt}(r^2+6r-2)=0$
using quadratic formula we get
$r=-3+\sqrt{11},\quad r=-3-\sqrt{11}$

just seeing if I going down the right road🕶

 

[HOI]

I think you are completely misunderstanding the problem!
You are trying to solve this initial value problem
(You have a typo, the characteristic equation is $r^2- 6r- 2= 0$ but you have $r^2- xr- 2= 0$.)
but you are not asked to do that. You are asked to convert this second order differential equation to two rirst order equations.

There are many ways to do that. The simplest is to define y= x'. The x''= y' and the equation so that the equation x''- 6x'- 2x= 0 becomes y'- 6y- 2x= 0 or y'= 6y+ 2x.

The two equations are
x'= y and
y'= 6y+ 2x.

 

[karush]

I think you are completely misunderstanding the problem!
You are trying to solve this initial value problem
(You have a typo, the characteristic equation is $r^2- 6r- 2= 0$ but you have $r^2- xr- 2= 0$.)
but you are not asked to do that. You are asked to convert this second order differential equation to two rirst order equations.

There are many ways to do that. The simplest is to define y= x'. The x''= y' and the equation so that the equation x''- 6x'- 2x= 0 becomes y'- 6y- 2x= 0 or y'= 6y+ 2x.

The two equations are
x'= y and
y'= 6y+ 2x.

so what do we do with the initial values?

 

[HOI]

You now have two functions x(t) and y(t). You were told in the original problem that x(0)= 1 and that x'(0)= 1.
So what are x(0) and y(0) for the pair of first order equations?

 

[Prove It]

$\tiny{2.1.5.1}$
Change the second-order initial-value problem into a system of equations
$x''+6x'-2x= 0\quad x(0)=1\quad x'(0)=1$
ok my first step was to do this
$e^{rt}(r^2+6r-2)=0$
using quadratic formula we get
$r=-3+\sqrt{11},\quad r=-3-\sqrt{11}$

just seeing if I going down the right road🕶

You haven't converted into a system of equations...

Let $\displaystyle u = x $ and $\displaystyle v = x' $, then your system of equations is

$\displaystyle \begin{align*} u' &= v \\ v' &= 2\,u - 6\,v \end{align*} $

 

[karush]

You now have two functions x(t) and y(t). You were told in the original problem that x(0)= 1 and that x'(0)= 1.
So what are x(0) and y(0) for the pair of first order equations?

so $x'(0)=1=y$ and $x(0)=\int y dy =\dfrac{y^2}{2}=1$ then $y=\sqrt{2}$

kinda maybe

textbook source

 

[HOI]

so $x'(0)=1=y$ and $x(0)=\int y dy =\dfrac{y^2}{2}=1$ then $y=\sqrt{2}$

kinda maybe

textbook source

NO! You are still completely misunderstanding. I recommend that you talk to your teacher about this.

Your original problem was to convert the second order equation to a pair of first order equations.

The original second order equation is x''- 6x'- 2x= 0 with initial conditions x(0)= 1 and x'(0)= 1.
I suggested defining y= x'(t) so that the equation becomes y'- 6y- 2x=0 or y'= 2x+ 6y.

So your two equation are x'= y and y'= 2x+ 6y. The initial conditions are x(0)= 1 and y(0)= x'(0()= 1.

 

[karush]

void