1. Linear Differential Equations with Constant Coefficients:
f(t): Input
u(t): Output (response)
Example:
Homogeneous solution f(t)= u(t)=est
With Matlab:
Characteristic Equation:
s3est + 4s2est + 14sest + 20est= 0
a=[1,4,14,20];roots(a)
s3 + 4s2 + 14s + 20 = 0
Eigenvalues: -13i, - 2
uh(t) = C1e(-1+3i)t + C2 e(-1-3i)t + A2e-2t
uh(t) = A1e-tcos(3t-φ)+A2e-2t

2. uh(t) = A1e-tcos(3t-φ)+A2e-2t
Initial conditions: at t=0
-1.2 = A1 cosφ + A2
2.5 = -A1 cosφ +3A1 sinφ -2A2
-3.1= -8A1 cosφ - 6A1 sinφ + 4A2
A1, A2 and φ can be found by Newton-Raphson method.

3. Laplace Transform:

4. Laplace Transform of the Derivatives :

5. (shift in time or delay):

6. Laplace transform of the solution due to the initial conditions:
Initial conditions: at t=0

7. Partial fraction expansion:
With Matlab;
num=[-1.2,-2.3,-9.9]; den=[1,4,14,20]; [r,p,k]=residue(num,den)
r(1)= i, r(2)= i, r(3)=-1.01

8. Homogeneous solution :
uh(t) = A1e-tcos(3t-φ)+A2e-2t
With Matlab;
z= i
A1=2*abs(z)
fi=angle(z)

9. At t=0 are given. Find θ(t).
EXAMPLES:
The equation of the motion for the unforced motion of a simple pendulum is given as: m g θ
Joint friction, B L
m=2 kg
B=4 Nms/rad
L=2 m
At t= are given. Find θ(t).
Applying the Laplace transform,

10. [r,p,k]=residue(num,den)
EXAMPLES:
Laplace transform of the homogenous solution (due to the initial conditions)
Eigenvalues
The system is stable because the real parts of all the roots are negative.
clc;clear
num=[4 10];
den=[ ];
[r,p,k]=residue(num,den)
r(2)
A=2*abs(r(2))
Fi=angle(r(2)) Re
0.25
0.2556
Img

11. EXAMPLES:
clc;clear
dt=0.1418;
ts=25.149;
t=0:dt:ts;
tetat=0.7151*exp(-0.25*t).*cos(2.2006*t );
plot(t,tetat)