Thomas Calculus 13th [Solutions]

690 Chapter 9 First-Order Differential Equations
(d) Consider trajectory ( x , y ) m , a .
n b
For a b y m nx y n x
y e Kx e e K , taking the limit of both
by m
e x
a n x M
a
y
sides lim e
y
y M y m
lim K K . Thus,
x represents the equation any
b y m
x m/ n e x x m/
n
M
b y
n x
x
e M x e
y a/ b y a/
b
solution trajectory must satisfy if the trajectory approaches the rest point asymptotically.
(e) Pick initial condition y a
0 . Then, from the figure at right,
b
M
a
y m
0
f ( y0
) M y implies x y
M
M n x b y and thus
e e y
x
e
m
nx
x
M x.
From the figure for g( x ), there exists a unique
m
0
x m
0 satisfying x M .
n
n x x That is, for each y a there
e
b
a
m
y M y x
is a unique x satisfying . Thus, there can exist
b y M n x
e x e
only one trajectory solution approaching m , a . (You can
n b
think of the point ( x0 , y 0)
as the initial condition for that
trajectory.)
(f ) Likewise there exists a unique trajectory when a
M
a
y m
0
y 0 . Again, f ( y
b
0) M y implies x y
M
M e
n x e
b y0
y
m
and thus x M .
n x x From the figure for g( x ), there exists a unique x m
0 satisfying M .
e
n
x That is,
e
a
for each y a
y M y m
there is a unique x satisfying x . Thus, there can exist only one trajectory
b
b y M n x
e x e
solution approaching m , a .
n b
8. Let y dy dz
dx dx
z y , y F ( x , y , y ), we can write it as the
following system of first order differential equations: dy
dx
z
dz
dx
F( x, y, z)
( n) ( n 1)
dy
In general, for the nth order differential equation given by y F x, y, y , y , , y , let z1
y
dx
dz1
dz2
z1 y , let z
dx
2 z1 y
dx
z2 y , , let zn 1 zn 2
( n 1)
y zn
1
( n)
y . This gives us
dy
the following system of first order differential equations:
dx
z1
dz1
dx
z2
dz2
dx
z3
a
x
m
x
n x
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