1st Order ODE Time-Domain Methods (PDF)

Time-domain methods for first order systems ENGS 22 Reading material: §§2.5.2, 2.5.3, and pages 48-49. This short note is meant to provide a reasonably complete account of standard time-domain treatment of first order inhomogeneous linear ordinary differential equations with constant coefficeints....

Time-domain methods for first order systems ENGS 22 Reading material: §§2.5.2, 2.5.3, and pages 48-49. This short note is meant to provide a reasonably complete account of standard time-domain treatment of first order inhomogeneous linear ordinary differential equations with constant coefficeints. Given a first-order linear time-invariant system in its canonical form ẋ + ax = b, (1) in which a ∈ R and b is, by and large, a function of time t. For every prescribed initial condition x(0), the solution of Equation (1) may be written as a linear superposition of its general solution xg and particular solution xp . 1. General solution: Set the inhomogeneous term b = 0, then Equation (1) becomes ẋ + ax = 0 with initial condition x(0). (a) x(t) = 0: it does satisfy the homogeneous equation, namely the value of x(t) remains zero for all t ≥ 0. However, it is indeed a solution if and only if the initial condition is consistent, i.e. i. x(0) = 0: if x(t) is initially zero, it will remain zero forever. In this case, x(t) = 0 is a valid general solution. ii. x(0) ̸= 0: this is a contradiction; the solution states that x(t) has to vanish for all t ≥ 0 but the initial condition is non-zero. Thus, the solution x(t) = 0 cannot satisfy such an initial condition and fails to be a valid solution. In this case, however, the fault is not on the form of solution; rather, it is the inconsistent inivial condition x(0) ̸= 0 that is ill-imposed. (b) x(t) ̸= 0: apply separation of variables, for example, to obtain dx = −adt, x which becomes, upon integration on both sides, ln |x| = −at + C1 for an arbitrary constant C1 ∈ R. By exponentiation on both sides, the above expression reads |x| = C2 e−at for C2 = eC1 > 0. The actual solution x(t) can now be written as x(t) = ±C2 e−at . Notice that since C2 > 0, ±C2 ∈ (−∞, 0) ∪ (0, +∞); denote C = ±C2 ̸= 0, then x(t) = Ce−at , in which the remaining arbitrary constant C ̸= 0 can be determined by the initial condition x(0): i. x(0) = 0: this is an inconsistent choice of initial condition for, according to the form of solution, x(0) = Ce−a·0 = C ̸= 0. In this case, solution that satisfies the zero initial condition does not exist. 1 Time-domain methods for first order systems ENGS 22 ii. x(0) ̸= 0: By evaluating the solution at t = 0, x(0) = Ce−a·0 = C = x(0) ̸= 0; hence, the general solution that satisfies the non-zero initial condition is given by x(t) = x(0)e−at . (2) Now, it should be noted that the consistent solutions in these two cases can actually be combined together to produce a uniform result for the general solution. Namely, (a) x(0) = 0: the solution that satisfies the zero initial condition is simply x(t) = 0, or, x(t) = Ce−at with C = x(0)= 0. (b) x(0) ̸= 0: the solution that satisfies the non-zero initial condition is given by x(t) = Ce−at for C = x(0)̸= 0. More compactly, the general solution of Equation (1) can be written as xg (t) = Ce−at (3) for C = x(0)∈ R. Remark: It is important to realize that for a homogeneous equation, only the general solution, xg (t), is present; the arbitrary constant C therein can then be determined by the initial condition satisfied by xg (t). When the inhomogenous term is taken into account as shown below, it will be the full solution, x(t) = xg (t) + xp (t), that has to satisfy the initial condition, not just the general solution of the homogeneous part. 2. Particular solution: it is usually difficult to represent the particular solution in closed form, if at all possible, for a given inhomogenous term b(t). For illustrative purposes, only the simplest case b = constant is considered here. It is then trivial to verify that b a satisfies Equation (1) by disregarding the initial condition. xp (t) = Altogether, to find the complete solution to Equation (1), as in the Remark above, it is necessary to first write b x(t) = xg (t) + xp (t) = Ce−at + a with an arbitrary constant C, and then apply the initial condition x(0) to the full solution x(t) to determine C. In greater detail, evaluate x(t) at t = 0: x(0) = xg (0) + xp (0) = Ce−a·0 + b b =C+ , a a from which C can be solved to be b C = x(0) − . a Hemce, the full solution to Equation (1) can finally be written as  b  −at b x(t) = x(0) − e + a | {za } |{z} transient = steady−state  b x(0)e−at + 1 − e−at | {z } |a {z } free response forced response Compare this solution with Equations (2.3.5) and (2.3.6) on page 48. 2