### LECTURE 4: LINEAR ALGEBRA: 11.2UE2 Case II: b2 −4ac = 0

```LECTURE 4: LINEAR ALGEBRA: 11.2UE2
SIMON MALHAM
Case II: b2 − 4ac = 0, “The critically damped case”. If b2 − 4ac =
0, there is one real, repeated root to the auxillary equation, namely
b
λ1 = λ2 = − .
2a
Hence we have only one solution, which is
b
y(t) = C1 eλ1 t = C1 e− 2a t .
However, we know there must be another independent solution. It’s
not obvious what that might be, but let’s make the educated guess
y(t) = C2 teλ1 t
b
where λ1 is the same as above, i.e. λ1 = − 2a
.
Substituting this guess for the second solution into our second order
differential equation,
⇒a
d2 y
dy
+ b + cy = a (C2 λ21 teλ1 t + C2 (2λ1 )eλ1 t )
2
dt
dt
+ b (C2 eλ1 t + C2 λ1 teλt )
+ c C2 teλ1 t
= C2 eλ1 t t (aλ21 + bλ1 + c) + (2aλ1 + b)
= 0,
since we note that aλ21 + bλ1 + c = 0 and 2aλ1 + b = 0 because
λ1 = −
b
.
2a
b
Thus C2 te− 2a t is another solution (which is clearly not a multiple of
the first solution).
Important. Principle of Superposition ⇒ the general solution is
b
y(t) = (C1 + C2 t)e− 2a t .
1
2
MALHAM
In appearance, the solutions for the critically damped case look very
much like those in figure 3 for the over-damped case.
[[Refer to handout on the Euler formula before proceeding.]]
Case III: b2 − 4ac < 0, “The underdamped case”. If b2 − 4ac < 0,
there are two complex roots to the auxillary equation, namely
p
λ1 = p + iq, where q = |b2 − 4ac|/2a,
λ2 = p − iq,
p = −b/2a.
Hence the general solution takes the form
y(t) = A1 e(p+iq)t + A2 e(p−iq)t ,
where A1 and A2 are arbitrary (complex) constants.
Note that
y(t) = ept A1 eiqt + A2 e−iqt .
Hence using the Euler formula, eiφ = cos φ + i sin φ,
y(t) = ept (A1 (cos qt + i sin qt) + A2 (cos qt − i sin qt))
⇒ y(t) = ept ((A1 + A2 ) cos qt + i(A1 − A2 ) sin qt)) .
Our solution must be real as we have a real displacement of the spring.
⇒ suppose
C1 − iC2
,
2
C1 + iC2
A2 =
,
2
where C1 and C2 are two arbitrary, real constants.⇒
A1 =
A1 + A2 = C1 ,
i(A1 − A2 ) = C2 .
Important. Hence the general solution is
y(t) = ept (C1 cos qt + C2 sin qt).
LECTURE 4
3
Figure 1. Case III: b2 − 4ac < 0, “The underdamped
case”. For example, when the mass is immersed in light
oil or air. The graph of the solution oscillates about the
equilibrium position, between the ‘exponential envelopes’
which are the two curves y(t) = ept and y(t) = −ept . Note
that for the spring p = −b/2a < 0.
Since for the case of the spring
b
C
p = − < 0, and b =
> 0,
2a
m
we see that the mass will oscillate about the equilibrium position; the
amplitude of the oscillations, decay exponentially in time (see figure 4).
References
[1] Boyce, W.E. & DiPrima, R.C. 1996 Elementary differential equations and
boundary-value problems, John Wiley.
[2] Hughes-Hallet, D. et al. 1998 Calculus, John Wiley & Sons, Inc. .
```
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}