Writing the equation d² dt2 n the form of the system dx x = V, =) 3) e) X= dt x³ + x², v = x³ + x² x = x(t), x = x(t), v = v(t), (1 (2 1) Find all the stationary points (x, v) (the points where d = 0, d = 0) 5) 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, v² +W(x) = const for each solution (x(t), v(t)) to system (2), here the potential energy is given by the antiderivative of -x³ – x¹, x4 x5
Writing the equation d² dt2 n the form of the system dx x = V, =) 3) e) X= dt x³ + x², v = x³ + x² x = x(t), x = x(t), v = v(t), (1 (2 1) Find all the stationary points (x, v) (the points where d = 0, d = 0) 5) 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, v² +W(x) = const for each solution (x(t), v(t)) to system (2), here the potential energy is given by the antiderivative of -x³ – x¹, x4 x5
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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