Eigenvalue Method as a diferential equation!

[wolly]

How did m+nt appear and how did n=0 and the other things appear? Also who is $$μ$$?
I'm talking about an Eigenvalue Method as a differential equation and I don't know why this is solved like this.Can someone help me?

 

[blamocur]

I don't understand the context, and I cannot read Romanian, but here is what I think about your first question:

$m+nt$ appears because $\lambda_1$, which is equal to -3, has multiplicity of 2.

 

[wolly]

I don't understand the context, and I cannot read Romanian, but here is what I think about your first question:

$m+nt$ appears because $\lambda_1$, which is equal to -3, has multiplicity of 2.

And n=0 c=-a-b r=d=s
IT says that IT results from Y'=AY?

 

[wolly]

I don't understand the context, and I cannot read Romanian, but here is what I think about your first question:

$m+nt$ appears because $\lambda_1$, which is equal to -3, has multiplicity of 2.

Do You have Something like a PDF that can show me how to solve this?

 

[wolly]

And n=0 c=-a-b r=d=s
IT says that IT results from Y'=AY?

@blamocur How did these 3 solutions come from?

@logistic_guy A Little help here!

 

[blamocur]

And n=0 c=-a-b r=d=s

Not sure about $n$, but the rest is derived from the fact that (a,b,c) and (d,r,s) are eigenvectors of A.

 

[blamocur]

Do You have Something like a PDF that can show me how to solve this?

No, I don't have it. But it looks like pretty standard stuff about ODEs with constant coefficients. Don't you have a textbook? Where does this problem come from? Is it part of your homework?

 

[wolly]

No, I don't have it. But it looks like pretty standard stuff about ODEs with constant coefficients. Don't you have a textbook? Where does this problem come from? Is it part of your homework?

IT comes from a math college contest exercise book and I know how to solve some parts but not all of them!
I am referring to the eigenvalue and eigenvector Matrix Equations

 

[blamocur]

I am referring to the eigenvalue and eigenvector Matrix Equations

You need to learn the subject in a more systematic way. Asking for help on a forum is no substitute for studying.

 

[wolly]

I found a pdf of how to solve these. I forgot to add them
@blamocur

 

[fresh_42]

The idea is as follows. If we have $(A-\lambda I)(x)=0,$ i.e. $Ax=\lambda x$ then we solve $y'=\lambda y$ and obtain $y=me^{\lambda t}.$ However, we actually have for $\lambda=-3$ that
$$(A-\lambda I)^2(x)=0.$$ That means that either $(A-\lambda I)(x)=0$ is already zero, in which case we get the solution $y=me^{\lambda t},$ or, and this is different, that $(A-\lambda I)((A-\lambda I)(x))=0.$ Now
$$(A-\lambda I)(A-\lambda I)=A^2-2\lambda A +\lambda^2 I=0$$and we have to solve $y''-2\lambda y'+\lambda^2y=0.$ Let's see if $y=nte^{\lambda t}$ is a solution for this equation.

$$\begin{array}{lll} y&=n te^{\lambda t}\\ y'&=ne^{\lambda t}+n\lambda te^{\lambda t}=n e^{\lambda t}+\lambda y\\ y''&=n\lambda e^{\lambda t}+\lambda y'=2n\lambda e^{\lambda t}+\lambda^2 y \end{array}$$$$\begin{array}{lll} y''-2\lambda y'+\lambda^2 y&=2n\lambda e^{\lambda t}+\lambda^2 y-2\lambda(n e^{\lambda t}+\lambda y)+\lambda^2 y=0 \end{array}$$
Hence, it is a solution. Combining both, we get
$$y=me^{\lambda t}+n te^{\lambda t}=(m+nt)e^{\lambda t}.$$This is the solution space for the generalized eigenspace $\lambda=\lambda_1=-3$ with multiplicity two.
If we finally add the solution space for the generalized eigenspace $\lambda=\lambda_2=0$ with multiplicity one, then we get
$$y=(m+nt)e^{\lambda_1 t}+pe^{\lambda_2 t}=(m+nt)e^{-3t}+pe^{0\cdot t}=(m+nt)e^{-3t}+p.$$The parameters $m,n,p$ result from the initial values for the differential equation.