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Anonymous Mon 05 Aug 03:27:28 2019 No.10864711 View Reply Original Report

Quoted By: >>10864738 >>10864824 >>10864828

Is it possible to solve every differential equation system by elimination? I keep solving pic related but keep getting c1cost+c2sint over and over. for both x(t) and y(t) but that is not correct. anyone good with DiffEQ's want to tell me what I might be doing wrong?

Anonymous

Anonymous Mon 05 Aug 2019 03:39:13 No.10864738 Report

Quoted By:

>>10864711
Uh, what do you think Row Reduced Echelon form is sweetie

Anonymous

Anonymous Mon 05 Aug 2019 03:40:11 No.10864743 Report

Quoted By:

idk :(

Anonymous

Anonymous Mon 05 Aug 2019 03:43:51 No.10864755 Report

Quoted By:

nothing I have read has said elimination should not be applicable to certain equations but this MUST be true. someone here must know this?

Anonymous

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Anonymous Mon 05 Aug 2019 03:54:10 No.10864776 Report

Quoted By:

>tfw 120 IQ bigboy :(

Anonymous

Anonymous Mon 05 Aug 2019 04:12:17 No.10864824 Report

Quoted By: >>10864832

>>10864711
you start with the system of equations in your pic.

Then using elimination methods (adding, subtracting and scalar multiplication of rows) you make sure each equation has only one derivative function in it....so you drive at:
$\frac{1}{3}x - \frac{1}{3}y = x'$
$\frac{-2}{3}x -\frac{1}{3}y = y'$

New you write it in metrix form.
$\begin{bmatrix} \frac{1}{3} & \frac{-1}{3} \\ \frac{-2}{3} & \frac{-1}{3} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} =\frac{d}{dt} \begin{bmatrix} x\\ y \end{bmatrix}$

Solve for the Eigen-values and Eigen-vectors of that matrix. and plug them into the solution:
$\begin{bmatrix} x(t)\\ y(t) \end{bmatrix} = \begin{bmatrix} a_1\\ a_2 \end{bmatrix} e^{\lembda_1 t} + \begin{bmatrix} b_1\\ b_2 \end{bmatrix} e^{\lambda_2 t}$ .

Anonymous

Anonymous Mon 05 Aug 2019 04:13:44 No.10864828 Report

Quoted By: >>10864847 >>10864890 >>10864896

>>10864711
you start with the system of equations in your pic.

Then using elimination methods (adding, subtracting and scalar multiplication of rows) you make sure each equation has only one derivative function in it....so you drive at:
$\frac{1}{3}x - \frac{1}{3}y = x'$
$\frac{-2}{3}x -\frac{1}{3}y = y'$

New you write it in matrix form.
$\begin{bmatrix} \frac{1}{3} & \frac{-1}{3} \\ \frac{-2}{3} & \frac{-1}{3} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} =\frac{d}{dt} \begin{bmatrix} x\\ y \end{bmatrix}$

Solve for the Eigen-values and Eigen-vectors of that matrix. and plug them into the solution:
$\begin{bmatrix} x(t)\\ y(t) \end{bmatrix} = \begin{bmatrix} a_1\\ a_2 \end{bmatrix} e^{\lambda_1 t} + \begin{bmatrix} b_1\\ b_2 \end{bmatrix} e^{\lambda_2 t}$ .

Anonymous

Anonymous Mon 05 Aug 2019 04:14:58 No.10864832 Report

Quoted By: >>10864856

>>10864824
where did those 1/3 co-efficients come from?

Anonymous

Anonymous Mon 05 Aug 2019 04:20:00 No.10864847 Report

Quoted By:

>>10864828
Step 1: teke 2 time the first row, and add it to the second.
Step 2: dived the resulting second row by -3
Stap 3: Add the resulting second row to the first row.
Step 4: rearrange it such. that you have derivative functions on one side (primed functions) and non derivatives on the other side.

Anonymous

Anonymous Mon 05 Aug 2019 04:21:38 No.10864856 Report

Quoted By: >>10864864

>>10864832
Step 1: teke 2 time the first row, and add it to the second.
Step 2: dived the resulting second row by -3
Stap 3: Add the resulting second row to the first row.
Step 4: rearrange it such. that you have derivative functions on one side (primed functions) and non derivatives on the other side.

Anonymous

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Anonymous Mon 05 Aug 2019 04:25:20 No.10864864 Report

Quoted By: >>10864889

>>10864856
I don't know man will this give you this answer? because that's what the calculator says it is and it checks

Anonymous

Anonymous Mon 05 Aug 2019 04:37:40 No.10864889 Report

Quoted By:

>>10864864
That's interesting. Maybe I made a mistake somewhere in the elimination. You should be able to use elimination for that type of equation.

http://www.math.psu.edu/tseng/class/Math251/Notes-LinearSystems.pdf

Anonymous

Anonymous Mon 05 Aug 2019 04:38:05 No.10864890 Report

Quoted By: >>10864908

>>10864828
$ \\ x'-x-y'=0 \\ 2x'-y'-y=0 \\ \\ x+y'= x' \\ 2x'-y=y' \\ \\ x+y'= x' \\ 2(x+y')-y=y' \\ \\ x+y'= x' \\ 2x+2y'-y=y' \\ ok\\ \frac{1}{3}x-\frac{1}{3}y=x' \\ \frac{-2}{3}x-\frac{1}{3}y=y' \\ huh? $

Anonymous

Anonymous Mon 05 Aug 2019 04:39:53 No.10864896 Report

Quoted By:

>>10864828
$ x'-x-y'=0 \\ 2x'-y'-y=0 \\ ok\\ x+y'= x' \\ 2x'-y=y' \\ ok\\ x+y'= x' \\ 2(x+y')-y=y' \\ ok\\ x+y'= x' \\ 2x+2y'-y=y' \\ ok\\ \frac{1}{3}x-\frac{1}{3}y=x' \\ \frac{-2}{3}x-\frac{1}{3}y=y' \\ huh? $

Anonymous

Anonymous Mon 05 Aug 2019 04:44:21 No.10864908 Report

Quoted By:

>>10864890
nobody:

him: 1/3

Anonymous

Anonymous Mon 05 Aug 2019 05:16:48 No.10864965 Report

Quoted By:

bump

Anonymous

Anonymous Mon 05 Aug 2019 13:35:50 No.10865607 Report

Quoted By:

Ok, i’m back again. I have must have messed up in my elimination process. I did it again. And got the new equation

$ \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \frac{d}{dt} \begin{bmatrix} x\\ y \end{bmatrix} $
This leads to a characteristic equation:
$\lambda^2 + 1 = 0$

Which, when solved, gives complex numbers +/- i. Which means you get the sin(t) and cos(t).

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