Question

given a DE system and first having to find the general solution and then one with given starting values and thus having to find the coefficients. I'm unsure how to do the second part for this DE system. Specifics and what I have found as solution in the following post.

Answer 1

\[x"=-40x+8y\]
\[y"=12x-60y\]

\[\left[\begin{matrix}-40 & 8 \\ 12 & -60\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}x" \\ y"\end{matrix}\right)\]

eigenvalues are -36 and -64.
for eigenvalue -36 -> eigenvector (2,1)
for eigenvalue -64 -> eigenvector (1,-3)

I originally thought the general solution would be
\[\left(\begin{matrix}x \\ y\end{matrix}\right)=c _{1}\left(\begin{matrix}2 \\ 1\end{matrix}\right)e ^{-36t}+c2\left(\begin{matrix}1 \\ -3\end{matrix}\right)e ^{-64t}\]

But it seems to me that follows the procedure as if I'm finding the solution for (x', y') instead of (x", y").

I've read a comment somewhere that because it's the second derivative I would need to take the square of the eigenvalues. That would be 6i and 8i. But then do I only need to plug these in for the power of e, or do I need to use the square also to find the eigenvectors?

Answer 2

The problem regarding finding particular solutions for given starting values is that these are the starting values:

x(0)=19, x'(o)=12
y(0)=3, y'(0)=6

Answer 3

here is wolfram solution if its any help
http://www.wolframalpha.com/input/?i=x%27%27%3D-40x%2B8y%2C+y%27%27+%3D+12x-60y

seems you are on right track of taking sqrt of eigenvalues, then find new eigenvectors

Answer 4

I've used a solution

\[\left(\begin{matrix}x \\ y\end{matrix}\right)=c _{1}\left(\begin{matrix}2 \\ 1\end{matrix}\right)e ^{6it}+c _{2}\left(\begin{matrix}1 \\ -3\end{matrix}\right)e ^{8it}\]

to calculate the coefficients for x0 and y0


\[\left(\begin{matrix}x \\y\end{matrix}\right)=\frac{ 13 }{ 7 }\left(\begin{matrix}2 \\ 1\end{matrix}\right)e ^{i6t} + \frac{ 60 }{ 7 }\left(\begin{matrix}1 \\ -3\end{matrix}\right)e ^{i8t}\]

Then I took the derivative of the general solution (or what I think might be the general solution)
\[\left(\begin{matrix}x' \\ y'\end{matrix}\right)=6i c _{1}\left(\begin{matrix}2 \\ 1\end{matrix}\right)e ^{i6t}+8ic _{2}\left(\begin{matrix}1 \\ -3\end{matrix}\right)e ^{i8t}\]
and found this

\[\left(\begin{matrix}x' \\ y'\end{matrix}\right)=6\left(\begin{matrix}2 \\ 1\end{matrix}\right)e ^{i6t}+0\]

Answer 5

ouch, that looks horrible... an application of the law of misery and headaches

Answer 6

Ok, so it basically comes down to that the method of doing this is really just saying: \[\Large x= C_1 e^{rt} \ \ \ \ \ and \ \ \ \ \ y=C_2e^{rt}\]When we take the second derivative we will just end up with an r^2 term out front of each, correct? \[\Large x''= r^2*C_1 e^{rt} \ \ \ \ \ and \ \ \ \ \ y''=r^2*C_2e^{rt}\]And in fact, we can see that this is just a multiple of the original equation, yes? So let's plug these suckers back in! \[\Large x''= r^2*x \ \ \ \ \ and \ \ \ \ \ y''=r^2*y\]

Now if we look back we have: \[\Large -40x+8y=r^2x \\ \Large 12 x-60y=r^2y\] Can we express this system of equations as a matrix equation? You bet! \[\Large \left[\begin{matrix}-40 & 8 \\ 12 & 60\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=r^2 \left(\begin{matrix}x \\ y\end{matrix}\right)\]

Does what I'm saying make sense so far? This should probably give you some sort of tingly, satisfying feeling of understanding if you're able to guess what I'm about to do next. If not, that's fine. When you're ready to keep going or if you want to try to anticipate where I'm going, then go for it.

Answer 7

Well that would mean that the matrix = r²

(sorry, laptop froze)

Answer 8

so the square of the matrix = r

Answer 9

So do I have to take the square of every number in the matrix?

