Lecture Notes in Differential Equations - Bruce E. Shapiro

33
Multiplying equation 4.68 by µ gives
d (
t 4 y ) (
= t 4 y ′ + 4y dt
t
)
= (4t)(t 4 ) = 4t 5 (4.70)
Integrating,
∫ d
dt (t4 y)dt =
∫
4t 5 dt (4.71)
t 4 y = 2tt
3 + C (4.72)
y = 2t2
3 + C t 4 (4.73)
The last example is interesting because it demonstrates a differential equation
whose solution will behave radically differently depending on the initial
value.
Example 4.6. Solve 4.67 with initial conditions of (a) y(1) = 0; (b) y(1) =
1; and (c) y(1) = 2/3.
(a) With y(1) = 0 in equation 4.73
0 = (2)(1)
3
+ C 1 = 2 3 + C (4.74)
hence C = −2/3, and
(b) With y(1) = 1,
hence C = 1/3, and
(c) With y(1) = 2/3,
hence C = 0 and
1 = (2)(1)
3
y = 2t2
3 − 2
3t 4 (4.75)
+ C 1 = 2 3 + C (4.76)
y = 2t2
3 + 1
3t 4 (4.77)
2
3 = (2)(1) + C 3 1 = 2 3 + C (4.78)
y = 2t2
3
(4.79)