Another repeated roots problem

[Jamin2112]

Homework Statement

Find the solution of the initial value problem.

x'=[3 9; -1 -3]x, x(0)=(2 4)^T

Homework Equations

A good guess should be x=te^rt$ + ert#

The Attempt at a Solution

But ...

the only root is r=0, which gives me a solution x⁽¹⁾(t)= C₁(-3 1)^T. Using the guess above, stuff goes wrong because I end up with tA$ + A# -$ = 0, which just gives me #=(-3 1)^T as well.

 

[gabbagabbahey]

A good guess should be x=te^rt$ + ert#

Good..

the only root is r=0, which gives me a solution x⁽¹⁾(t)= C₁(-3 1)^T. Using the guess above, stuff goes wrong because I end up with tA$ + A# -$ = 0, which just gives me #=(-3 1)^T as well.

I don't see how you are getting this. If your repeated root is r=0, then your solution is x=te^0t$ + e^0t#=t$+#...substituting your initial condition gives you (2,4)^T=0$+#=#

 

[Jamin2112]

Good..



I don't see how you are getting this. If your repeated root is r=0, then your solution is x=te^0t$ + e^0t#=t$+#...substituting your initial condition gives you (2,4)^T=0$+#=#

Then I'd have a solution of the form x=C₁(-3 1)^T + C₂(2 4)^T, which is different from the solution in the back of the book.

 

[vela]

You need to give us a bit more detail. What you've written doesn't really make sense, and it's hard to figure out what you're doing since you've omitted all the steps.

the only root is r=0, which gives me a solution x⁽¹⁾(t)= C₁(-3 1)^T. Using the guess above, stuff goes wrong because I end up with tA$ + A# -$ = 0, which just gives me #=(-3 1)^T as well.

What do A, $, and # stand for? Is A a matrix? Similarly, what exactly are C₁ and C₂?

Then I'd have a solution of the form x=C₁(-3 1)^T + C₂(2 4)^T, which is different from the solution in the back of the book.

I'm still not sure how you came up with the vector (-3 1)^T, why the constant vectors are multiplied by C₁ and C₂, and where the t and the exponential went.

 

[Jamin2112]

You need to give us a bit more detail. What you've written doesn't really make sense, and it's hard to figure out what you're doing since you've omitted all the steps.

What do A, $, and # stand for? Is A a matrix? Similarly, what exactly are C₁ and C₂?

I'm still not sure how you came up with the vector (-3 1)^T, why the constant vectors are multiplied by C₁ and C₂, and where the t and the exponential went.

$,# are vectors with constant entries
A is a matrix.
Boldness means a non-scalar, in my personal math language.

The problem I'm having comes from the fact that 0 is an eigenvalue.

Say there was one eigenvalue a≠0. One solution would be e^at$, where $ is the eigenvector associated with a. The I'd guess a second solution x⁽²⁾(t)=te^at$ + e^at#.

-----> x'= (ate^at+e^at)$ + ae^at#.

And plugging this back into x'=Ax,

(ate^at+e^at)$ + ae^at# = te^atA$ + e^atA#,

which is true if

te^at(A-aI)$=0
and
e^at(A-aI)#=e^at$.

e^at is never equal to zero, so to simplify,

t(A-aI)$=0
and
(A-aI)#=$.

Notice that I can solve for $, since there's the first equation is an eigenvalue problem. In the particular problem this thread is about, a=0 and $=(-3 1)^T. Plug that into the second equation and you can solve for #. But since a=0, we just get A#=$ ------> #₁ = -3#₂ - 1; #₂=#₂ ------> # = (-3 1)^T + k(-1 0)^T------> x= C₁te^at(-3 1)^T + C₂e^at [ (-3 1)^T + k(-1 0)^T ]

 

[vela]

Thanks for the explanation. Sorry, my brain is a bit slow this morning/afternoon. I think you're just getting the constants messed up a bit. First, the k is in the wrong place. You should have #=k(-3 1)^T+(-1 0)^T, so your solution should be

x = C₁[(-3 1)^Tte^at+[k(-3 1)^T+(-1 0)^T]e^at] + C₂(-3 1)^Te^at

so when you rearrange everything, you get

x = C₁[(-3 1)^Tte^at+(-1 0)^Te^at] + (kC₁+C₂)(-3 1)^Te^at

Because the C's are arbitrary, you can just set k=0, which you could have done right from the start to simplify the subsequent algebra. So finally, you get

x = C₁[(-3 1)^Tte^at+(-1 0)^Te^at] + C₂(-3 1)^Te^at

to which you now apply the initial conditions.

 

[Jamin2112]

Thanks for the explanation. Sorry, my brain is a bit slow this morning/afternoon. I think you're just getting the constants messed up a bit. First, the k is in the wrong place. You should have #=k(-3 1)^T+(-1 0)^T, so your solution should be

x = C₁[(-3 1)^Tte^at+[k(-3 1)^T+(-1 0)^T]e^at] + C₂(-3 1)^Te^at

so when you rearrange everything, you get

x = C₁[(-3 1)^Tte^at+(-1 0)^Te^at] + (kC₁+C₂)(-3 1)^Te^at

Because the C's are arbitrary, you can just set k=0, which you could have done right from the start to simplify the subsequent algebra. So finally, you get

x = C₁[(-3 1)^Tte^at+(-1 0)^Te^at] + C₂(-3 1)^Te^at

to which you now apply the initial conditions.

But since the initial condition is at t₀=0, the constant that goes with the t term will disappear, and there won't be a way to solve for it.

The answer in the back of the book is x=2(1 2)^T + 14(-3 1)^Tt, so there is a t term---somehow.

By the way, thank you for your cooperation. I have a midterm Wednesday, so I'm just making sure I do the homework questions correctly.

 

[vela]

The other term multiplying C₁ won't disappear, though, so you'll still have two equations and two unknowns.