/sci/ - Science & Math

[Anonymous No.10864711]

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 No.10864738]

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

[Anonymous No.10864743]

idk :(

[Anonymous No.10864755]

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 No.10864776]

>tfw 120 IQ bigboy :(

[Anonymous No.10864824]

>>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:
[math]\frac{1}{3}x - \frac{1}{3}y = x' [/math]
[math]\frac{-2}{3}x -\frac{1}{3}y = y' [/math]

New you write it in metrix form.
[math] \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}[/math]

Solve for the Eigen-values and Eigen-vectors of that matrix. and plug them into the solution:
[math] \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}[/math].

[Anonymous No.10864828]

>>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:
[math]\frac{1}{3}x - \frac{1}{3}y = x' [/math]
[math]\frac{-2}{3}x -\frac{1}{3}y = y' [/math]

New you write it in matrix form.
[math] \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}[/math]

Solve for the Eigen-values and Eigen-vectors of that matrix. and plug them into the solution:
[math] \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}[/math].

[Anonymous No.10864832]

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

[Anonymous No.10864847]

>>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 No.10864856]

>>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 No.10864864]

>>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 No.10864889]

>>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 No.10864890]

>>10864828
[math]
\\
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?
[/math]

[Anonymous No.10864896]

>>10864828
[math]
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?
[/math]

[Anonymous No.10864908]

>>10864890
nobody:

him: 1/3

[Anonymous No.10864965]

bump

[Anonymous No.10865607]

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

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

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