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)
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