Wednesday, August 7, 2013

Solutions to Higher Order Differential Equations (2)


1) d4y/ dt4 –2d3y/ dt3  –7 d2y/ dt2  + 20  dy/dt  –12 y = 0

 
Rewrite as: yiv  – 2y”’  - 7y” + 20y’ – 12 y = 0

 
The auxiliary equation here is:  m4 – 2m3 – 7m2 + 20m – 12 = 0


This can be factored (i.e.  guess or use synthetic division) to get:

(m-1) (m- 2)2 (m-3)


Yielding the roots: m = 1, m = 2 (twice) and m = 3


So, the general solution is:


y= c1 exp (x) + (c2 + c3) exp(2x) + c4 exp (-3x)



2) y”’ + 3y” – 4y = 0


The auxiliary equation here is:


m3  + 3m2 – 4 = 0


Use synthetic division, careful inspection (or guessing) to obtain the factors:


(m-1) (m2  + 4m + 4) = (m – 1) (m + 2)2

Yielding the roots: m = 1, m = 2 (twice), so the general solution is:


y = c1 exp (x) + c2 exp(-2x) +   c3(x exp (-2x))


(Note that the absence of a 1st derivative term,  i.e. dy/dt or y’, means one of the roots will be in terms of x exp (r2x)  where r2 is the 2nd of the twinned roots)

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