1) Find the general solution of: y”’ – 3y” + 7y’ – 5y = 0
The auxiliary
equation is:
m3 - 3m2 + 7m - 5 = 0
Then the root m =
1 can be extracted using synthetic division so:
1….-3….+7…..-5
(1)
……1….-2……5
----------------------
1…..-2….5……0
And the resulting quadratic is:
m2 -2m + 5 = 0
Solve using the quadratic formula to obtain:
m = 1 + 2i
with a = 1 and b
= 2.
The general solution can then be written:
y = c1 exp (x) +
exp (x)[ c2 cos 2x + c3 sin 2x]
2)Find the general solution of: yiv +
18y” + 81 = 0
The auxiliary
equation is:
m4 + 18m2 + 81 = 0
It is fairly easy to factor this to get:
(m2 + 9)2 = 0
And each (m2 + 9) is easily factored in turn to get:
(m + 3i)(m – 3i) = 0
i.e. m2 + 3mi – 3mi -9i2 = m2 -9 (-1) = m2 + 9
The roots are doubly repeated and are: m= + 3i and + 3i
Hence a = 0 and b = 3 each time so the general solution can be written:
y = (c1 + c2x) cos 3x + (c3 + c4 x) sin 3x
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