MHB Solving System of ODEs: Matrix Form, Eigenvalues/Vectors

[shamieh]

Getting stuck on something I think that could be trivial. Maybe someone can see my mistake.

consider the system: $x' = -2x + y$ and $y' = 2x - 3y$

a) Write the system in matrix form
my solution

$\overrightarrow{X} = (^x_y)$

so: $X' = (^{x'}_{y'})$

so $A = $ \begin{bmatrix} -2 & 1 \\ 2 & -3 \end{bmatrix}

b) Find the eigenvalues for the system.
my solution
$\lambda_1 = -1$ and $\lambda_2 = -4$

c) Find an eigenvector for the smallest eigenvalue.
my solution ( Here is where I have either made a algebra mistake or I don't know how to proceed)
$(I-A)X \rightarrow (-I-A)X =$ (\begin{bmatrix} -1 & 0 \\ 0 & - 1 \end{bmatrix} $-$ \begin{bmatrix} -2 & 1 \\ 2 & - 3 \end{bmatrix} ) $X$

Then I am getting: $v_1 - v_2 = 0$ and $-2v_1 + 2v_2 = 0 $ which can be rewritten as one equation $\rightarrow -2v_1+2v_2=0$ so it seems the only numbers I can plug in are: $v_1 = v_2 = -1$ ? But then won't I have like $(^{-1}_{-1})$ and I thought you couldn't do that? Is this the repeated root thing and if so how do I proceed? Or have i made a algebra mistake.

Thanks in advance.

 

[I like Serena]

Hi shamieh,

You are correct. The eigenvector for -1 is $\binom {-1}{-1}$. Or alternatively $\binom 11$. If you multiply A with it, you'll get its negative equivalent, as expected.
What do you mean with a repeated root?

 

[shamieh]

Ignore my comment about the repeated root, I'm thinking too hard .. So similarly would the eigenvector for the largest value be $(^1_{-2})$ ?

- - - Updated - - -

so I have two more questions with regards to this problem.

e) Write the general solution to the system in matrix form.
I got: $X(t) = C_1(^1_1)e^{-t} + C_2 (^1_{-2})e^{-4t}$but part f) says : Write the general solution to the system in scalar form.. What exactly does that mean? How do I do that?

 

[I like Serena]

Ignore my comment about the repeated root, I'm thinking too hard .. So similarly would the eigenvector for the largest value be $(^1_{-2})$ ?

Yup. (Nod)

so I have two more questions with regards to this problem.

e) Write the general solution to the system in matrix form.
I got: $X(t) = C_1(^1_1)e^{-t} + C_2 (^1_{-2})e^{-4t}$

Did you check if it fits into the original DE?
If it fits, it must be right.

but part f) says : Write the general solution to the system in scalar form.. What exactly does that mean? How do I do that?

The solution you have in e) is an equation containing vectors.
So I believe you're supposed to rewrite it into a set of equations that do not contain vectors:
\begin{cases}
x_1(t) = C_1 e^{-t} + C_2 e^{-4t}\\
x_2(t) = C_1 e^{-t} -4 C_2 e^{-4t}
\end{cases}
See? Only scalars.

 

[shamieh]

Yup. (Nod)
Did you check if it fits into the original DE?
If it fits, it must be right.

The solution you have in e) is an equation containing vectors.
So I believe you're supposed to rewrite it into a set of equations that do not contain vectors:
\begin{cases}
x_1(t) = C_1 e^{-t} + C_2 e^{-4t}\\
x_2(t) = C_1 e^{-t} -4 C_2 e^{-4t}
\end{cases}
See? Only scalars.

But do you think I should write it as: $X(t) = 2C_1 e^{-t} - C_2 e^{-4t}$ ? Also why do you have a 4 in front of the $C_2$ in your $x_2(t)$ eqn ?

 

[I like Serena]

But do you think I should write it as: $X(t) = 2C_1 e^{-t} - C_2 e^{-4t}$ ?

Huh?
X(t) is a vector while the right hand side is a scalar. That can't be right.
And where do you coefficients come from?

Also why do you have a 4 in front of the $C_2$ in your $x_2(t)$ eqn ?

Sorry. That should be a 2.