Question

It's been a long while since I've done some exercises on vectors, and I could use a refreshener on "how to". In this case I got the following x'= matrix . x. Then two vectors are given and I have to verify whether they are solutions for the system of DEs (expressed as a matrix). I'll post the matrix and vectors as a comment after this question post, but I prefer help on "how to" rather than a calculated answer.

Answer 1

System of DEs: \[x'=\left[\begin{matrix}4 & -3 \\ 6 & -7\end{matrix}\right]x\]

Verify whether the vectors

\[x _{1}=\left(\begin{matrix}3e ^{2t} \\ 2e ^{2t}\end{matrix}\right)\]

and

\[x _{2}=\left(\begin{matrix}e ^{-5t} \\ 3e ^{-5t}\end{matrix}\right)\]

are solutions for the system.

Answer 2

Take x1. Calculate x1'.

Answer 3

Plug x1 and x1' into the system. You get an identity?

Answer 4

So when I calculate x'1 I get

\[x _{1}^{'}=\left(\begin{matrix}6e ^{2t} \\ 4e ^{2t}\end{matrix}\right)\]

Answer 5

Yes.

Answer 6

Not sure how to write the matrix as DEs. They are not given except as the matrix.

Answer 7

Our system is of the form \[X' = AX\]

Answer 8

oh yeah... lol

Answer 9

Identity isn't that a diagonal matrix with only 1s on the diagonal and the rest zero?

Answer 10

We can think of it like this too\[\left(\begin{matrix}x_{11}' \\ x_{12}'\end{matrix}\right) = \left[\begin{matrix}4 & -3 \\ 6 & -7\end{matrix}\right]\left(\begin{matrix}x_{11} \\ x_{12}\end{matrix}\right)\]

Answer 11

By identity, I mean something like x = x.

Answer 12

When you calculate AX, you will get a vector.

Answer 13

yeah, x'

Answer 14

Does this vector equal X' ?

Answer 15

But I just calculated x'1

Answer 16

So x_1 is a solution then. Is x_2 a solution?

Answer 17

If I can calculate it, it's a solution then?

Answer 18

No. Sorry, I must have misunderstood. You calculated x_1', so now calculate Ax

Answer 19

Where A is the coefficient matrix you were given.

Answer 20

And without using x1 for x?

Answer 21

With using x1.

Answer 22

We are checking if x1 and x1' satisfies the equation, much like you would with a single DE

Answer 23

Hmmm, I'm missing something here... I calculated x'1 by multiplying A with x1. And now I have to calculate Ax using x1. Isn't that doing the same thing twice?

Answer 24

Lol.

Answer 25

You have just shown that x1 is a solution.

Answer 26

By calculate x1'

Answer 27

I meant take x1, and take its derivative.

Answer 28

So we would take the derivative of 3e^(2t) and 2e^(2t)

Answer 29

so I derive 3e^2t and 2e^2t

Answer 30

Yes.

Answer 31

And if that equals Ax, then x is a solution.

Answer 32

I can see that on the spot, yeah, the derivative of x1 is x'1

Answer 33

Yeah, just take [x1]d/dx, plug in x1 for x, and you see that they are the same immediately.

Answer 34

yup d/dt of (3e^2t) = 6e^2t and d/dt of (2e^2t)=4e^2t

Answer 35

So is x_2 a solution?

Answer 36

So, by using both methods I prove I have the same result, and thus it's a solution.

Answer 37

I'll check

Answer 38

It's a solution because you derive an identity.

Answer 39

The second vector isn't a solution I think

x'2 calculated by multiplying the matrix with x2 =

\[\left(\begin{matrix}e ^{-5t} \\ -3e ^{-5t}\end{matrix}\right)\]

but the derivative of x2=

\[\left(\begin{matrix}-5e^{-5t} \\ -15e ^{-5t}\end{matrix}\right)\]

Answer 40

I think you should get <e^-5t, -15e^-5t> after the product, but nonetheless, it isn't a solution.

Answer 41

Yeah, right, I multiplied the wrong elements for the second row

Answer 42

Ok, thanks for the help... I have to test other stuff in this exercise but I'll try to see whether I can find it on my own first. Next step is using the Wronskian to prove the solutions are lineair independent.

Answer 43

Fun stuff. Good luck.

Answer 44

Hehehe. After systems of DEs I can start on multivariate analyses. I do find it fun stuff though... especially when you figure things out :-)

Answer 45

BTW I just checked: x2 is a solution. I made the same calculation mistake for the first row as well. As I said, a bit rusty on the matrixes.

Answer 46

Yes, you are right. 4 - 9. I didn't catch it at first either.

Answer 47

A good exercise would be to find the eigenvalues of A and compute the solutions.

Answer 48

Yeah, that's what I thought I was supposed to do at first.