Lecture Notes in Differential Equations - Bruce E. Shapiro

210 LESSON 24. VARIATION OF PARAMETERS
Next, we need to find a particular solution; we can do this using the variation
of parameters formula. Since a(t) = t 2 and f(t) = 2t, we have
∫
∫
y2 (t)f(t)
y P = −y 1 (t)
a(t)W (t) dt + y y1 (t)f(t)
2(t)
dt (24.49)
a(t)W (t)
We previously had W = C; but here we need an exact value. To get
the exact value of C, we calculate the Wronskina from the now-known
homogeneous solutions,
C = W =
∣ y ∣ ∣
1 y 2
∣∣∣ y 1 ′ y 2
′ ∣ = t 2 1/t ∣∣∣
2t −1/t 2 = −3 (24.50)
Using y 1 = t 2 , y 2 = 1/t, f(t) = 2t, and a(t) = t 2 ,
∫
y P = −t 2 2t
t · t 2 · −3 dt + 1 ∫
t2 · 2t
t t 2 dt (24.51)
· −3
∫
= 2t2 t −2 dt − 2 ∫
tdt (24.52)
3
3t
= 2t2
3 · −1 − 2 t 3t · t2 (24.53)
2
= − 2t
3 − t = −t (24.54)
3
Hence y P = −t and therefore
y = y H + y P = C 1 t 2 + C 2 t − 1 t
(24.55)
From the first initial condition we have 0 = C 1 + C 2 − 1 or
To use the second condition we need the derivative,
C 1 + C 2 = 1. (24.56)
y ′ = 2C 1 t − C 2
t 2 − 1 (24.57)
hence the second condition gives 1 = 2C 1 − C 2 − 1 or
2C 1 − C 2 = 2. (24.58)
Solving for C 1 and C 2 gives C 1 = 1, hence C 2 = 0, so that the complete
solution of the initial value problem is
y = t 2 − 1 t
(24.59)