(1.2) Initial-Value Problems
INTRODUCTION:
We are often interested in problems in which we seek a solution 𝑦(𝑥) of a differential equation so that
𝑦(𝑥) satisfies prescribed side conditions—that is, conditions imposed on the unknown 𝑦(𝑥) or its
derivatives. On some interval 𝐼 containing 𝑥0 the problem
𝑆𝑜𝑙𝑣𝑒:
𝑑𝑛 𝑦
= 𝑓(𝑥, 𝑦, 𝑦 ′ , … , 𝑦 (𝑛−1) )
𝑑𝑥 𝑛
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜:
𝑦(𝑥0 ) = 𝑦0 , 𝑦 ′ (𝑥0 ) = 𝑦1 , … , 𝑦 (𝑛−1) (𝑥0 ) = 𝑦𝑛−1 ,
… … … … … … … … … … (1)
where 𝑦0 , 𝑦1 , … , 𝑦𝑛−1 are arbitrarily specified real constants, is called an initial-value problem
(IVP). The values of 𝑦(𝑥) and its first 𝑛 − 1 derivatives at a single point 𝑥0 , 𝑦(𝑥0 ) = 𝑦0 , 𝑦 ′ (𝑥0 ) =
𝑦1 , … , 𝑦 (𝑛−1) (𝑥0 ) = 𝑦𝑛−1 , are called initial conditions.
FIRST- AND SECOND-ORDER IVPS:
The problem given in (1) is also called an 𝒏th-order initial-value problem. For example,
𝑆𝑜𝑙𝑣𝑒:
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜:
and
𝑆𝑜𝑙𝑣𝑒:
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜:
𝑑𝑦
= 𝑓 (𝑥, 𝑦)
𝑑𝑥
… … … … … … … … … … (2)
𝑦(𝑥0 ) = 𝑦0
𝑑2 𝑦
= 𝑓 (𝑥, 𝑦, 𝑦 ′ )
2
𝑑𝑥
𝑦(𝑥0 ) = 𝑦0 , 𝑦 ′ (𝑥0 ) = 𝑦1
… … … … … … … … … … (3)
are first- and second-order initial-value problems, respectively.
THEOREM 1.2.1 (Existence of a Unique Solution)
Let 𝑅 be a rectangular region in the 𝑥𝑦-plane defined by 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑 that contains the
point (𝑥0 , 𝑦0 ) in its interior. If 𝑓 (𝑥, 𝑦) and 𝜕𝑓 ⁄𝜕𝑦 are continuous on 𝑅 , then there exists some interval
𝐼0 : (𝑥0 − ℎ, 𝑥0 + ℎ), ℎ > 0, contained in [𝑎, 𝑏], and a unique function 𝑦(𝑥), defined on 𝐼0 , that is a
solution of the initial-value problem (2).
𝑑𝑦
Roughly speaking, if we have the IVP
= 𝑓 (𝑥, 𝑦), 𝑦(𝑥0 ) = 𝑦0 , then we have 3 cases explain the
𝑑𝑥
theorem which are follows:
 If 𝑓 (𝑥, 𝑦) and 𝜕𝑓/𝜕𝑦 are continuous on a common domain 𝐼
⇒ the solution of the IVP exists and unique.
 If 𝑓 (𝑥, 𝑦) is only continuous on a domain 𝐼
⇒ the solution of the IVP exists only. This means that the IVP has infinite solutions
 If 𝑓 (𝑥, 𝑦) is discontinuous on a domain 𝐼
⇒ the solution of the IVP does not exist.
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(1.2) Initial-Value Problems
EXAMPLE 1 (Second-Order IVP)
In Example 4 of Section 1.1, we saw that
𝑥 = 𝑐1 cos 4𝑡 + 𝑐2 sin 4𝑡
is a two-parameter family of solutions of
𝑥 ′′ + 16𝑥 = 0.
Find a solution of the initial-value problem
𝜋
𝜋
𝑥 ′′ + 16𝑥 = 0, 𝑥 ( ) = −2, 𝑥 ′ ( ) = 1.
2
2
… … … … … … … … … … (4)
SOLUTION:
We first find 𝑥 ′ and 𝑥 ′′ using the general solution 𝑥 = 𝑐1 cos 4𝑡 + 𝑐2 sin 4𝑡 , we obtain
𝑥 ′ = −4𝑐1 sin 4𝑡 + 4𝑐2 cos 4𝑡
and
𝜋
𝑥 ′′ = −16𝑐1 cos 4𝑡 − 16𝑐2 sin 4𝑡.
Then, we apply 𝑥 ( ) = −2 to the given family of solutions:
2
𝜋
𝜋
𝜋
𝑥 ( ) = 𝑐1 cos (4 ( )) + 𝑐2 sin (4 ( )) = 𝑐1 cos 2𝜋 + 𝑐2 sin 2𝜋 = 𝑐1 (1) + 𝑐2 (0) = 𝑐1 = −2.
2
2
2
𝜋
Hence, 𝑐1 = −2. Similarly, we next apply 𝑥 ′ ( ) = 1 to the two-parameter family
2
𝑥 ′ = −4𝑐1 sin 4𝑡 + 4𝑐2 cos 4𝑡.
So,
𝜋
𝜋
𝜋
𝑥 ′ ( ) = −4(−2) sin (4 ( )) + 4𝑐2 cos (4 ( )) = 8 sin 2𝜋 + 4𝑐2 cos 2𝜋 = 8(0) + 4𝑐2 (1) = 4𝑐2 = 1.
