Part II. Writing the equation d² dt2 in the form of the system dx d dt (c) (d) (e) = V, = x³ + x², X = V= 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 – x4, x4 10/2/20 dv dt x5 (1) (2) = 0).
Part II. Writing the equation d² dt2 in the form of the system dx d dt (c) (d) (e) = V, = x³ + x², X = V= 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 – x4, x4 10/2/20 dv dt x5 (1) (2) = 0).
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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