Use Theorem 5.4 to show that each of the following initial-value problems has a unique solution, and find the solution. y = y cos t, 0sIs1, y(0) = 1. a. b. y = y +re, Isıs2, y(1) = 0. c. y = -y+re, Isis2, y(1) = Vze. 4r'y d. y = 0sIs1, y(0) = 1.

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Chapter2: Second-order Linear Odes
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Use Theorem 5.4 to show that each of the following initial-value problems has a unique solution, and
find the solution.
y = y cos t, 0sIs1, y(0) = 1.
a.
b. y = y+fe,
y = -y+re, 1siS2, y(1) = Jze.
1<1<2, y(1) = 0.
2
с.
4r'y
d. y = 1+'
y(0) = 1.
Transcribed Image Text:Use Theorem 5.4 to show that each of the following initial-value problems has a unique solution, and find the solution. y = y cos t, 0sIs1, y(0) = 1. a. b. y = y+fe, y = -y+re, 1siS2, y(1) = Jze. 1<1<2, y(1) = 0. 2 с. 4r'y d. y = 1+' y(0) = 1.
Theorem 5.4 Suppose that D = {(t,y) | a < t < b and -o < y < o} and that f(t, y) is continuous on
D. If f satisfies a Lipschitz condition on D in the variable y, then the initial-value problem
y'(t) = f(t, y), a <i < b, y(a) =
has a unique solution y(t) for a <t <b.
Transcribed Image Text:Theorem 5.4 Suppose that D = {(t,y) | a < t < b and -o < y < o} and that f(t, y) is continuous on D. If f satisfies a Lipschitz condition on D in the variable y, then the initial-value problem y'(t) = f(t, y), a <i < b, y(a) = has a unique solution y(t) for a <t <b.
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