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

• MHB
• karush
In summary, the task was to convert the second-order initial-value problem into a system of equations. This was done by defining $y$ as $x'$ and using it to create two first-order equations. The initial conditions were then applied to the new equations, resulting in $x(0)=1$ and $y(0)=1$.
karush
Gold Member
MHB
$\tiny{2.1.5.1}$
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 roadđź•¶

Last edited:
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.

Country Boy said:
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?

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?

karush said:
$\tiny{2.1.5.1}$
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 roadđź•¶

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*}

Country Boy said:
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

karush said:
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:

1. What is a second-order IVP?

A second-order IVP (initial value problem) is a type of differential equation that involves a second derivative of a function, along with its initial conditions. It is typically written in the form y'' = f(x, y, y') with initial conditions y(x0) = y0 and y'(x0) = y'0.

2. Why would I want to change a second-order IVP into a system of equations?

Converting a second-order IVP into a system of equations can make it easier to solve, especially if the equation is nonlinear or has complicated initial conditions. It also allows for the use of matrix methods, which can be more efficient for solving systems of equations.

3. How do I change a second-order IVP into a system of equations?

To change a second-order IVP into a system of equations, you will need to introduce a new variable, typically denoted by u, that represents the first derivative of the original function y. This will result in a system of two first-order differential equations, which can then be solved using standard methods.

4. Are there any advantages to using a system of equations instead of a single second-order IVP?

Yes, there are several advantages to using a system of equations. As mentioned before, it can make the equation easier to solve and allows for the use of matrix methods. It also allows for more flexibility in terms of initial conditions, as the system can be easily modified to accommodate different starting points.

5. Can I convert any second-order IVP into a system of equations?

In theory, yes, any second-order IVP can be converted into a system of equations. However, the process may not always be straightforward and may require some manipulation of the original equation. It is important to carefully consider the advantages and disadvantages before deciding to convert a second-order IVP into a system of equations.

• Linear and Abstract Algebra
Replies
2
Views
997
• Linear and Abstract Algebra
Replies
4
Views
996
• Differential Equations
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
14
Views
1K
• Linear and Abstract Algebra
Replies
5
Views
1K
• Differential Equations
Replies
4
Views
989
• Differential Equations
Replies
2
Views
821
• Differential Equations
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
3
Views
1K
• Differential Equations
Replies
3
Views
1K