Differential equation with repeated roots

[hops1]

Hi,
I'm somewhat new here, only posted a few times, and would like some help from you guys here if possible
I'm stuck with a problem on the topic mentioned.

x'=Ax

A is a 2*2 matrix
A =
[-5 1]
[-1 -3]

Now I managed to find the eigenvalues which is -4, repeated twice (multiplicity 2)
And the eigenvector as well which is

[1]
[1]

Now I'm really confused as to what to do next, I know the equation is e^-4t
But I am having trouble deriving the homogenous and particular solutions to the problem.
Would appreciate any input

 

[tiny-tim]

hi hops1!

[-5 1]
[-1 3]
…
Now I'm really confused as to what to do next, I know the equation is e^-4t
But I am having trouble deriving the homogenous and particular solutions to the problem.
Would appreciate any input

(don't you mean
[-5 1]
[-1 -3] ? )

for a repeated root, the solutions are usually e^At and te^At ​

 

[hops1]

hi hops1!


(don't you mean
[-5 1]
[-1 -3] ? )

for a repeated root, the solutions are usually e^At and te^At ​

You are correct thanks, but what I'm actually having trouble with is that the second vector should come out to be [t;1]
I don't understand how that comes about.
The first colution I understand is C1[1;1]e^-4t, but for the second solution I had the feeling that I should use te^At, but to get the second vector, that's what's got me confused

 

[tiny-tim]

hi hops1!

try an easier one …

[1 2]
[0 1]

 

[hops1]

Ok thanks, I tried the one you posted.
I hope I did it right
x'=Ax

I split it into a system of equations
x₁' = x₁ + 2x₂
x₂' = x₂

x₂ could be found easily,
since x₂' - x₂ = 0
Thus follows x₂ = C₂e^t

Then to find x₁ -> x₁' - x₁ = 2x₂
I first find the homogenous solution thus x₁' - x₁ = 0
Which yields the same solution as x₂, so x₁ = C₁e^t

Now to find the particular solution
x₁' - x₁ = 2x₂
Tried Ate^t

Ae^t + Ate^t - Ate^t = 2C₂e^t
Solving this yields A = 2C₂

Thus
x₁ = C₁e^t + 2C₂te^t
x₂ = C₂e^t

x = C₁ [1;0] e^t + C₂[2t;1] e^t

 

[tiny-tim]

i'm really not following your reasoning

and if we put C₁ = 0, C₂ = 1, into your solution, it gives x = [(t/2);1] e^t, which doesn't work

 

[hops1]

i'm really not following your reasoning

and if we put C₁ = 0, C₂ = 1, into your solution, it gives x = [(t/2);1] e^t, which doesn't work

I'm worse of than I thought.

Well I think e^t is correct.
I'm guessing I messed up in the particular solution for x₁ ??I think I found the error in my calculation, will try to run through it again

 

[hops1]

I think I got it now, I made an error in the particular solution

So it should be
C₁ [1;0]e^t + C₂ [2t;1] e^t

 

[hops1]

Now I went back to my initial problem.
I'm stuck when trying to get the particular solution, since if you rearrange the matrix
x₁' + 5x₁ = x₂
x₂' + 3x₁ = -x₁

So let's say I take the top equation.

x₁' + 5x₁ = x₂

I choose a particular solution to test say Cte^-4t
I end up with
Ce^-4t-4Cte^-4t + 5Cte^-4t = x₂
But I have yet to solve x₂ , and doing the same for x₂ would give
Ce^-4t-4Cte^-4t + 3Cte^-4t = x₁
And except for combining the Cte^-4t in each individual equation I don't see where to go next from there.

 

[tiny-tim]

(just got up :zzz:)

i don't understand where you think a particular solution comes into it

x' - Ax = 0 is a homogeneous equation (it has no t)

x' - Ax = B(t) is not, and its solutions are all the solutions to the homogeneous equation plus one particular solution to the whole equation

when the RHS is 0, no particular solution is needed

(because every homogeneous solution is a particular solution)

 

[hops1]

(just got up :zzz:)

i don't understand where you think a particular solution comes into it

x' - Ax = 0 is a homogeneous equation (it has no t)

x' - Ax = B(t) is not, and its solutions are all the solutions to the homogeneous equation plus one particular solution to the whole equation

when the RHS is 0, no particular solution is needed

(because every homogeneous solution is a particular solution)

I understand that the Homogenous equation doesn't require the t.
I have already gotten the homogenous solution. That's where I get the e^-4t from.
So found the homogenous solution already using the eigenvalues of the system

But the equation given as
x₁' + 5x₁ = x₂
requires the solution to be a sum of the homogenous solution and the particular solution, since it's not set equal to 0.

