ODE: System of Linear Equations usuing Diff. Operator

[RJLiberator]

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

Answer: x = 5ce^(-2t)
y = -3ce^(-2t)

Homework Equations

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.

 

[eys_physics]

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.

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.

 

[Mark44]

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?

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

I hope it's obvious that these two equations are equivalent.

 

[lurflurf]

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
$$(\mathrm{D}+2)(x+y)=0$$

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.