Question

linear system of 1st order DE's

Answer 1

say i have a system of linear DE's
\(\dot x = x + y\)
\(\dot y = x - y\)

just made that up so the numbers may be inconvenient; and to be clear, \( \dot x, \dot y\) means \( \frac{dx}{dt}, \frac{dy}{dt}\) ; and \(x = x(t)\), \(y = y(t)\)

then
\[\left[\begin{matrix}\dot x \\ \dot y\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & -1\end{matrix}\right] \left[\begin{matrix}x \\ y\end{matrix}\right]\]

with eigenvalues \( \; \pm \sqrt{2}\)

but what if it's
\(\dot x = x + y \color{red}{+2}\)
\(\dot y = x - y \color{red}{+3}\)

ie just adding in those constants??

i can't remember but i think i need to use superposition, ie from
\[\left[\begin{matrix}\dot x \\ \dot y\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & -1\end{matrix}\right] \left[\begin{matrix}x \\ y\end{matrix}\right] + \left[\begin{matrix}2 \\ 3 \end{matrix}\right]\]

so separately, i find the particular solution
\[\left[\begin{matrix}\dot x \\ \dot y\end{matrix}\right] = \left[\begin{matrix}2 \\ 3 \end{matrix}\right]\]

\[\rightarrow \left[\begin{matrix} x \\ y\end{matrix}\right]_p = t \left[\begin{matrix}2 \\ 3 \end{matrix}\right] + \left[\begin{matrix}C_x \\ C_y \end{matrix}\right]\]

**i'm looking at finding fixed points; and sketching the phase diagram**.

so, for the phase diagram, i then find a way to build that extra "particular" bit in?

doesn't feel right though. isn't there a better way to do this?!?! have googled like crazy and nothing there that i can find.

many thanks in advance 😋

Answer 2

Wow Thats quit hard!

Answer 3

undetermined coefficients should work smoothly in finidng a particular solution :

let \(\vec{c}=\begin{pmatrix}c_1\\c_2\end{pmatrix}\) be a particular solution. plugging this into the system gives
\[\left[\begin{matrix} 0\\ 0\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & -1\end{matrix}\right] \left[\begin{matrix}c_1 \\ c_2\end{matrix}\right] + \left[\begin{matrix}2 \\ 3 \end{matrix}\right] \]
we can solve \(\vec{c}\) as the matrix is invertible..

Answer 4

thanks ganesh

i eventually did this by ID'ing the fixed points

\(\dot x = x + y +2 = 0 \\
\dot y = x - y +3 = 0 \)

\(\implies x,y = -\frac{5}{2}, \frac{1}{2}\)

then re-centering round the fixed point gives \(X = x+\frac{5}{2}, Y = y -\frac{1}{2}; \quad \dot X = \dot x, \; \dot Y = \dot y \)

so
\(\dot X = X - \frac{5}{2} + Y + \frac{1}{2} +2 = X+Y= 0 \\
\dot Y = X - \frac{5}{2} - (Y + \frac{1}{2} ) +3 = X-Y=0 \)


so we have the same eigenvalues.....

and hey presto 💥 your solution yields
\(c_1 + c_2 = 2\\c_1 - c_2 = 3\)
\(c_1, c_2 = \frac{5}{2}, -\frac{1}{2} \)


the method i am using is intended to unravel phase diagrams for non linear. i think i have overlooked the linear approach, which is not good ;-(

i think i am explaining myself really badly here, but your post was a tremendous help. i will try make this clearer tomorrow when i have more time to sift through it.....

Answer 5

i have to close this to open another but i still intend to come back to it......if that's ok.