2
2
2
1
Thus, 𝑐2 = . Hence,
4
1
𝑥 = −2 cos 4𝑡 + sin 4𝑡
4
is a solution of (4).
EXAMPLE 2
According to the Existence and Uniqueness Theorem, determine whether the IVP
has a unique solution.
SOLUTION:
𝑑𝑦
= 𝑥𝑦1/2 ,
𝑑𝑥
𝑦 (0) = 0
1
1 − 𝑓(𝑥, 𝑦) = 𝑥𝑦 2 = 𝑥 √𝑦 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ [0, ∞)}.
𝜕𝑓
𝑥
𝑥
2−
= 1=
is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ (0, ∞)}.
𝜕𝑦 2𝑦 2 2√𝑦
Since (0,0) is not in the region defined by 𝑥 ∈ ℝ and 𝑦 > 0, this shows that the IVP has no
unique solution but it has infinite solutions.
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(1.2) Initial-Value Problems
EXAMPLE 3
According to the Existence and Uniqueness Theorem, determine whether the IVP
has a unique solution.
SOLUTION:
𝑑𝑦
= 𝑥𝑦1/2 ,
𝑑𝑥
1
𝑦 (2) = 1
1 − 𝑓(𝑥, 𝑦) = 𝑥𝑦 2 = 𝑥 √𝑦 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ [0, ∞)}.
𝜕𝑓
𝑥
𝑥
2−
= 1=
is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ (0, ∞)}.
𝜕𝑦
2
𝑦
√
2𝑦 2
Since (2,1) is in the region defined by 𝑥 ∈ ℝ and 𝑦 > 0, this shows that the IVP has a unique
solution.
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(1.2) Initial-Value Problems
Exercises 1.2: Page 17
(𝟏𝟖) Determine a region of the 𝑥𝑦-plane for which the given differential equation would have a
unique solution whose graph passes through a point (𝑥0 , 𝑦0 ) in the region.
SOLUTION:
𝑑𝑦
= √𝑥𝑦
𝑑𝑥
1 − 𝑓(𝑥, 𝑦) = √𝑥𝑦 is continuous on either {(𝑥, 𝑦): 𝑥 ∈ [0, ∞) and 𝑦 ∈ [0, ∞)} or
{(𝑥, 𝑦): 𝑥 ∈ (−∞, 0] and 𝑦 ∈ (−∞, 0]}.
2−
𝜕𝑓 1 𝑥
= √
𝜕𝑦 2 𝑦
is continuous on either {(𝑥, 𝑦): 𝑥 ∈ [0, ∞) and 𝑦 ∈ (0, ∞)} or
{(𝑥, 𝑦): 𝑥 ∈ (−∞, 0] and 𝑦 ∈ (−∞, 0)}.
Thus, the differential equation will have a unique solution in any region where 𝑥 > 0 and 𝑦 > 0
or where 𝑥 < 0 and 𝑦 < 0.
------------------------------------------------------------------------------------------------------------------(𝟐𝟕) Determine whether Theorem 1.2.1 guarantees that the differential equation
𝑦 ′ = √𝑦 2 − 9
possesses a unique solution through the given point (2, −3).
SOLUTION:
1 − 𝑓(𝑥, 𝑦) = √𝑦 2 − 9 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ (−∞, −3] ∪ [3, ∞)}.
𝜕𝑓
2𝑦
𝑦
2−
=
=
is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ (−∞, −3) ∪ (3, ∞)}.
𝜕𝑦 2√𝑦 2 − 9 √𝑦 2 − 9
Since (2, −3) is not in either of the regions defined by 𝑦 < −3 or 𝑦 > 3, so there is no
guarantee of a unique solution through (2, −3).
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(1.2) Initial-Value Problems
Exercises 1.2: Page 17 (Homework)
(𝟔) In the following problem, 𝑦 = 1/(𝑥 2 + 𝑐) is a one-parameter family of solutions of the first-
order DE 𝑦 ′ + 2𝑥𝑦 2 = 0. Find a solution of the first-order IVP consisting of this differential
equation and the given initial condition. Give the largest interval 𝐼 over which the solution is
defined.
1
𝑦 ( ) = −4
2
------------------------------------------------------------------------------------------------------------------(𝟗) In the following problem, 𝑥 = 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡 is a two-parameter family of solutions of the
second-order DE 𝑥 ′′ + 𝑥 = 0. Find a solution of the second-order IVP consisting of this
differential equation and the given initial conditions.
𝜋
1
𝜋
𝑥 ( ) = , 𝑥′ ( ) = 0
6
2
6
------------------------------------------------------------------------------------------------------------------In the following problems, determine a region of the 𝑥𝑦-plane for which the given differential
equation would have a unique solution whose graph passes through a point (𝑥0 , 𝑦0 ) in the region.
𝑑𝑦
= 𝑦 2/3
𝑑𝑥
(𝟐𝟐) (1 + 𝑦 3 )𝑦 ′ = 𝑥 2
------------------------------------------------------------------------------------------------------------------(𝟏𝟕)
(𝟐𝟓) In the following problem, determine whether Theorem 1.2.1 guarantees that the differential
equation
𝑦 ′ = √𝑦 2 − 9
possesses a unique solution through the given point (1, 4).
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