Question

pls.
solve this linear system ODE
dy/dt=x-2y
dy/dt=x-y

Answer 1

hmm... can you make my life easier and tell me which topic this is.. I forgot all about ODEs.. and I only took this course last semester..

Answer 2

lookin in my textbook for similar problems.. unfortunately this seems to be the chapter that my professor skipped so that we had time to do some partial DEs..

Answer 3

ARMISTRE!! Help us out! I don't know how to do eigenvalues...

Answer 4

This is quite a large problem to do online. I'm going to assume your system of equations is actually\[\frac{dx}{dt}=x-2y\]\[\frac{dy}{dt}=x-y\]

Answer 5

The system may be written in matrix form as,\[x'=Ax\]where the x is a vector, (x,y). We start by assuming a solution of the form,\[x=ve^{\lambda t}\]where v is a vector and the exponential is a scalar. Taking the derivative, you obtain,\[x'=\lambda v e^{\lambda t}\](where we assume v is independent of t). Substituting this into the d.e.,\[\lambda v e^{\lambda t}=A v e^{\lambda t} \rightarrow (A-\lambda I)v e^{\lambda t}=0 \rightarrow (A-\lambda I)v=0\]We find the eigenvalues and eigenvectors by evaluating the determinant of (A-lambda I):

Answer 6

Solve\[\det\left[ \begin{array}c 1-lambda & -2\\1 & -1-lambda\\\end{array} \right]=0\]gives\[-(1-\lambda^2)+2=0 \rightarrow \lambda = \pm i\]

Answer 7

Now we use these to find the eigenvectors. For lambda_1 = i, we're solving for v_1, v_2 in the vector, v:\[\left[ \begin{array}c 1-i & -2\\1 & -1-i\\\end{array} \right]\left[ \begin{array}c v_1\\ v_2\\\end{array} \right]=0\]

Answer 8

The first eigenvector is then\[v_1=\left[ \begin{array}c 1+i\\ 1\\\end{array} \right]\]

Answer 9

For the second eigenvalue, lambda = -i, substituting again and solving, you get\[v_2=\left[ \begin{array}c 1-i\\ 1\\\end{array} \right]\]

Answer 10

Looking back at our assumed form for the solution, we get, \[x_1=e^{i t}\left[ \begin{array}c 1+i\\ 1\\\end{array} \right], x_2=e^{-i t}\left[ \begin{array}c 1-i\\ 1\\\end{array} \right]\]as the pair of fundamental solutions. To obtain a set of real-values functions, we must find the real and imaginary parts of either x_1 or x_2 (the real and imaginary parts taken from either fundamental solution will form a pair of linearly independent solutions).

Answer 11

\[x_1=\left[ \begin{array}c x\\ y\\\end{array} \right]=(\cos t + i \sin t )\left[ \begin{array}c 1+i\\ 1\\\end{array} \right]\]Evaluating each of the x and y components independently,

Answer 12

\[x=(\cos t - \sin t )+i(\cos t + \sin t )\]\[y=\cos t + i \sin t \]

Answer 13

The solution set is then\[u(t)=\left[ \begin{array}c cos t - sin t\\ cos t\\\end{array} \right]\]and\[v(t)=\left[ \begin{array}c cos t + sin t\\ sin t\\\end{array} \right]\]

Answer 14

The solution is then \[\left[ \begin{array}c x\\ y\\\end{array} \right]=c_1u(t)+c_2v(t)\]where u and v are the above vectors.