Part II. Writing the equation d² dt2 in the form of the system d dt X = U, X = x³ + x², x = x(t), dv = x³ + x² 4 dt x = x(t), v=v(t), == (1) (a) Find all the stationary points (x, v) (the points where d = 0, d = 0). dx dt (b) Find the corresponding linear system near each critical point. (c) (d) Find the corresponding linear system near each critical point. Draw a phase portrait of the system near each critical point. Draw a phase portrait taking into account the energy conservation, for each solution (x(t), v(t)) to system (2), (e) + W(x) = const where the potential energy is given by the antiderivative of -x³-x¹, x4 W(x) 4 x5 5 (2)
Part II. Writing the equation d² dt2 in the form of the system d dt X = U, X = x³ + x², x = x(t), dv = x³ + x² 4 dt x = x(t), v=v(t), == (1) (a) Find all the stationary points (x, v) (the points where d = 0, d = 0). dx dt (b) Find the corresponding linear system near each critical point. (c) (d) Find the corresponding linear system near each critical point. Draw a phase portrait of the system near each critical point. Draw a phase portrait taking into account the energy conservation, for each solution (x(t), v(t)) to system (2), (e) + W(x) = const where the potential energy is given by the antiderivative of -x³-x¹, x4 W(x) 4 x5 5 (2)
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:Part II.
Writing the equation
d²
dt2
in the form of the system
dx
d
dt
(c)
(d)
(e)
= V₂
·V=
= x³ + x²,
X =
x³ + x²
W(x):
x = x (t),
=
x = x(t), v = v(t),
(a) Find all the stationary points (x, v) (the points where d = 0,
dt
(b)
Find the corresponding linear system near each critical point.
Find the corresponding linear system near each critical point.
Draw a phase portrait of the system near each critical point.
Draw a phase portrait taking into account the energy conservation,
12/201²
v² + W(x) = const for each solution (x(t), v(t)) to system (2),
where the potential energy is given by the antiderivative of −x³ – x¹,
x4
10/2/20
dv
dt
x5
(1)
(2)
= 0).
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