homework 05 soutions - Section 2.3 13 14 Equation(12 is dv...

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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 y-yo:\/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(0.478854-0.103827t) 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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