Lecture Notes in Differential Equations - Bruce E. Shapiro

208 LESSON 24. VARIATION OF PARAMETERS
and consequently
Differentiating,
y ′ = 3u(t)e 3t + 2v(t)e 2t (24.27)
y ′′ = 3u ′ (t)e 3t + 9u(t)e 3t + 2v ′ (t)e 2t + 4v(t)e 2t (24.28)
Substituting into the differential equation y ′′ − 5y ′ + 6y = t,
t =(3u ′ (t)e 3t + 9u(t)e 3t + 2v ′ (t)e 2t + 4v(t)e 2t ) − 5(3u(t)e 3t + 2v(t)e 2t )
+ 6(u(t)e 3t + v(t)e 2t ) (24.29)
=3u(t) ′ e 3t + 2v(t) ′ e 2t (24.30)
Combining our results gives (24.26) and (24.30) gives
u ′ (t)e 3t + v ′ (t)e 2t = 0 (24.31)
3u(t) ′ e 3t + 2v ′ (t)e 2t = t (24.32)
Multiplying equation (24.31) by 3 and subtracting equation (24.32) from
the result,
v ′ (t)e 2t = −t (24.33)
We can solve this by multiplying through by e −2t and integrating,
∫
v(t) = −
(
te −2t dt = − − t 2 − 1 ) ( t
e −2t =
4
2 + 1 e
4)
−2t (24.34)
because ∫ te at dt = [ t/a − 1/a 2] e at .
Multiplying equation (24.31) by 2 and subtracting equation (24.32) from
the result,
u ′ (t)e 3t = t (24.35)
which we can solve by multiplying through by e −3t and integrating:
∫
(
u(t) = te −3t dt = − t 3 − 1 ) ( t
e −3t = −
9
3 + 1 e
9)
−3t (24.36)
Thus from (24.24)
y = u(t)e 3t + v(t)e 2t (24.37)
( ( t
= −
3 + 1 ) (( t
e
9) −3t e 3t +
2 4) + 1 )
e −2t e 2t (24.38)
= − t 3 − 1 9 + t 2 + 1 4
= t 6 + 5 36
(24.39)
(24.40)