Question

Help please!
find the general solution of the following system of equations:
dx/dt = 2y - 4x
dy/dt = 2y - (5/2)x

Answer 1

@jim_thompson5910 @radar would love some guidance if someone's free please? @Zarkon

Answer 2

@satellite73 @Luis_Rivera calling in the big guns please...?
@asnaseer @Agent_Sniffles

Answer 3

Sorry, out of my league.

Answer 4

@katlin95 @SWAG @timo86m @goformit100
...
@genius12 @Compassionate @cshalvey
would any of you guys have an idea where to start...?

Answer 5

Sorry I do not know :/

Answer 6

what if you divide all of it by dt and take integral? maye idk lol.

Answer 7

\[\begin{cases}
x'=2y-4x\\
y'=2y-\frac{5}{2}x
\end{cases}\]
As a matrix equation, you'd write
\[\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}-4&2\\-\frac{5}{2}&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\]
Find the eigenvalues:
\[\begin{align*}\begin{vmatrix}-4-\lambda&2\\
-\frac{5}{2}&2-\lambda\end{vmatrix}=0~&\Rightarrow~\lambda^2+2\lambda - 3=0\\
~&\Rightarrow~ (\lambda+3)(\lambda-1)=0\\
~&\Rightarrow~\lambda_1=-3, \lambda_2=1 \end{align*}\]

Answer 8

Now for the eigenvectors.

Take \(\lambda_1=-3\). Let \(\vec{b_1}=\begin{pmatrix}b_1\\b_2\end{pmatrix}\) be the associated eigenvector. Now you solve for \(\vec{b_1}\):
\[\begin{pmatrix}
-4-(-3)&2\\ -\frac{5}{2}&2-(-3)
\end{pmatrix}\begin{pmatrix}b_1\\b_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\\
\begin{pmatrix}
-1&2\\ -\frac{5}{2}&5
\end{pmatrix}\begin{pmatrix}b_1\\b_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
So, you get the following equations:
\[b_1=2b_2\\
\frac{5}{2}b_1=5b_2~\Rightarrow~b_1=2b_2\]
Now you choose any non-zero values of \(b_1\) and \(b_2\) that work. It looks like the simplest one is when \(b_2=1\), making \(b_1=2\).

So, the first eigenvector is \(\vec{b_1}=\begin{pmatrix}2\\1\end{pmatrix}\).

Answer 9

Similarly, take \(\lambda_1=1\). Let \(\vec{b_2}=\begin{pmatrix}b_1\\b_2\end{pmatrix}\) be the associated eigenvector.
\[\begin{pmatrix}
-4-1&2\\ -\frac{5}{2}&2-1
\end{pmatrix}\begin{pmatrix}b_1\\b_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\\
\begin{pmatrix}
-5&2\\ -\frac{5}{2}&1
\end{pmatrix}\begin{pmatrix}b_1\\b_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
You have
\[5b_1=2b_2\\
\frac{5}{2}b_1=b_2\]
(Again, they're the same equation.) Let \(b_1=2\), making \(b_2=5\). So, your second eigenvector is \(\vec{b_2}=\begin{pmatrix}2\\5\end{pmatrix}\).

Answer 10

The solution to the system of ODEs is then
\[\large\begin{pmatrix}x\\y\end{pmatrix}=C_1e^{\lambda_1}\vec{b_1}+C_2e^{\lambda_2}\vec{b_2}\]

Answer 11

... holy cr@p !
that is awesome! thank you so much!

Answer 12

look, what you wrote above now makes the lectures seem more understandable, so thanks, I'm going to try the next one solo

ok, so just to clarify in my mind:
we can only work out the general solution (not the exact solution) to the equation because we dont have initial values, yeah...? @SithsAndGiggles

Answer 13

Yep. If you were given something like \(\begin{pmatrix}x(0)\\y(0)\end{pmatrix}=\begin{pmatrix}2\\0\end{pmatrix},\) only then could you solve for \(C_1\) and \(C_2\).

And you're welcome :)