Answer 10

Wait not quite, now we have the matrix equation: \[\Large \left[\begin{matrix}-40 & 8 \\ 12 & 60\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=r^2 \left(\begin{matrix}x \\ y\end{matrix}\right)\] So we can put the identity matrix in here, it won't change anything. \[\Large \left[\begin{matrix}-40 & 8 \\ 12 & 60\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=r^2 \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)\] take the r^2 inside and we get\[\Large \left[\begin{matrix}-40 & 8 \\ 12 & 60\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)= \left[\begin{matrix}r^2 & 0 \\ 0 & r^2\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)\] Now we can subtract the right from the left or the other way around, it doesn't matter really.\[\Large (\left[\begin{matrix}-40 & 8 \\ 12 & 60\end{matrix}\right]- \left[\begin{matrix}r^2 & 0 \\ 0 & r^2\end{matrix}\right])\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)\] Now we subtract the matrices \[\Large \left[\begin{matrix}-40-r^2 & 8 \\ 12 & 60-r^2\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]Now see how this is really the same as if we were just looking at our eigenvalue squared?

Now the only way we can have a matrix multiplied by a vector equal the zero vector is if the columns are multiples of each other. In other words, the columns are linearly dependent. These types of matrices don't have inverses, so that must mean that the determinant is zero of this matrix.\[\Large \det \left[\begin{matrix}-40-r^2 & 8 \\ 12 & 60-r^2\end{matrix}\right] = 0\]

Answer 11

sorry, laptop froze again, and I hadn't noticed... Ok, I get what you're doing.

Answer 12

Ok so hopefully that helps you get to your equations correctly, now I also realize you have one other issue, and that is how to find the particular solution to the DEs with your initial conditions right?

Answer 13

So my eigenvectors are correct, but to plug in r for the general solution I need to plug in the sqaure of my eigenvalues?

Answer 14

\[\lambda = r ^{2}\] right... use lambda for my eigenvectors, but only r to plug in e^rt

Answer 15

Yeah, so really it looks like you need the square root of your eigenvalues?

Answer 16

yeah, that would give

\[\left(\begin{matrix}x \\ y\end{matrix}\right) = c _{1} \left(\begin{matrix}2 \\ 1\end{matrix}\right)e ^{i6t} + c _{2}\left(\begin{matrix}1 \\ -3\end{matrix}\right)e ^{i8t}\]

Answer 17

and I can apply Euler's formula on this

Answer 18

that would make it

\[\left(\begin{matrix}x \\ y\end{matrix}\right)=c _{1}\left(\begin{matrix}2 \\ 1\end{matrix}\right)\left[ \cos (6t)+isin (6t)\right]+c2\left(\begin{matrix}1 \\ -3\end{matrix}\right)\left[ \cos(8t)+isin(8t) \right]\]

Answer 19

you also need to account for the negative case
r^2 = -36,-64
r = +-6i , +-8i

\[\left(\begin{matrix}x \\ y\end{matrix}\right) = c_1 \left(\begin{matrix}2 \\ 1\end{matrix}\right)e^{i 6t}+ c_2 \left(\begin{matrix}2 \\ 1\end{matrix}\right)e^{-i 6t}+ c_3 \left(\begin{matrix}1 \\ -3\end{matrix}\right)e^{i 8t}+ c_4 \left(\begin{matrix}1 \\ -3\end{matrix}\right)e^{-i 8t}\]
then
\[\left(\begin{matrix}x' \\ y'\end{matrix}\right) = i 6c_1\left(\begin{matrix}2 \\ 1\end{matrix}\right)e^{i 6t}-i 6c_2\left(\begin{matrix}2 \\ 1\end{matrix}\right)e^{-i 6t}+i 8c_3\left(\begin{matrix}1 \\ -3\end{matrix}\right)e^{i 8t}-i 8c_4\left(\begin{matrix}1 \\ -3\end{matrix}\right)e^{-i 8t}\]
apply initial values
solve system of of 4 equations for constants
\[\left[\begin{matrix}2 & 2 &1&1 \\ 1 & 1&-3&-3 \\ 12i&-12i&8i&-8i \\ 6i&-6i&-24i&24i\end{matrix}\right] \left(\begin{matrix}c_1 \\ c_2\\c_3\\c_4\end{matrix}\right)=\left(\begin{matrix}19 \\ 3\\12\\6\end{matrix}\right)\]

once you know constants you can use Eulers to simplify and write solution in terms of sin/cos
the "i" terms should cancel out

final solution
http://www.wolframalpha.com/input/?i=x%27%27%3D-40x%2B8y%2C+y%27%27+%3D+12x-60y%2Cx%280%29%3D19%2Cy%280%29%3D3%2Cx%27%280%29+%3D12%2Cy%27%280%29+%3D6

Answer 20

You are right @dumbcow. I had forgotten about the negative root.

Answer 21

:)