1) Find the general solution of: y”’ – 3y” + 7y’ – 5y = 0

The auxiliary
equation is:

*m*^{3}- 3m^{2}+ 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:

*m*^{2}-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: y

^{iv}+ 18y” + 81 = 0
The auxiliary
equation is:

*m*^{4}+ 18m^{2}+ 81 = 0
It is fairly easy to factor this to get:
(m

^{2}+ 9)^{2}= 0
And each (m

^{2}+ 9) is easily factored in turn to get:
(m + 3i)(m – 3i) = 0

i.e. m

^{2}+ 3mi – 3mi -9i^{2}= m^{2}-9 (-1) = m^{2}+ 9
The roots are

**and are: m=***doubly repeated*__+__3i and__+__3iHence 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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