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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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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.
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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