## Wednesday, August 21, 2013

### Solution to DE Problem with Undetermined Coefficients

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

The auxiliary equation is:

m3  - 3m2 + 3 m – 1 = 0

This can be factored on inspection to give:

(m – 1)(m – 1) (m – 1) = 0

With the triplicate repeated solution: m = 1, 1, 1.

Then the complementary function is:

y c = (c o + c1x + c2 x2)  exp (x)

Normally, the assumed form of particular integral would be:

y p  = (A + Bx) exp (x)

However, these are already contained in    y p  so it’s necessary to multiply by  x3, i.e. the exponent must be one greater than that peculiar to y c . So:

y p  = (A x3 +B x4 ) exp (x)

y’ p  =  (3A x2 + 4B x3 + A x3 + B x4  ) exp (x) = [3A x2 + (4B + A) x3 + B x4 ) exp (x)

y” p  = [ 6A x + (12B  + 6A) x2 + (8B + A) x3 + B x4 ) exp (x)

y”’ p  = [ 6A  + (24B + 18A) x + (36B + 9A) x2 + (12B + A) x3 + B x4 ] exp (x)

Now, substitute each of the above into the DE and collect like terms to get:

(6A  + 24 Bx) exp (x) = (0 + 48x) exp (x)

Then:  6A = 0 and 24B = 48, or A = 0 and B = 2

The general solution is then:

y  = (c o + c1x + c2 x2 +  2 x4 ) exp (x)