Solutions to Higher Order Differential Equations (2)

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

 

Rewrite as: y^iv  – 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 1^st 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)