Problem #4: Existence and Uniqueness Theorem for reference: Let an(x), an-1(x), ..., a₁(x), ao(x) and g(x) be continuous on an interval I, and let an(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 dn-ly dx-1 dx dx y(x) = yo, y'(x) = y₁, ..., y(n-¹)(x₁) = Yn-1, exists on the interval and is unique. d" y dy an(x) + an-1(x) · + ... + a₁(x)ª² + aŋ(x) y = g(x), (a) Find the largest interval on which the above theorem guarantees that the following initial value problem has a unique solution. (x+8) y'"' + (x² − 25) y" + 18y = ¹10, y(0) = 2, y'(0) = 6, y″(0) = 10 (b) Find the largest interval on which the above theorem guarantees that the following initial value problem has a unique solution. (x − 8) y'"' + (x² − 25) y" + 18y=-¹₁0 y(0) = 2, y'(0) = 6, y″(0) = 10 (A) (-10, ∞) (B) (8,00) (C) (-8, 10) (D) (-8,00) (E) (-00,-8) (F) (-0, 8) (G) (-10, 8) (H) (-∞0, 10) (I) (8,10) (J) (-10,-8) (K) (-∞, -10) (L) (10,00) Problem #4(a): Select Part (a) choices. (A) (8,10) (I) (-10, ∞) (J) (-∞,-10) (K) (-10,-8) (L) (-∞, 10) (B) (10, ∞) (C) (-10, 8) (D) (-8, 10) (E) (-0, 8) (F) (-8, ∞) (G) (-∞, -8) (H) (8,00) Problem #4(b): [Select Part (b) choices.
Problem #4: Existence and Uniqueness Theorem for reference: Let an(x), an-1(x), ..., a₁(x), ao(x) and g(x) be continuous on an interval I, and let an(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 dn-ly dx-1 dx dx y(x) = yo, y'(x) = y₁, ..., y(n-¹)(x₁) = Yn-1, exists on the interval and is unique. d" y dy an(x) + an-1(x) · + ... + a₁(x)ª² + aŋ(x) y = g(x), (a) Find the largest interval on which the above theorem guarantees that the following initial value problem has a unique solution. (x+8) y'"' + (x² − 25) y" + 18y = ¹10, y(0) = 2, y'(0) = 6, y″(0) = 10 (b) Find the largest interval on which the above theorem guarantees that the following initial value problem has a unique solution. (x − 8) y'"' + (x² − 25) y" + 18y=-¹₁0 y(0) = 2, y'(0) = 6, y″(0) = 10 (A) (-10, ∞) (B) (8,00) (C) (-8, 10) (D) (-8,00) (E) (-00,-8) (F) (-0, 8) (G) (-10, 8) (H) (-∞0, 10) (I) (8,10) (J) (-10,-8) (K) (-∞, -10) (L) (10,00) Problem #4(a): Select Part (a) choices. (A) (8,10) (I) (-10, ∞) (J) (-∞,-10) (K) (-10,-8) (L) (-∞, 10) (B) (10, ∞) (C) (-10, 8) (D) (-8, 10) (E) (-0, 8) (F) (-8, ∞) (G) (-∞, -8) (H) (8,00) Problem #4(b): [Select Part (b) choices.
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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Transcribed Image Text:Problem #4: Existence and Uniqueness Theorem for reference:
Let an(x), an-1(x), ..., a₁(x), ao(x) and g(x) be continuous on an interval I, and let an(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
dly
dx-1
dx
y(x) = yo, y'(x) = y₁, ..., y(n-¹)(x₁) = Yn-1,
exists on the interval and is unique.
d" y
an(x) + an-1(x)
(a) Find the largest interval on which the above theorem guarantees that the following initial value problem has a
unique solution.
(x+8) y'"' + (x² − 25) y" + 18y = ¹10, y(0) = 2, y'(0) = 6, y″(0) = 10
(b) Find the largest interval on which the above theorem guarantees that the following initial value problem has a
unique solution.
(x − 8) y'"' + (x² − 25) y" + 18y=-¹₁0 y(0) = 2, y'(0) = 6, y″(0) = 10
dy
· + ... + a₁(x)ª² + aŋ(x) y = g(x),
dx
(A) (-10, ∞) (B) (8,00) (C) (-8, 10) (D) (-8,00) (E) (-00,-8) (F) (-0, 8) (G) (-10, 8) (H) (-∞0, 10)
(I) (8,10) (J) (-10,-8) (K) (-∞, -10) (L) (10,00)
Problem #4(a): Select Part (a) choices.
(A) (8,10)
(I) (-10, ∞)
Problem #4(b): [Select
(B) (10, ∞) (C) (-10, 8) (D) (-8, 10) (E) (-∞, 8) (F) (-8, ∞) (G) (-∞, -8) (H) (8,00)
(J) (-∞,-10) (K) (-10,-8) (L) (-∞, 10)
Part (b) choices.
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