differential equations of 2005th power

[alext87]

Can anybody explain how to and solve the equation: d2005y/dx2005 - y = 0 for y?

 

I have tried ancillary equs but are there complex roots. I know the ans it y=exp(x) but need to be able to show it.

[timo]

Plug it in?

[CPL.Luke]

what do you mean by d2005y/dx2005?

 

do you mean dy/dx-y=0?

 

the problem above is seperable and thus integrable. however since you know the solution is e^x you can just plug t in to verify the solution, and because its a linear first orer equation any scalar multiple of e^x will be a solution, and there is only one basis function.

[Tannin]

We shall prove in general that the equation [math]\frac{d^n y}{dx^n}-y=0[/math] has [math]y=e^x[/math] as it's solution.

The proof is based on the mathematical induction. As usual it is done in two stages:

1. Show for n=1

[math]\frac{dy}{dx}=y[/math]

As CPL.Luke has pointed out, the solution is [math]e^x[/math]

2. Assuming that for arbitrary natural [math]n[/math] the solution is [math]e^x[/math], i.e. [math] \frac{d^n y}{dx^n}=y[/math] we shall show that this is true for [math]n+1[/math]

[math]\frac{d^{n+1} y}{dx^{n+1}}=\frac{d}{dx} \left( \frac{d^n y}{dx^n} \right) =\frac{dy}{dx}=y [/math]

We are substituting the assumption instead of the parentheses.

The proof is done.

[the tree]

I would assume that he means [math]\frac{d^{2005}x}{dy^{2005}}-y[/math]. The 2005^th differencial of e^x is e^x (no surprises there), this can be assumed so nothing needs to be shown.

 

Edit: meh, Tannin beat me to it.

[uncool]

The people before me have given one possible answer to the equation; now the problem is to find all answers.

Let us assume for the moment that y will be positive. Let us then assume that y = e^(ax) where a is an integer, as that is the form that will work best for the equation.

Then the equation reduces to:

e^(ax)*a^(2005) - e^(ax) = 0. As e^(ax) =/= 0, we can remove it. Therefore, we are left with a^2005 - 1 = 0 - that is, a is one of the 2005th roots of unity.

As we now have 2005 unique answers and the equation is a linear equation of order 2005 (or however you say it), we now have all the possible answers - they are linear combinations of the 2005 answers given.

=Uncool-

[woelen]

There is a general solution for homogeneous linear differential equations of order N. Such an equation has N characteristic solutions (the eigenvalues of the system of N first order differential equations, which can be derived), which simply are the zeros of the N-th degree characteristic polynomial.

 

Just an example:

 

Suppose we have:

 

8*d5y/dt5 + 12*d2y/dt2 - 8*dy/dt + 7y = 0.

 

Then we solve the equation 8λ^5 + 12λ^2 - 8λ + 7 = 0

 

This yields 5 solutions, λ1, λ2, .. λ5.

 

Now the solution of the differential equation can be written as

 

y(t) = c1*exp(λ1*t) + c2*exp(λ2*t) + c3*exp(λ3*t) + c4*exp(λ4*t) + c5*exp(λ5*t)

 

A unique solution is obtained by imposing sufficient initial conditions. For an N-th degree equation, N initial conditions must be specified, i.e. the value of y at time 0, but also the value of the first N-1 derivatives of y at time 0. If all are specified, then one can uniquely determine c1 ... cN, corresponding to those initial conditions.

 

The situation becomes a little more complicated, if some of the solutions λi are equal (multiple zeros of the characteristic polynomial), in that case, terms of the form t*exp(λi*t) also appear, or even higher powers of t, if the multiplicity of the solutions is higher than 2.

 

In the original problem we have to solve the equation d2005y/dt2005 - y = 0. The characteristic equation for this is λ^2005 - 1 = 0. This has 2005 distinct zeros, one of which equals 1, the other 2004 are complex and are on the unit circle in the complex plane. The solution now can be written as

 

c1*exp(λ1*t) + c2*exp(λ2*t) + .... c2005*exp(λ2005*t)

 

Here, λ1 can be regarded equal to 1.

 

In order to find a unique particular solution, you now need 2005 (!!!) initial values before the solution is fixed. For the solution to be equal to exp(t), without other terms we require that at time 0 all initial values of y(0) and also the first 2004 derivatives of y are equal to 1. If one of the derivatives of y(t) is not equal to 1 at time 0, then the solution will be more complicated and will also contain other exponentials with the other λi's in it.

[CPL.Luke]

in other words, its just a question of solving the charecteristic equation x^2005-1=0, because there is no generic equation for solving a polynomial of this order were left to approximations.

 

I'd imagine that newton's method could produce some values sufficiently close to x for most purposes, and you could get an approximate general solution based on that.

