Convert the equation to a first order system.

[killersanta]

Homework Statement

Convert the equation to a first order system.

d^2y/dt^2 + 3dy/dt + 2y = 0

The Attempt at a Solution

Set d^2y/dt^2 = dv/dt

dy/dt = V

So, I now have: dv/dt + 3v + 2y = 0

Now, I'm not sure what to do with that Y, In my notes we only did examples with the last number not having an variable.

 

[tiny-tim]

hi killersanta!

(try using the X² icon just above the Reply box )

put d/dy = D …

then it's (D² + 3D + 2)y = 0

 

[HallsofIvy]

You have exactly what you want. Yes, dv/dt+ 3v+ 2y= 0 so your two equations are
dv/dt= -3v- 2y and dy/dt= v.

 

[killersanta]

hi killersanta!

(try using the X² icon just above the Reply box )

put d/dy = D …

then it's (D² + 3D + 2)y = 0

What reply box?
______________________________________

Now, I'm having troubles again.

dv/dt = -3v -2y
dy/dt=v

A = [-3 -2 ]
[1 0]

Det: [-3-Lamba -2 ] = 0
[1 0-lamba ]

(-3 - lamba)(0-lamba) -(-2)(1) = 0
Lamba² -3 lamba + 2 =0

(lamba -1)(lamba-2)=0
Lamba = 1, lamba = 2

Find V1 using Lamba =1:

[-3-1 -2|0]
[1 0-1 |0]
________________
[-4 -2|0]
[1 -1|0]
_________________
Swap rows
[1 -1|0]
[-4 -2|0]

This is where I am having troubles, no matter what I do, I can't get the bottom rows all zeros...

 

[tiny-tim]

(just got up :zzz: …)

try putting z = (D + 1)y

(and by the Reply box I mean the box on the Reply page that you get to if you click the "QUOTE" button )

 

[HallsofIvy]

What reply box?
______________________________________

Now, I'm having troubles again.

dv/dt = -3v -2y
dy/dt=v

A = [-3 -2 ]
[1 0]

Det: [-3-Lamba -2 ] = 0
[1 0-lamba ]

(-3 - lamba)(0-lamba) -(-2)(1) = 0
Lamba² -3 lamba + 2 =0

This is incorrect. (-3- lambda)(-lambda)= lambda²+ 3\lambda+ 2= 0.

(lamba -1)(lamba-2)=0
Lamba = 1, lamba = 2

Find V1 using Lamba =1:

[-3-1 -2|0]
[1 0-1 |0]
________________
[-4 -2|0]
[1 -1|0]
_________________
Swap rows
[1 -1|0]
[-4 -2|0]

This is where I am having troubles, no matter what I do, I can't get the bottom rows all zeros...

That's because 1 is NOT an eigenvalue.