MHB System of differential equations involving matrices

[Guest2]

Let $\mathbf{A} = \begin{pmatrix} 9 & 2 \\ 1 & 8 \end{pmatrix}.$ Obtain the general solution $\mathbf{y}(t)$ of the system of differential equations $\displaystyle \frac{d\mathbf{y}}{dt} = \mathbf{Ay}$:

$\begin{cases} \frac{dy_1}{dt} = 9y_1+2y_2 \\ \frac{dy_2}{dt} = y_1+8y_2 \end{cases}$

and find the unique solution satisfying $y_1(0) = 1$ and $y_2(0) = 5$.​

I've found that $\mathbf{y} = ke^{\mathbf{A}t}$, where $k \in \mathbb{R}$. That also eigenvalues of $\mathbf{A}$ are $\lambda = 7,10$. But I do not know how to proceed.

 

[topsquark]

Re: System of ifferential equations involving matrices

Let $\mathbf{A} = \begin{pmatrix} 9 & 2 \\ 1 & 8 \end{pmatrix}.$ Obtain the general solution $\mathbf{y}(t)$ of the system of differential equations $\displaystyle \frac{d\mathbf{y}}{dt} = \mathbf{Ay}$:

$\begin{cases} \frac{dy_1}{dt} = 9y_1+2y_2 \\ \frac{dy_2}{dt} = y_1+8y_2 \end{cases}$

and find the unique solution satisfying $y_1(0) = 1$ and $y_2(0) = 5$.​

I've found that $\mathbf{y} = ke^{\mathbf{A}t}$, where $k \in \mathbb{R}$. That also eigenvalues of $\mathbf{A}$ are $\lambda = 7,10$. But I do not know how to proceed.

Well, you have non-zero eigenvalues for A. That means you can diagonalize it. What happens to your two equations if you change the basis so that A is diagonal?

-Dan

 

[Guest2]

Re: System of ifferential equations involving matrices

Well, you have non-zero eigenvalues for A. That means you can diagonalize it. What happens to your two equations if you change the basis so that A is diagonal?

-Dan

I haven't obtained the two equations yet. They're asking me to show it. I'm at the step where I've $\displaystyle \frac{d\mathbf{y}}{dt} = \mathbf{Ay} \implies \mathbf{y} = ke^{\mathbf{A}t}$.

 

[topsquark]

Let $\mathbf{A} = \begin{pmatrix} 9 & 2 \\ 1 & 8 \end{pmatrix}.$ Obtain the general solution $\mathbf{y}(t)$ of the system of differential equations $\displaystyle \frac{d\mathbf{y}}{dt} = \mathbf{Ay}$:

$\begin{cases} \frac{dy_1}{dt} = 9y_1+2y_2 \\ \frac{dy_2}{dt} = y_1+8y_2 \end{cases}$

and find the unique solution satisfying $y_1(0) = 1$ and $y_2(0) = 5$.​

Ignore the y = ke^{At} thing for a moment.

This is the long way around for this level of problem, but it generalizes nicely.

What I want you to do is diagonalize A. You have non-zero eigenvalues, so we can do the following. Find a transformation matrix S to act on you original equations as so:
[math]S \vec{y}' = SA \vec{y} = (S A S^{-1}) (S \vec{y} )[/math]

such that
[math]S A S^{-1} = \left ( \begin{matrix} 7 & 0 \\ 0 & 10 \end{matrix} \right )[/math]

One possibility is [math]S = \left ( \begin{matrix} 1 & -2 \\ 1 & 1 \end{matrix} \right )[/math]

Thus
[math]S \vec{y}' = \left ( \begin{matrix} 7 & 0 \\ 0 & 10 \end{matrix} \right ) S(\vec{y})[/math]

Call [math]S \vec{y} = Y[/math] and thus we also have [math]S \vec{y}' = Y'[/math]

So you need to solve
[math]\left ( \begin{matrix} Y_1' \\ Y_2 ' \end{matrix} \right ) = \left ( \begin{matrix} 7 & 0 \\ 0 & 10 \end{matrix} \right ) \left ( \begin{matrix} Y_1 \\ Y_2 \end{matrix} \right )[/math]

This decouples nicely. Once you have Y_1 and Y_2 then you can solve the simultaneous equations problem to find y_1 and y_2.

-Dan

 

[Guest2]

...

I was actually trying to get the system of equations in the cases, i.e. the one I was supposed solve (I completely misread the problem)! (Rofl)

Thank you. I'll study your post.