# ODE: System of Linear Equations usuing Diff. Operator

• RJLiberator
In summary: The key distinction for me in this thread was the relationship of 3x+5y=0. I now see how my answer can be corrected to the final answer as presented in the book.
RJLiberator
Gold Member

## Homework Statement

This is an ordinary differential equation using the differential operator.

Given the system:
d^2x/dt - x + d^2y/dt^2 + y = 0
and
dx/dt + 2x + dy/dt + 2y = 0

find x and y equation

y = -3ce^(-2t)

## The Attempt at a Solution

We change it into a differential operator equation...

[D^2-1]x+[D^2+1]y = 0
[D+2]x +[D+2]y = 0

We simply cancel values out until we are left with

3x+5y = 0
[D+2]x +[D+2]y = 0

Here we have x = -5y/3 and y = -3x/5

We substitute into the second equation and we can easily find that

x = ce^(-2t)
y = ce^(-2t)

My question is: How does the answer book know that there is a 5 for the x equation and a -3 for the y equation. It seems to me that the constant "c" handles this for me.
Am I doing something wrong to get a less precise answer? Or did I work out this problem correctly.

Thank you.

RJLiberator said:
We simply cancel values out until we are left with

3x+5y = 0
[D+2]x +[D+2]y = 0

What do you mean by "cancel values". This is a bit sloppy notation. Remember that ##D## is an operator.
Your first equation is telling you that ##3x=-5y##, which is explaining the factors 5 and -3.

RJLiberator said:
x = ce^(-2t)
y = ce^(-2t)

No, this is wrong, since it violates the first equation.

I also recommend that you use Latex to write your equations. It makes them much easier to read.

Last edited by a moderator:
RJLiberator
RJLiberator said:
Given the system:
d^2x/dt - x + d^2y/dt^2 + y = 0
and
dx/dt + 2x + dy/dt + 2y = 0

find x and y equation

## The Attempt at a Solution

We change it into a differential operator equation...

[D^2-1]x+[D^2+1]y = 0
[D+2]x +[D+2]y = 0

We simply cancel values out until we are left with

3x+5y = 0
[D+2]x +[D+2]y = 0
Yeah, I'm not following what you did, either.How did you go from the 2nd-order DE to 3x + 5y = 0?
RJLiberator said:
Here we have x = -5y/3 and y = -3x/5
I hope it's obvious that these two equations are equivalent.

RJLiberator
Mark44 said:
Yeah, I'm not following what you did, either.How did you go from the 2nd-order DE to 3x + 5y = 0?
$$(\mathrm{D}^2-1)x+(\mathrm{D}^2+1)y=0\\ (\mathrm{D}^2-4)(x+y)+3x+5y=0\\ (\mathrm{D}-2)(\mathrm{D}+2)(x+y)+3x+5y=0\\ (\mathrm{D}-2)0+3x+5y=0\\ 3x+5y=0$$
since the second equation is
$$(\mathrm{D}+2)(x+y)=0$$

RJLiberator said:
x = ce^(-2t)
y = ce^(-2t)

My question is: How does the answer book know that there is a 5 for the x equation and a -3 for the y equation. It seems to me that the constant "c" handles this for me.
.
you should have two constants
x = c1e^(-2t)
y = c2e^(-2t)

related by
3x+5y=0

RJLiberator
The key distinction for me in this thread was the relationship of 3x+5y=0. I now see how my answer can be corrected to the final answer as presented in the book.

Thank you guys for the help.

I was very close :p.

## 1. What is an ODE?

An ODE, or ordinary differential equation, is a mathematical equation that describes how a variable changes over time. It involves the derivative of a function with respect to one or more independent variables.

## 2. What is a system of linear equations?

A system of linear equations is a set of equations where each equation is a linear combination of the same variables. The goal is to find a set of values for the variables that satisfies all of the equations simultaneously.

## 3. How is a differential operator used in a system of linear equations?

A differential operator is used to represent the derivative of a function in a system of linear equations. It allows us to express the equations in a concise and efficient way, making it easier to solve the system.

## 4. What are some common applications of ODEs and linear equations?

ODEs and linear equations are used in many fields of science and engineering to model and solve problems. Some common applications include predicting population growth, analyzing chemical reactions, and designing electrical circuits.

## 5. What are some methods for solving a system of linear equations using a differential operator?

There are several methods for solving a system of linear equations using a differential operator, including substitution, elimination, and Gaussian elimination. These methods involve manipulating the equations to eliminate variables and find a solution that satisfies all of the equations.

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