Thread: Solving First order differential equations using matrix methods

1. Solving First order differential equations using matrix methods

I have been trying to solve this equation using matrix methods to find the general solution for x1(t) and x2(t)
(There should be a dot above the x1, x2 to show first order)

x1 = 4x1+ 7x2

x2 = -
6x1-9x2

Now i have been shown how to do a second order question so i have used a similar method with this which gets me to this point.

det (a-lambda^2I) = (4-lambda^2)(-9-lambda^2)+42
(lambda^2+2)(lambda^2+3)

lambda^2 = -2 and lambda^2=-3

This is where its different to the question i have been through

As this would mean lambda is j1.414 and j1.732

The second order question i have been through comes out with lambda +or- 2j and +or- j which is then obvious how to continue.

Could somone point me in the right direction or to where ive gone wrong?

I have an exam on wednesday and realy need to be able to work with any combination of question they give me, It could be first or second order.
I didnt think it mattered which kind i was until differentiating the general solution part.

Thank you!

2. Check this out.

3. Ive just been reading through that thread and i can't understand unfortunatly, How would i go through my question step by step? are my eigen values correct?

4. You're setting up the wrong characteristic equation. Your system looks like this:

$\dot{\vec{x}}=A\vec{x},$ where

$A=\begin{bmatrix}4 &7\\-6 &-9\end{bmatrix}.$

The solution to the system is $\vec{x}(t)=e^{At}\vec{x}(0).$

We now have to make sense of the exponential. You use the series definition:

$e^{At}=I+tA+\frac{(tA)^{2}}{2}+\frac{(tA)^{3}}{6}+ \dots$

It's the usual series expansion of the exponential function. We can easily find the nth power of $A$ if we can find an invertible $P$ such that $A=PDP^{-1},$ where $D$ is diagonal. For then

$A^{n}=(PDP^{-1})(PDP^{-1})\dots(PDP^{-1})=PDD\dots DP^{-1}=PD^{n}P^{-1}.$

Taking the nth power of a diagonal matrix is the same as taking the nth power of the diagonal elements, component-wise. Solving $A=PDP^{-1}$ is the diagonalization problem. To diagonalize, assuming it's possible to do this, you find the eigenvalues and eigenvectors. The eigenvalues make up the diagonal of $D$, and the eigenvectors make up the columns of $P$. Hence the importance of the eigenvalues for a system of ODE's.

Now, to find the eigenvalues, you have to set $\det(A-\lambda I)=0,$ not $\det(A-\lambda^{2}I)=0.$ That's where you first went wrong. So carry that correction through, and see what you come up with.

5. Ok ive done it using dec (A-lambdaI)=0 which would give me Lambda= -2 and -3
Then putting that back into the matrix i get.

For lambda -2 6x1=-7x2
Lambda -3 x1=-x2

x1 = 6(Ae^-2jt + Be^-3jt)
x2 = -7(Ae^-2jt -Be^-3jt)

to be honest im lost, i do not understand the exponential part.

6. So your eigenvectors turn out to be

$\vec{x}_{-2}=\begin{bmatrix}1\\ -6/7\end{bmatrix},$ and

$\vec{x}_{-3}=\begin{bmatrix}1\\ -1\end{bmatrix}.$

Therefore, your invertible $P$ can be written as

$P=\begin{bmatrix}1 &1\\-6/7 &-1\end{bmatrix}.$

To form your final solution, you write

$\vec{x}(t)=e^{At}\vec{x}(0)=Pe^{Dt}P^{-1}\vec{x}(0).$

But $D=\begin{bmatrix}-2 &0\\0 &-3\end{bmatrix}.$ So its exponential

$e^{Dt}=\begin{bmatrix}e^{-2t} &0\\0 &e^{-3t}\end{bmatrix}.$

Can you finish from here?

7. I dont understand why there is a 1 in the first vector bracket? i only get either 6 and -7 or 7 and -6.
Then i realy do not understand where the final solution comes from I can see that D and e^Dt are from the two eigen values so do you then work the determinant for that out?

Please forgive my lack of understanding but i realy am trying to understand.

8. I dont understand why there is a 1 in the first vector bracket? i only get either 6 and -7 or 7 and -6.
Any nonzero scalar multiple of an eigenvector is an eigenvector. Why is that, might you ask? Well, recall that a vector $\vec{x}$ is an eigenvector of the matrix $A$ if and only if $\vec{x}\not=0$ and $A\vec{x}=\lambda \vec{x}$ for some scalar $\lambda,$ which is called the eigenvalue corresponding to $\vec{x}$. So, suppose that $\vec{x}$ is an eigenvector of the matrix $A$ with corresponding eigenvalue $\lambda$. Let $y$ be a nonzero scalar. Then $A(y\vec{x})=yA\vec{x}=y\lambda\vec{x}=\lambda(y\ve c{x}).$ Since $A(y\vec{x})=\lambda(y\vec{x}),$ and $y\vec{x}\not=0$ (since $y\not=0$ and $\vec{x}\not=0$), it follows by definition that $y\vec{x}$ is an eigenvector of matrix $A$.

Now, if you look at my $\vec{x}_{-2},$ and multiply by $7$, you might get a more recognizable eigenvector. Does that make sense?

I can see that D and e^Dt are from the two eigenvalues; so do you then work the determinant for that out?
I don't compute the determinant, I just compute the exponential $e^{Dt}.$

I'm not sure I understand what you're not understanding. The solution to this problem: pepper me with questions. Anything at all you don't understand, ask about. Ok?