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Consider the initial value problem (* + t)v + v = c", v(2) = 3r. According to the Existence and Uniqueness Theorem, what is the largest interval in which a unique solution is guaranteed to exist? (a) (4, 00) (b) (-0, 4) (c) (0,4) (d) (1,4)

Consider the initial value problem (* + t)v + v = c", v(2) = 3r. According to the Existence and Uniqueness Theorem, what is the largest interval in which a unique solution is guaranteed to exist? (a) (4, 00) (b) (-0, 4) (c) (0,4) (d) (1,4)

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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12:02 Consider the initial value problem 1 y = e (t – 4) ' (t2 + t)y/ + y(2) = 37. According to the Existence and Uniqueness Theorem, what is the largest interval in which a unique solution is guaranteed to exist? (a) (4, 00) (b) (-00, 4) (c) (0, 4) (а) (1,4) Consider the initial value problem (t2 – 4)y/ + 3y = In |5 – t|, y(3) = 0. According to the Existence and Uniqueness Theorem, what is the largest interval which a unique solution is guaranteed to exist? (a) (2, 00) (b) (-2, 2) (c) (2,5) (d) (-00, 5) The initial value problem (4 – t)y/ + In(t)y = sin(t), y(1.33) = 3.14159 is certain to have a unique continuous solution on the interval. (а) (-4,4) (b) (-2,2) (c) (0, 2) (d) (0,3.14159) Consider the initial value problem (t2 – 9)y/ + (2t sin t)y = In(1 - t), v(0) = 4. According to the Existence and Uniqueness Theorem, what is the largest interval in which a unique solution guaranteed to exist? (a) (-3, 3) (b) (3, оо) (c) (1,3) (d) (-3, 1) 5-8 Without solving the initial value problem, what is the largest interval in which a unique solution is guaranteed to exist for each initial condition? (a) y(x) = 7, (b) y(1) =-9, (c) y(-4) = e. 5. (1+5)y' + (-8)(t -1), 1-3 (1- 6)(t +1) 6. fy+ 1-2 y= sec(t/3) 1+3 7. (f + 41 – 5)y' + tan(2r)y =r² – 16 8. (4 - )y + In(6 - 1)y =e" 9. Find the general solution of the following differential equation, and show that both initial conditions y(1)=1 and y(-1)=-3 result in an identical particular solution. Does this fact violate the existence and uniqueness theorem? fy + 2ty = 2 This problem has been solved! See the answer
Transcribed Image Text:12:02 Consider the initial value problem 1 y = e (t – 4) ' (t2 + t)y/ + y(2) = 37. According to the Existence and Uniqueness Theorem, what is the largest interval in which a unique solution is guaranteed to exist? (a) (4, 00) (b) (-00, 4) (c) (0, 4) (а) (1,4) Consider the initial value problem (t2 – 4)y/ + 3y = In |5 – t|, y(3) = 0. According to the Existence and Uniqueness Theorem, what is the largest interval which a unique solution is guaranteed to exist? (a) (2, 00) (b) (-2, 2) (c) (2,5) (d) (-00, 5) The initial value problem (4 – t)y/ + In(t)y = sin(t), y(1.33) = 3.14159 is certain to have a unique continuous solution on the interval. (а) (-4,4) (b) (-2,2) (c) (0, 2) (d) (0,3.14159) Consider the initial value problem (t2 – 9)y/ + (2t sin t)y = In(1 - t), v(0) = 4. According to the Existence and Uniqueness Theorem, what is the largest interval in which a unique solution guaranteed to exist? (a) (-3, 3) (b) (3, оо) (c) (1,3) (d) (-3, 1) 5-8 Without solving the initial value problem, what is the largest interval in which a unique solution is guaranteed to exist for each initial condition? (a) y(x) = 7, (b) y(1) =-9, (c) y(-4) = e. 5. (1+5)y' + (-8)(t -1), 1-3 (1- 6)(t +1) 6. fy+ 1-2 y= sec(t/3) 1+3 7. (f + 41 – 5)y' + tan(2r)y =r² – 16 8. (4 - )y + In(6 - 1)y =e" 9. Find the general solution of the following differential equation, and show that both initial conditions y(1)=1 and y(-1)=-3 result in an identical particular solution. Does this fact violate the existence and uniqueness theorem? fy + 2ty = 2 This problem has been solved! See the answer
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