# [SOLVED]097 Change the second-order IVP into a system of equations

#### karush

##### Well-known member
$\tiny{2.1.5.1}$
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
$e^{rt}(r^2+6r-2)=0$
$r=-3+\sqrt{11},\quad r=-3-\sqrt{11}$

just seeing if I going down the right road

Last edited:

#### Country Boy

##### Well-known member
MHB Math Helper
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

##### Well-known member
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?

#### Country Boy

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
$\tiny{2.1.5.1}$
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
$e^{rt}(r^2+6r-2)=0$
$r=-3+\sqrt{11},\quad r=-3-\sqrt{11}$

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

##### Well-known member
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

#### Country Boy

##### Well-known member
MHB Math Helper
so $x'(0)=1=y$ and $x(0)=\int y dy =\dfrac{y^2}{2}=1$ then $y=\sqrt{2}$

kinda maybe

textbook source

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.

void

Last edited: