## Thursday, August 1, 2013

### Tricks to Solving Higher Order Differential Equations (1)

Readers who followed the last two blogs on rewriting higher order DEs may have come away with the impression that all you need to know is a reduction template and it’s pretty easy to do. This, however, would be an error, since most higher order DEs (maybe 90%) are bloody hard to work out and often require numerical methods or much more difficult ones than we’ve been accustomed to.

Take the case of the 2nd order equations rewritten as linear systems.  We found that instead of dealing with:   y"  - 3y' + 4y = sin 3t

We would have the system:

dx1/dt = x2

dx2/dt = -4x1 + 3x2  + sin 3t

Math savvy readers will immediately note the non- homogenous system with a forcing term F(t) = sin 3t) in the bottom. They then likely proceeded in the usual way solving the homogeneous system first from (A - lI)D = 0 =

[(0 - l)  …..1]
[-4 ……..(3-  l)]

=  l2 – 3l  - 4   or:  (l +1) (l - 4) = 0

yielding two eigenvalues:  l1  = -1  and l2  = 4

One ultimately finds the pair of homogenous solns.

x1= c1 exp (-t) + c2 exp (4t)

x2 = -4c1 exp (-t) - c2 exp (4t)

(But the forcing term sin 3t must still be dealt with – as per methods, i.e. undetermined coefficients,  we previously used).
In reality, one would not even attempt to solve most of these DEs this way but instead would use something called a “Laplace transform”. Then each of the separate (linear) equations would be subject to this transform, i.e. £  {dx1/dt} and . £  {dx2/dt}   to solve.

In this case one would find:

£  {dx1/dt}  =  £  {x2} and £  {dx2/dt}   = £  {-4x1 + 3x2  + sin 3t}

Whence also:  £  {x1(t)}  =  x1(s) and £  {x2(t)}  =  x2(s)

The first equation can then be rewritten: sx1(s) = x2(s)
(Note the derivative has an ‘s’ factor preceding it)

and the second:

sx2(s) = -4x1(s) + 3x2(s) +   3/ (s2 + 32)

where:   3/ (s2 + 32)
is the Laplace transform of sin 3t (the details of how this is found are way too complex to go into, but readers can probably find them in any good mathematical physics or  advanced calc text).
Use some algebra on the x2(s) form:

x2(s) [s – 3]  = -4x1(s) +   3/ (s2 + 32)

and meanwhile: x2(s) = sx1(s)

The approach then is to solve this as a simultaneous pair of equations in two unknowns, x1(s) and x2(s) and then look for what we call the “inverse” transform to obtain the actual solution.  Of course, it is also possible to attack the original 2nd order DE directly using Laplace transforms: y"  - 3y' + 4y = sin 3t

Then the Laplace transformed equation would appear:

s2 y(s)  - 3 s y(s) + 4y(s) =  3/ (s2 + 32)

Exercises for the Math Maven:

1) For the 2nd problem of the 'rewrite' DE set (of the July 29 blog) put the 2nd order DE and its (simplified, normal form) linear pair in Laplace transform mode.

2) Propose a way that you'd get to the inverse Laplace transform, if you approached the 2nd order problem directly, instead of reducing it to normal form.