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Main Question or Discussion Point
(2x-4y+5)y` +x-2y+3=0
I've tried shifting the coordinate axis, but it doesn't work!
It's also not exact...
I've tried shifting the coordinate axis, but it doesn't work!
It's also not exact...
For:GCT said:are you sure its not exact?
Edit: The linear solutions are NOT singular solutions. See discussion belowHurkyl said:Educated guessing yields a simpler solution than last time: if you try y = mx + b, you get [itex]y = x/2 + 11/8[/itex]. (The other linear solution is singular! How amusing to have a singular singular solution. At least I feel that these linear solutions are singular in some sense)
I notice that this one is halfway between the two lines...
Sure looks like it Hurkyl although c2 and d2 being zero affects what c1 and d1 can be which in turn limits what the initial condition can be. I'll work it through. Thanks!Hurkyl said:Then I suppose it's not singular, then.
Doesn't my linear solution arise when c2 = d2 = 0?
Alright it works out. Through the parametric analysis, I got for the constants in the case y(0)=a:saltydog said:Sure looks like it Hurkyl although c2 and d2 being zero affects what c1 and d1 can be which in turn limits what the initial condition can be. I'll work it through. Thanks!
Yes you do.Also, when AD-BC=0 as the case above, then we really have no eigenvalues to determine its behavior.
Is the negative exponent a typo? Should it not be positive? Also, I'll spend time with your analysis. Thanks!Hurkyl said:Working it out, we have two families:
[x, y] = (At + B) [2, 1]
and
[x, y] = B [2, 1] + C exp(-4 t) [2, -1]
(You notice that the equation is of the form x^T M dx for some matrix M?)
Thanks for the clairification Hurkyl. I'll spend time with it. Also, I tell you what, this:Hurkyl said:No, the negative is correct. My parametrization has the opposite orientation!
(Notice our original ODE's are off by a factor of -1)
matrix form is easiersaltydog said:Thanks for the clairification Hurkyl. I'll spend time with it. Also, I tell you what, this:
[tex]\frac{dy}{dt}=Ax+By+E[/tex]
[tex]\frac{dx}{dt}=Cx+Dy+F[/tex]
should be comprehensively catagorized and I suspect it is somewhere already. That is, what does it do as a function of the parameters? Surely someone has already worked this out. I just don't have the reference.
Jesus Lurflurf. Why am I not surprised . . . Can you kindly help me with a matter of notation:lurflurf said:matrix form is easier
x'=Ax+b
related homogeneous problem
x'=Ax
solution
x=exp(At)x0
variation of parameters
x=exp(At)u
u'=exp(-At)b
[tex]x=e^{At}x_0+e^{At}\int e^{-At}b dt[/tex]
also undetermined coeficents works well
for b constant
b'=0
Thanks a lot Qbert. I'm not supprised neither but I digress. It's all very interesting to me, and I wish to better obtain a handle on the global dynamics of the system as a whole.qbert said:it's more like
[tex]
\left( \begin{array}{c} x'(t) \\ y'(t) \end{array} \right) =
\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]
\left( \begin{array}{c} x(t) \\ y(t) \end{array} \right) +
\left( \begin{array}{c} b_1 \\ b_2 \end{array} \right)
[/tex]
gives
[tex] \mathbf{x'} = \mathbf{Ax + b} [/tex]
given some matrix A then exp(A) is
[tex] I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \ldots [/tex]
which I always thought was a singularly unhelpful way to define anything.
It's a little easier if A is diagonal. Then the elements of A^{n}
are just a_{ii}^{n}. That is, if A is diagonal you just
exponentiate along the main diagonal.
example:
[tex]
\exp{\left[ \begin{array}{cc} a & 0 \\ 0 & d \end{array} \right] } =
\left[ \begin{array}{cc} e^a & 0 \\ 0 & e^d \end{array} \right]
[/tex]
If you can diagonalize A. (That is if there is some matrix such that
TDT^{-1} = A). Then you can get exp(A) fairly easy.
A^{n} = (TDT^{-1}) (TDT^{-1})...(TDT^{-1})
= (TD^{n}T^{-1})
[tex] \exp{A} = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \ldots [/tex]
[tex] = TT^{-1} + TDT^{-1} + \frac{1}{2!}(TDT^{-1})^2 + \frac{1}{3!}(TDT^{-1})^3 + \ldots [/tex]
[tex] = TT^{-1} + TDT^{-1} + (T \frac{1}{2!}D^2T^{-1}) + (T\frac{1}{3!}D^3T^{-1}) + \ldots [/tex]
[tex] = T(I + D + \frac{1}{2!}D^2 + \frac{1}{3!}D^3 + \ldots)
T^{-1} [/tex]
[tex] \exp{A} = Te^DT^{-1} [/tex]
Of course, it just so happens if you make a matrix whose columns
are the eigenvectors of A you get T. (I'll leave the proof to whomever
is interested.)
If A is invertible and b is constant you can solve the
x' = Ax + b system super easy.
note x' = A(x + A^{-1}b)
make a linear change of variables
y = x + A^{-1}b,
then y' = x' and the system
is y' = Ay
whose solution is y = exp(At) c
where c is a column vector of arbitrary constants
and
x = y - A^{-1}b = exp(At) c - A^{-1}b
Now the questions comes up, under what
conditions can you diagonalize A? what should
you do if you can't? what if A is not invertible?
etc.