Lecture Notes in Differential Equations - Bruce E. Shapiro

110 LESSON 13. UNIQUENESS OF SOLUTIONS*
From the initial condition,
When we solve for C we get two possible values:
1 = 1 4 (1 + C)2 (13.8)
(1 + C) 2 = 4 (13.9)
1 + C = ± √ 4 = ±2 (13.10)
C = ±2 − 1 = 1 or − 3 (13.11)
Using these in equation (13.7) gives two possible solutions:
y 1 = 1 4 (t + 1)2 (13.12)
y 2 = 1 4 (t − 3)2 (13.13)
It is easily verified that both of these satisfy the initial value problem. In
fact, there are other solutions that also satisfy the initial value problem,
that we cannot obtain by the straightforward method of integration given
above. For example, if a < −1 then
⎧
1
⎪⎨ 4 (t − a)2 , t ≤ a
y a = 0, a ≤ t ≤ −1
(13.14)
⎪⎩
1
4 (t + 1)2 , t ≥ −1
is also a solution (you should verify this by (a) showing that it satisfies the
initial condition; (b) differentiating each piece and showing that it satisfies
the differential equation independently of the other two pieces; and then
showing that (c) the function is continuous at t = a and t = −1.
2.
1.5
1.
0.5
0.
10 5 0 5 10
The different non-unique solutions are illustrated in the figure above; they
all pass through the point (1,1), and hence satisfy the initial condition.