## Monday, August 19, 2013

### Tricks to Solving Higher Order DEs (4): Undetermined Coefficients

We now come to the last part of the tricks employed in solving higher order linear differential equations – this one based on the use of undetermined coefficients. Though the method is not universally applicable, it can be used if the right hand side, f(x), contains only terms which have a finite number of linearly independent derivatives, i.e. xn, exp (mx), sin bx and cos bx or products of these.  The general procedure is to assume a particular integral y p of a form similar to the right member f(x) and containing undetermined coefficients for each term. The necessary derivatives of  y p  are then obtained and substituted into the given differential equation. This results in an identity in the independent variable, and thus the coefficients of like terms can be equated and the values of the undetermined coefficients found from the resulting system of linear equations.

Example: Find the general solution of:

y” – y’ -6y = 6x3 + 26 sin 2x

The auxiliary equation is:

m2 – m – 6 = 0

This can be factored to give:  (m-3)(m +2) = 0

So the roots are m = 3 and m = - 2

The complementary function solution is:

y c = c1 exp (3x) + c2 exp (-2x)

Now, let the form of the particular integral, y p,  be:

y p  = A x3 +B x2 + Cx + D + Esin 2x + F cos 2x

in which we choose a 3rd degree polynomial to correspond to the x3   term in f(x) and a linear combination of sin 2x and cos 2x to correspond to sin 2x in f(x). Obtaining the first and second derivatives of y p  :

y’ p  =  3A x2 + 2B x + C + 2E cos2x -2F sin 2x

y” p  =  6A x + 2B  - 4Esin 2x -4F cos2x

Substitute these into the original differential equation to get:

6A x + 2B  - 4Esin 2x -4F cos2x – [3A x2 + 2B x + C + 2E cos2x -2F sin 2x] –
6[A x3 +B x2 + Cx + D + Esin 2x + F cos 2x] = 6 x3 + 26 sin 2x

Collect like terms on the left side to get:

-    6 x3 – (3A + 6B) x2 + (6A – 2B – 6C)x + (2B – C – 6D) – (4E – 2F + 6E)sin 2x – (4F + 2E + 6F) cos 2x = 6 x3 + 26 sin 2x

Equating coefficients of like terms we get:

-6A = 6

-3A – 6B = 0

6A – 2B – 6C = 0

2B – C – 6D = 0

-10E + 2F = 26

-10F – 2E = 0

The solution to the above system yields:

A = -1, B = ½

C = -7/6, D = 13/36, E = - 5/2 and F = ½

Then inserting the coefficients into the particular integral:

y p  = -  x3 + ½  x2  - 7x/ 6  + 13/36  - 5 sin 2x/ 2  + ½  cos 2x

The general solution is then the combination of  y c  and y p   or:

y = = c1 exp (3x) + c2 exp (-2x) -  x3 + ½  x2  - 7x/ 6  + 13/36  - 5 sin 2x/ 2  + ½  cos 2x

Problem for the Math Maven:

Find the general solution of: y”’ – 3y” + 3y’ – y = 48 x exp (x)