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(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...
GCT said:are you sure its not exact?
Hurkyl 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...
Hurkyl said:Then I suppose it's not singular, then.
Doesn't my linear solution arise when c2 = d2 = 0?
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!
Also, when AD-BC=0 as the case above, then we really have no eigenvalues to determine its behavior.
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?)
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.
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
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 An
are just aiin. 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.
An = (TDT-1) (TDT-1)...(TDT-1)
= (TDnT-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-1b)
make a linear change of variables
y = x + A-1b,
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-1b = exp(At) c - A-1b
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.
alist = {u, v} /.
Solve[{s u == 2 u - 4 v + 5/s, s v - 1 == - u + 2 v - 3/s}, {u, v}]
x = InverseLaplaceTransform[alist[[1, 1]], s, t]
y = InverseLaplaceTransform[alist[[1, 2]], s, t]
A non-exact O.D.E. (ordinary differential equation) is a type of mathematical equation that involves the derivatives of a function with respect to one or more independent variables. Unlike exact O.D.E.s, non-exact O.D.E.s cannot be solved by simply integrating both sides of the equation. Instead, additional techniques such as coordinate axis shift may be needed to find a solution.
Coordinate axis shift is a technique used to transform a non-exact O.D.E. into an exact one. This involves changing the variables in the equation to eliminate any terms that are not exact differentials. By doing so, the equation becomes easier to solve using standard integration techniques.
The steps involved in solving a non-exact O.D.E. with coordinate axis shift are:
No, not all non-exact O.D.E.s can be solved using coordinate axis shift. This technique is only applicable when the non-exact terms in the equation can be transformed into exact ones. If this is not possible, other methods such as the use of integrating factors may be needed.
Yes, there are certain limitations to using coordinate axis shift. This technique may not work for all types of non-exact O.D.E.s and may only be applicable in specific cases. It also requires a good understanding of the underlying mathematical concepts and may not always result in a closed-form solution.