Rayleigh's equation is y” +4 (½(1¹)² − 1) y′ + y = 0, where µ is a constant. Show that differenti- ation of this equation and setting y' = z reduces Rayleigh's equation to the Van der Pol equation. y" +μ(y²-1)y + y = 0.

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
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4.
(a) Rayleigh's equation is y" + 4 ((1′)² − 1) y′ + y = 0, where µ is a constant. Show that differenti-
ation of this equation and setting y' = z reduces Rayleigh's equation to the Van der Pol equation,
y" +μ(y² - 1)y + y = 0.
(b) The Van der Pol equation is equivalent to the system u' = v and v' = u(1-²)v-u. Show that
(0,0) is the only critical point of the system. Determine the nature and stability of the critical point
when μ< 2, μ = 2 and μ > 2.
Transcribed Image Text:4. (a) Rayleigh's equation is y" + 4 ((1′)² − 1) y′ + y = 0, where µ is a constant. Show that differenti- ation of this equation and setting y' = z reduces Rayleigh's equation to the Van der Pol equation, y" +μ(y² - 1)y + y = 0. (b) The Van der Pol equation is equivalent to the system u' = v and v' = u(1-²)v-u. Show that (0,0) is the only critical point of the system. Determine the nature and stability of the critical point when μ< 2, μ = 2 and μ > 2.
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