 

[tiny-tim]

… But the equation given as
x₁' + 5x₁ = x₂
requires the solution to be a sum of the homogenous solution and the particular solution, since it's not set equal to 0.

ohh, i see what you mean now …

you're treating x₂ as if it's a known function of t, instead of as a variable

as you say, the problem with that is it's not a known function (unless of course you make a guess as to what x₂ is)

So attained the homogenous solution already using the eigenvalues of the system

technically, no, the homogeneous solution to x₁' + 5x₁ = x₂ is x₁ = Ce^-5t, isn't it?

can you adapt the solution you already found in the easier case? ([1 2][0 1])

 

[hops1]

ohh, i see what you mean now …

you're treating x₂ as if it's a known function of t, instead of as a variable

as you say, the problem with that is it's not a known function (unless of course you make a guess as to what x₂ is)technically, no, the homogeneous solution to x₁' + 5x₁ = x₂ is x₁ = Ce^-5t, isn't it?

can you adapt the solution you already found in the easier case? ([1 2][0 1])

Actually I wanted to ask about that as well.
Why does the eigenvalue give a different solution to the equation than if I solve the individual homogenous equation.
That's really been confusing me

That's why I'm having trouble adapting the other solution as well.

 

[tiny-tim]

sorry, got to rush out

absolutely no time

 

[hops1]

sorry, got to rush out

absolutely no time

No prob,
I'll see how far I can get in the mean time.
Hope some more people can give a little input as well

 

[tiny-tim]

(i'm back )

ok, the reason why your solution doesn't work (though it did work for the simpler case)

is that the "plain" eigenvector is neither x₁ nor x₂ (in the simpler case, it was)

first you need to find that eigenvector, ie the one that corresponds to e^-4t (with no extra t) …

just solve for Ax = -4x ​

 

[hops1]

(i'm back )

ok, the reason why your solution doesn't work (though it did work for the simpler case)

is that the "plain" eigenvector is neither x₁ nor x₂ (in the simpler case, it was)

first you need to find that eigenvector, ie the one that corresponds to e^-4t (with no extra t) …

just solve for Ax = -4x ​

Hi, thanks for taking your time with me on this.
I understand now why it wouldn't work since, the two equations are dependent on each other pretty much.
So the eigenvector that corresponds to the given eigenvalue is
[1]
[1]

I actually managed to find some notes online regarding this.
Stating that I should solve
(A-rI)w=v
Where r is my eigenvalue and v is my eigenvector and solve the equations to find the w vector which would then be a generalized eigenvector for the maxtrix A.
So this would give me a particular solution with "t" multiplied the original eigenvector plus a term without "t" multiplied by the vector gotten from solving the above equation.
Using the above method I got this answer after going through the calculations

x= C1Ve^(-4t) + C2Vte^(-4t) + C2We^(-4t)
Where V = [1;1] and W = [-1;0]

So I think I managed to get the answer using this method.
Our lecturer had told us about the other method I was trying to do, but when explaining gave a simple example much similar to what you had given me, but then gave the other one in a mock exam we had and said we were supposed to use the method for the simpler version. Anyway the solution he gave turned out to be incorrect as well after closer inspection.

 

[tiny-tim]

So the eigenvector that corresponds to the given eigenvalue is
[1]
[1]

yes, Ve^-4t, where V = [1 1], is a solution

x= C1Ve^(-4t) + C2Vte^(-4t) + C2We^(-4t)
Where V = [1;1] and W = [-1;0]

yes, or you could write it

x = C₁Ve^-4t + C₂Ue^-4t
Where V = [1;1] and U = [t-1;t]

 

[hops1]

yes, Ve^-4t, where V = [1 1], is a solution



yes, or you could write it

x = C₁Ve^-4t + C₂Ue^-4t
Where V = [1;1] and U = [t-1;t]

Thanks a million man.