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Unformatted text preview: Q11 (7.5 points): Find the general solutions of the following two unrelated differential
equations (3.5 Points for DE.1 and 4.0 Points for DE.2):
0 DE.1:
′ DE.2: 2 2 1 0 Solution: DE.1: Method1: (substitution method)
DE.1 =>
(when x y 0, dividing the equation by x y) This equation is simply homogeneous and separable.
Introducing sub: v
y
n v , xv , plug this into DE1, We have v => Bernoulli’s equation 1
Let u = v v So that v = u , v’ = u u’
Plug these to equation v
u
ρ x
u=e e e
Q x e After back substitution, u e =x
C x dx x C , one gets the general solution: y x Method2: (separable method) 1 C C =
0 which means 0 0 => is the general solution of DE1.
Natrually, x + y = 0 and y =0 is included in DE.2: Method1: (Bernoulli's method)
y y y 0 (dividing the equation with 2x y) y y y n 1 y Let v = y So that y = v , y’ = v v’
′ Plug these to equation 2
We have x v
v
ρ x 2xv v 1 P(x) =
e e 2
0 0 (dividing the equation with x ) , Q(x) =
e Q x e v=e 1 =x
C x dx therefore the general solution is:
y x C Method2: (Simple construction of terms)
2 2 0
2 C x C 0
0 Q1‐2 (7.5 points): Solve DE: where, obviously, and are parameters independent of and . If 0
terms of and ? If so, do it. ∞ and , can you express and/or in Solution: : = e if y(0) = y If y(∞ If (β y => y
0 y 0 => α e (β ln 0 Q13 (7.5 points): A man with a parachute jumps out of a “frozen” helicopter at height .
During the fall, the man's drag coefficient is (with closed parachute) and
(with open
parachute) and air resistance is taken as proportional to velocity. The total weight of the
man and his parachute is . Take the initial velocity when he jumped to be
3 zero. Gravitational constant is . The man opens the parachute time after he jumps.
Please find the total falling time and the speed he hits the group. Can you adjust the to
enable the quickest fall and lightest hit on the ground?
Solution: This is a problem of Newtonian mechanics F F ma. Total weight of the man and his parachute m with acceleration a, gravitational force F
and a drag force F .
Let x be the distance above the earth’s surface, with positive direction downward,
a , where v is the velocity and F mg. Since the drag force is assumed to be proportional to velocity,
F kv with closed parachute and F nkv with open parachute The deployment occurs at time t ,
During initial fall with closed parachute,
m dv
dt mg kv, v 0 this is an IVP problem. Solve this separable ODE. We obtain, 0
dv g dt kv
m or
dv
kv
g
m dt, C ln g C
x 0 m C1 e g g. => v(t) =
v 0 => C m
k g 1
m2
k2 k g
e C e
kt
m , t = 0, v(0) = 0 , => x g
4 kt
m m
k g t m
k e kt
m C , m So the distance of close parachute fall is At g 1 , V( k m g 0 k e k0
m m2
k2 g, e
time, the differential equation becomes Starting with the opening parachute
m dv
dt mg nkv =>
m dv dx
dx dt mg nkv dv
v
dx mg nkv =>
m whose solution is v(t) = m
nk g C e m
nk g C e nkt
m k0 m => x nk g t m
nk g C 0 m
nk k0
m e 1
n
m C
x Which means g 1 , since v( en m · ng C
= nkt
m nk 1 k0
m e kt
m e e C x1 0
m C k m g 0 k e k0
m m2
k2 g One can easily find the constant C
H x m
nk g 1 C m
nk e k1
m x C x H m
k g m
0 k e m2 k0
m k2 g The falling time t is in principle solvable from the above equation, but it’s tedious and may
require numerical solution technique. (Reaching this point without the actual solutions for
C and t is sufficient to gain full marks.)
Therefore the total falling time is we can adjust the + , the speed he hits the ground is v( ) = to enable the quickest fall and lightest hit on the ground.
5 m
nk g C e nkt
m , ...
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This note was uploaded on 05/11/2009 for the course AMS 361 taught by Professor Staff during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 Staff

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