In Exercises 25-36, find the solution of the given initial value problem. y"-2y' +17y=0, y(0) = -2, y'(0)=3
In Exercises 25-36, find the solution of the given initial value problem. y"-2y' +17y=0, y(0) = -2, y'(0)=3
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

Transcribed Image Text:This information can be summarized as follows:
If p²-4q> 0, the characteristic equation has two distinct, real
roots A₁ and A₂. A fundamental set of solutions is
y₁ (t) = ₁ and 32(t) = e^2t
If p² - 4q = 0, the characteristic equation has one repeated real
root λ. A fundamental set of solutions is
.
₁ (t) = et and 2(t) = te
If p²-4g < 0, the characteristic equation has two complex con-
jugate roots a tib. A fundamental set of solutions is
y(t) = et cos bt and
y2(t) = eat sin bt.
In Exercises 25-36, find the solution of the given initial value
problem.
y"-2y+17y=0, y(0) = -2, y'(0)=3
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