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Unformatted text preview: Section 2.3 13. 14. Equation (12) is dv p
—=— 1 —2.
dt g(+gv) If u = vﬁ, then 53% = 5%, so the differential equation is
du
£217 2 “9” + “2)
This is separable:
du
1 +u2 z —‘/gpdt.
Since "0(0) 2 v0, 71(0) 2 voﬁ, so
”\fg du t
= «M dt.
[um/73’ 1 + "2 gp/O‘
By the hint,
tan—1(v\/E) —— tan"1(vo \/E) = ——t‘ /gp.
g 9
Therefore, v r: ﬁtaxﬂcl —— t‘/gp), where C1 = tan‘1(v0\/g) Equation (13) is _ ._ 2 ,_.
E—v——\/;tan(C1 tq/gp). This is separable:
dy = ﬁtmﬂCl M tdpg) dt. y t
/ dy = \/Q/ tan(C1 — t‘/pg) dt.
. W P . o :‘C ——t 1
yyo:\/E1n:e:(_.1_____. WI”.
0 cosC1 x/pg
y*y +11n
0 I) Since 31(0) = yo, By the hint , Therefore,
cos(01 — t‘ /pg) cos C1 15 Equation (15) is (11) p2
W449.) If u = 11%, then 131 = 5%, so the differential equation is
22% _  2
p dt 9(1 u )
This is separable:
du
1 _. 2 — —\/gpdt.
Since 11(0) : 00, 14(0) 2 voﬁ, so
”Mg du t
f 2 = ~4/gp/ dt.
”DJ—g 1 ~ u 0
By the hint,
talih—1 (11¢?) — tank—10mg) = —t,/gp.
Therefore, 1) z ﬂtanhmz — M973), where C2 = tanh“1(vO\/§). 16. Equation (16) is
_. __ _g , — ,/
(It v ‘ \/;tanh(C2 t 9P) This is separable: dy = ﬁtanMCg w tx/pg) dt. y t
/ dy 2 \/§ / tanh(C'2 — t,/pg) dt.
. yo P . 0 Since y(0) = 3/0, By the hint7
: :1 C —~t l
y_y0:_ gm __Lz____._.__ W9) _,__ p cosh Cg Mpg Therefore,
h C  tw/
yzyo—lln —————COS( 2 pg) .
p cosh C2 17. To solve the initial value problem 12’ = — 9.8 — 0.0011v2 , v(0) = 49 we write J_ﬂ_2 = 4.1.. [M = _ [0.1038274
9.8 + 0.0011v 1 + (0.010595 v) tan—1 (0.010595 v) = —0.103827t + C; v(0) = 49 implies C = 0.478854
v(t) = 94.3841 tan(0.4788540.103827t) Integration with y(0) = 0 gives
y(t) = 108.468 + 909.052 ln(cos(0.478854 — O. 1 03 827 t)) . We solve v(0) = 0 for t = 4.612, and then calculate y(4.612) = 108.468. Section 2.4 2. Iterative formula: yn+1 = y" + h(2y,,)
Approximate values 1.125 and 1.244; true value y(%) z 1.359 5. Iterative formula: yn+1 = y" + h(y,, — xn — 1)
Approximate values 0.938 and 0.889; true value y(§) z 0.851 ...
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 Spring '07
 TERRELL,R
 Math

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