first tier second tier third tier fourth tier

past issues

2004 2003 2002 2001 2000 1999 1998 1997 1996 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Welcome to NRICH.

dⁿy/dxⁿ=-ky

---

---

By Julian Pulman on Wednesday, May 08, 2002 - 07:49 pm:

dy/dx = -ky, when solved gives an Exponential Decay Relationship.
d²y/dx²=-ky, when solved gives a trigonometric relationship.

Both of these are revelent in modelling certain events, such as Radioactive Decay, and Harmonic Oscillators (respectively).
Does the solution to d³y/dx³ = -ky exist? and if so, is the solution of use in modelling some aspect of physics?
What about dⁿy/dxⁿ=-ky as n surpasses 3?
What methods can be used to solve such equations? I'm only aware of what I learnt in Further Maths, for 2nd order solutions.

Thanks,

Julian

---

By Dan Goodman on Thursday, May 09, 2002 - 01:09 am:

Do you know about complex numbers? If so, you'll know that the trigonometric solutions and exponential decay solutions are, in some sense, the same sort of thing.

For y⁽ⁿ⁾+ky=0 (here y⁽ⁿ⁾ means the nth derivative dⁿy/dxⁿ), we try a solution y=e^lx. Substitute this is and you get lⁿe^lx+ke^lx=0. Divide by e^lx to get lⁿ+k=0.

There are n solutions to this equation (if you use complex numbers), and each gives rise to a different solution. Linear combinations of these give you the general solution.

For example, if n=3 and k=-2 then we want to solve l³=2. One solution is obvious, l=2^1/3. So, if we check it we do get y=e^{2^(1/3)x} is a solution, because then y'=2^1/3y, y''=2^2/3y and y'''=2y, so y'''-2y=0.

If you know about complex numbers, you'll see the same thing works with the other two solutions l=2^1/3e^2ip/3 and l=2^1/3e^ip/3. This will give us things like e^{2^(1/3)cos(2p/3)x}cos(2^1/3sin(2p/3)x), yuch.

If you don't know about complex numbers, the thing to do is to try solutions like e^Axcos(Bx), e^Axsin(Bx) and simply e^Ax and see where you can get.

The same method works for general nth order homogeneous linear ordinary differential equations, i.e. yⁿ+a_n-1y^n-1+...+a₀y=0. You end up needing to solve the equation lⁿ+a_n-1l^n-1+...+a₁l+a₀=0, which is difficult in general but easy if all the a_i are 0 except a₀. In case you wanted to know the level of difficulty, at Cambridge this is taught in the first term of the first year of maths. So, pretty difficult but not too difficult (even if I've made it so by posting too late).

I don't know of any applications in physics I'm afraid, but no doubt there are some :-)

Sorry if that was a bit confused (or a bit too difficult), I'm up too late. Let me know if you didn't follow.

---

By Yatir Halevi on Friday, May 10, 2002 - 10:51 am:

for even n, the sinh and cosh functions fit.

Yatir

---

By Julian Pulman on Saturday, May 11, 2002 - 11:54 am:

Thanks everyone, I'm confident now I can use the same techniques outlined by Dan to solve the nth order ones. I'll get back if I have any trouble, thanks :-)

---