First-Order Differential Equations Review
We consider first-order differential equations of the form:
d x(t ) 1
+ x(t ) = f (t )
τ
dt
where f(t) is the forcing function. In general, the differential equation has two solutions:
(1)
1. complementary (or natural or homogeneous) solution, xC(t) (when f(t) = 0), and
2. particular (or forced or non-homogeneous) solution, xP(t) (when f(t) ≠ 0).
In our problems, f(t) is often a constant, and therefore, the overall solution to the differential
equation is
x(t ) = xC (t ) + x P (t ) = K 1 e − t / τ + K 2
The time constant, τ, for first-order electrical circuits is either τ = RC or τ = L/R.
Complementary Solution
The complementary solution is found by considering the homogeneous equation:
d x(t ) 1
+ x(t ) = 0
τ
dt
The complementary solution is the system’s natural response, which is
xC (t ) = K 1 e −t / τ
This result can be verified by substituting this answer into the differential equation, Eq. (3).
(2)
(3)
Particular Solution
The particular solution is found by considering the full (non-homogeneous) differential equation,
that is, Eq. (1). If the forcing function is a constant, then xP(t) is a constant (K2) also, and hence
dx P = 0 . Substituting x into the original differential equation, Eq. (1), yields:
P
dt
d xP 1
1
+ xP = 0 + xP = f
τ
dt τ
Reducing the above expression provides the particular solution, which is the forced response:
xP = K 2 = τ f
Note that, alternatively, K2 could have been found using Eq. (2) while considering the final value
reached by the variable of interest, and hence, in this case is also the steady-state response:
x(∞) = K 1 e −∞ + K 2 = K 2
Total Solution
The total solution is the sum of the complementary and particular solutions:
xT (t ) = xC (t ) + x P = K 1 e − t / τ + τ f
The coefficient term, K1, is found from the initial conditions (at t = 0).
(4)
Initial Conditions
Typically, the initial conditions are determined by considering that for dc steady-state conditions:
dv
(a) a capacitor is like an open circuit ( iC = C C = 0 ), and (b) an inductor appears as a short
dt
di
circuit ( v L = L L = 0 ). Further, the current through an inductor and the voltage across a
dt
capacitor cannot change instantaneously, for example, iL (0−) = iL (0+) and vC (0−) = vC (0+) .
DiffEQ.doc
November 2002
K.E. Holbert
Second-Order Differential Equations Review
The second-order differential equations of interest are of the form:
d2 y
dy
+ 2ς ω 0
+ ω 02 y (t ) = f (t )
(5)
2
dt
dt
where f(t) is the forcing function. Like the first-order case, this differential equation has two
solutions: (1) complementary (or natural) solution, yC(t) when f(t) = 0, and (2) particular (or
forced) solution, yP(t) when f(t) ≠ 0. In our problems, f(t) is often a constant, and therefore, the
overall solution to the differential equation is typically
y (t ) = y C (t ) + y P (t ) = K 1 e s1 t + K 2 e s2 t + K 3
(6)
Complementary Solution
The complementary solution is found by considering the homogeneous equation:
d2 y
dy
+a
+ b y (t ) = 0
(7)
2
dt
dt
where a = 2 ς ω 0 and b = ω 02 . The complementary solution is the system’s natural response:
y C (t ) = K 1 e s1 t + K 2 e s2 t
(8)
The natural frequencies, s1 and s2, are the roots of the characteristic equation:
s2 + a s + b = 0
(9)
These roots lead to three possible cases:
Damping
Natural Response
Roots (s1, s2)
ζ
Sum of two decaying exponentials
Overdamped
ζ > 1 Real and unequal
Critically damped Two faster decaying exponentials
ζ = 1 Real and equal
Exponentially damped sinusoid
ζ < 1 Complex conjugates Underdamped
Particular Solution
The particular solution is found by considering the full (non-homogeneous) differential equation,
Eq. (5). If the forcing function is a constant, then yP(t) is also a constant (K3), and hence
dy P = 0 = d 2 y P
. Substituting yP into the differential equation yields:
dt
dt 2
d2 y
dy
+a
+ b y (t ) = 0 + a 0 + b y P = f
2
dt
dt
Reducing the above expression provides the particular solution, which is the forced response
f
yP = K3 =
b
Alternatively, K3 could have been determined from Eq. (6) and the final value y(∞).
Initial Conditions
Initial conditions for y and its derivative are necessary to find K1 and K2; specifically needed are
. A numerical value for dy
is generally found by re-applying KVL or KCL at
y(0) and dy
dt
dt
t =0
t =0
t=0. Practically speaking, as the KVL or KCL is performed, the inductor voltage and capacitor
dv
current, respectively, are expressed as L di L
and C C
. The numerical value of dy
dt t = 0
dt t =0
dt t =0
is then equated to an analytical derivative of y(t) determined from Eq. (6). The initial conditions
for y(0) are found in the same manner as described for first-order differential equations.
DiffEQ.doc
November 2002
K.E. Holbert