Lecture Notes in Differential Equations - Bruce E. Shapiro

207
Substituting into equation (24.3)
y P = −y 1
∫
∫
y 2 f(t)
a(t)W (t) dt + y 2
y 1 f(t)
dt (24.18)
a(t)W (t)
which is the basic equation of the method of variation of parameters. It is
usually easier to reproduce the derivation, however, then it is to remember
the solution. This is illustrated in the following examples.
Example 24.1. Find the general solution to y ′′ − 5y ′ + 6y = e t
The characteristic polynomial r 2 − 5r + 6 = (r − 3)(r − 2) = 0, hence a
fundamental set of solutions is y 1 = e 3t , y 2 = e 2t . The Wronskian is
∣ W (t) =
e 3t e 2t ∣∣∣
∣ 3e 3t 2e 2t = −e 5t (24.19)
Since f(t) = e t and a(t) = 1,
∫ e
y P = − e 3t 2t e t ∫ e 3t
−e 5t dt + e t
e2t dt (24.20)
−e5t ∫
∫
=e 3t e −2t dt − e 2t e −t dt (24.21)
= − 1 2 et + e t = 1 2 et (24.22)
(We ignore any constants of integration in this method). Thus the general
solution is
y = y P + y H = 1 2 et + C 1 e 3t + C 2 e 2t (24.23)
Example 24.2. Find a particular solution to y ′′ −5y ′ +6y = t by repeating
the steps in the derivation of (24.18) rather than plugging in the general
formula.
From the previous example we have homogeneous solutions y 1 = e 3t and
y 2 = e 2t . Therefore we look for a solution of the form
Differentiating,
y = u(t)e 3t + v(t)e 2t (24.24)
y ′ = u ′ (t)e 3t + 3u(t)e 3t + v ′ (t)e 2t + 2v(t)e 2t (24.25)
Assuming that the sum of the terms with the derivatives of u and v is zero,
u ′ (t)e 3t + v ′ (t)e 2t = 0 (24.26)