 

may I ask why your interested in this particular equation?

[woelen]

In this particular case, the equation can be solved analytically, in general, polynomial equations cannot be solved analytically.

 

There are numerical methods for solving polynomial equations, but building a general numerical polynomial solver is an amazingly complicated matter. I have written some software (using Aberth's method, with extensions as used in MPSOLVE), which does the general thing, but it is a large piece of code (I estimate it it be a few 10000's lines of code).

[uncool]

The methods described above are not necessary once you notice that the equation is simply x^2005 = 1 - which has as roots the 2005th roots of unity. Basically, take e^(2*pi*i/2005), and all integer powers of this number are solutions.

=Uncool=

[CPL.Luke]

Really?

 

I haven't had to solve a polynomial of an order that high yet, although one book I read had an extensive section on newton's method and its use in solving polynomial equations.

 

out of curiosity how would you go about solving this equation? I know that we can get one solution spot on, and then reduce the order of the equation through division, but we'd still be left with a massive polynomial of the 2004th order.

 

oh yeah, e^-x is also a possible solution

[uncool]

Well, it's because this equation is one of the few that's really nice in that the solutions are "obvious." The solutions are of the form

e^(2*pi*i*k/2005), let k = 0, 1, 2, ..., 2004.

This works because each to the 2005 is e^(2*pi*i*k) = 1.

e^-x does not work as it is -y, not + y.

=Uncool-

[woelen]

The methods described above are not necessary once you notice that the equation is simply x^2005 = 1 - which has as roots the 2005th roots of unity. Basically, take e^(2*pi*i/2005), and all integer powers of this number are solutions.

=Uncool=

As I mentioned in my post, this particular equation is simple and can be solved analytically.

 

@CPL.luke: In general, polynomial solving is amazingly difficult, unless you know that your polynomial has special properties. If there is no a-priori knowledge, then you must assume all kind of possible nasties, such as clusters of (close) roots, multiple roots and general ill-condition of the polynomial. Newton's method only is useful, once you are close enough near the root. The method, I programmed, first attempts to find initial estimates of the roots, and once it is guaranteed that Newton's method converges, it is applied for accurately computing the root. For clusters of roots and multiple roots, much more complicated things must be done, such as shifting of variables and using derivatives of the polynomial in order to find good approximations of the roots. Also, when you have limited precision arithmetic (in computers this is the case), then the precision at which roots can be computed can be very low (e.g. for multiple roots of multiplicity M, one can only obtain appr. 1/M part of the number of digits of precision of the computer).

 

 

@CPL.luke: An equation of the form x^n = a is easy to solve.

 

Compute the n-th root of a, and call this r. Then all roots are of the form r*(exp(2*k*pi*i/n)) = r*cos(2*k*pi/n) + r*i*sin(2*k*pi/n).

 

For general polynomials of degree 4 and less, there are analytical solutions, although the solution for 4th degrees polynomials is very complicated and hardly has any practical use. For higher degree polynomials, only special cases can be solved analytically (such as x^n = a).

[CPL.Luke]

ah thanks, I don't have much experience with polynomials of a very high degree beyond a vector calculus book I have that was written by someone who spent a large amount of time solving them (there was a fractal pattern on the cover that was generated during an attempt to solve a 256th degree polynomial), but because it was a vector calculus book it only spent a little time on them.

[D H]

[math]e^{-x}[/math] is not a solution to [math]\frac {d^{2005}y}{dx^{2005}} - y = 0[/math], since -1 is not one of the [math]2005^{th}[/math] roots of unity.

 

This is easily checked by using [math]\frac {d^{\,r}e^{-x}}{dx^{r}} = (-1)^r e^{-x}[/math]

 

With this, [math]\frac {d^{2005}e^{-x}}{dx^{2005}} - e^{-x} = -2 e^{-x} \ne 0[/math].

[Ragib]

Polynomials with a degree higher than 4 can not be solved with a formula because someone proved a long time ago that when you produced a formula for them, it had an infinite amount of terms, which of course we cant do. And i would imagine these wouldn't converge, or else people would use them anyway, or at least computers would.

[woelen]

Polynomials of degree higher than 4 certainly can be solved by means of closed (finite) formulas, the only thing, which is proven is that there is no closed formula for general polynomials of degree higher than 4 in terms of roots of numbers (so-called radicals). When you are allowed to use special functions, or hypergeometric series, then one can determine solutions of higher degree polynomials. E.g. the quintic equation can be solved in terms of elliptic modular functions, much in the same way, as cubic equations can be solved in terms of trigonometric functions.

 

In practice, though, solving high degree polynomials is done numerically by means of advanced iterative methods.