Q1.2 find the biggest interval on which theorem 3.1.1 guarantees that the initial value problem has a unique a 1 (x + 5)y" + (x³²-4) y² + 174 = x + 1³ 4 (0)=3, y₁ (0) = 69 y" (0) = 6 options a) (-∞, -1) e) (-5, 1) i) (-1,5) answer var. problem has b find the biggest interval on which rantees that the g unique a)(-∞0,1) b) (0,5) f) (-∞, -5) j) (-5, -1) (۱ - ره-)(e i)(-5,∞0) following solution. Note: 1 (x - 57y" + (x²=-4)y" + 17y = = = = = 5 y(0) = 3₂ 41(0) = 6+9" (01-6 Answer Options c) (-∞0, 1) g) (1, -∞0) k) (1,5) b) (100) f)(1,5) j) (-1,5) d) (5,00) 4) (5, ∞0) 4) (-1,00) theorem 3.1.1 following initial value solution. c) (5,1) 9) (-1,00) K) (-∞,5) d) (5,∞0) W)(-∞,-5) 2)(-5, -1) Theorem 3.1.1 Existence of a Unique Solution Let an(x), an-1(x), ..., a₁(x), a (x), and g(x) be continuous on an interval I, and let a,(x) = 0 for every x in this interval. If x= xo is any point in this interval, then a solution y(x) of the initial-value problem (1) exists on the interval and is unique.
Q1.2 find the biggest interval on which theorem 3.1.1 guarantees that the initial value problem has a unique a 1 (x + 5)y" + (x³²-4) y² + 174 = x + 1³ 4 (0)=3, y₁ (0) = 69 y" (0) = 6 options a) (-∞, -1) e) (-5, 1) i) (-1,5) answer var. problem has b find the biggest interval on which rantees that the g unique a)(-∞0,1) b) (0,5) f) (-∞, -5) j) (-5, -1) (۱ - ره-)(e i)(-5,∞0) following solution. Note: 1 (x - 57y" + (x²=-4)y" + 17y = = = = = 5 y(0) = 3₂ 41(0) = 6+9" (01-6 Answer Options c) (-∞0, 1) g) (1, -∞0) k) (1,5) b) (100) f)(1,5) j) (-1,5) d) (5,00) 4) (5, ∞0) 4) (-1,00) theorem 3.1.1 following initial value solution. c) (5,1) 9) (-1,00) K) (-∞,5) d) (5,∞0) W)(-∞,-5) 2)(-5, -1) Theorem 3.1.1 Existence of a Unique Solution Let an(x), an-1(x), ..., a₁(x), a (x), and g(x) be continuous on an interval I, and let a,(x) = 0 for every x in this interval. If x= xo is any point in this interval, then a solution y(x) of the initial-value problem (1) exists on the interval and is unique.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Q1.2
a
find the biggest interval on which theorem 3.1.1
guarantees that the
initial value
problem has a
unique
1
(x + 5) y" + (x³²-4) y² + 17y= 2² +1² 460) = 3₂ y ₁ (0) = 69 4" (0)=6
' x +I³
options
a) (-∞, -1)
e) (-5,1)
i) (-1,5)
answer
var.
problem has
b find the biggest interval on which
rantees that the
g
unique
(۱ - ره-)(e
following
solution.
i)(-5, ∞0)
b) (0,5)
f) (-∞, -5)
j) (-5, -1) 1) (1,5)
Note:
c) (-∞0, 1)
g) (1, -∞)
1
(x - 57y" + (x²-4)y" + 17y = = = = = 54(0) = 3₂ y'(0) = 6+y" (01=6
+ 1
Answer Options
a) (-∞0,1)
b) (1,00)
f)(1,5)
j) (-1,5)
d) (5,00)
4) (5, ∞0)
4) (-1,00)
theorem 3.1.1
following initial value
solution.
c)(5,1)
9) (-1,00)
K) (-∞,5)
d) (5,∞0)
W)(-∞,-5)
2)(-5, -1)
Theorem 3.1.1
Existence of a Unique Solution
Let an(x), an-1(x), ..., a₁(x), a (x), and g(x) be continuous on an interval I, and let a,(x) = 0
for every x in this interval. If x= xo is any point in this interval, then a solution y(x) of the
initial-value problem (1) exists on the interval and is unique.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff1fdb7ad-6b81-4220-b2ec-026d3971ebe5%2Fdf71e5ae-3c75-4c78-bed3-85d0dafbac67%2F4ovvu2b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1.2
a
find the biggest interval on which theorem 3.1.1
guarantees that the
initial value
problem has a
unique
1
(x + 5) y" + (x³²-4) y² + 17y= 2² +1² 460) = 3₂ y ₁ (0) = 69 4" (0)=6
' x +I³
options
a) (-∞, -1)
e) (-5,1)
i) (-1,5)
answer
var.
problem has
b find the biggest interval on which
rantees that the
g
unique
(۱ - ره-)(e
following
solution.
i)(-5, ∞0)
b) (0,5)
f) (-∞, -5)
j) (-5, -1) 1) (1,5)
Note:
c) (-∞0, 1)
g) (1, -∞)
1
(x - 57y" + (x²-4)y" + 17y = = = = = 54(0) = 3₂ y'(0) = 6+y" (01=6
+ 1
Answer Options
a) (-∞0,1)
b) (1,00)
f)(1,5)
j) (-1,5)
d) (5,00)
4) (5, ∞0)
4) (-1,00)
theorem 3.1.1
following initial value
solution.
c)(5,1)
9) (-1,00)
K) (-∞,5)
d) (5,∞0)
W)(-∞,-5)
2)(-5, -1)
Theorem 3.1.1
Existence of a Unique Solution
Let an(x), an-1(x), ..., a₁(x), a (x), and g(x) be continuous on an interval I, and let a,(x) = 0
for every x in this interval. If x= xo is any point in this interval, then a solution y(x) of the
initial-value problem (1) exists on the interval and is unique.
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