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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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17 D SECOND-ORDER DIFFERENTIAL EQUATIONS<br />

17.1 Second-Order Linear Equations<br />

1. The auxiliary equation is r 2 - r- 6 = 0 => (r- 3)(r + 2) = 0 => r = 3, r = -2. Then by {8) the general s?lution<br />

isy = c1e 3 "' +cze- 2 "'.<br />

3. The auxiliary equation is r 2 + 16 = 0 => r = ± 4i. Then by ( 11 ) the general solution is<br />

y = e 0 "'(c1 cos4x + cz sin4x) = c1 cos4x + c2 sin4~.<br />

5. The auxiliary equation is 9r 2 - 12r· + 4 = 0 => (3r- 2) 2 = 0 => r = ~·Then by (10), the general solution is<br />

_1· The auxiliary equation is 2r·2 - r = r(2r - 1) = 0 => r = 0, r = t, soy = c1e 0 "' + cze"/ 2 = c1 + c 2 e"'1 2 .<br />

4± v'-36 .<br />

9. The auxiliary equation is r 2 - 4r + 13 = 0 => r =<br />

~ 2 ± 3t, soy = e 2 "'( c1 cos 3x + c 2 sin 3x ).<br />

11. The auxiliary equation is 2r 2 + 2r - 1 = 0 => r = - 2 ~ v'I2 = - ~ ± v;, so<br />

y = c 1 e( - l/2+VJ/2)t + cze( - l /2- VJ/2)t.<br />

2<br />

13. The auxiliary equation is 100r 2 + 200r + 101 = 0 => r =<br />

P = e-t (c 1 cos ( iot) + c2 sin U 0<br />

t)] .<br />

- 200 ± V - 400 1 .<br />

::;= -1 ± Tiit, so<br />

200<br />

15. The auxiliary equation is 5r 2 - 2r- 3 = (5r + 3)(r - 1) = 0 => r = _ 1 5,<br />

10<br />

r = 1, so the general solution is y = c 1 e- 3 x/S + c 2 e"' . We graph the basic<br />

solutions f(x) = e- S:r/ 5 , g(x) = e"' as well as y = e- 3 x/S + 2e"',<br />

y = e- 3 x /f> - e"', and.y = - 2e- 3 "'/ 5 - e"'. Each solution consists of a single<br />

continuous curve that approaches either 0 or ±oo as x _, ± oo.<br />

17. r 2 - 6r + 8 = (r - 4)(r- 2) = 0, so r = 4, r· = 2 and the general solution is y = c1e 4 "' + c 2 e 2 "'. Then<br />

·y' = 4c1e 4 "' + 2cze 2 "', so y(O) = 2 => c1 + cz = 2 and y'(O) = 2 => 4cl + 2cz = 2, giving c1 = - 1 and c 2 = 3.<br />

Thus the solution to the initial-value problem is y = 3e 2 "' -<br />

e 4 "'.<br />

19. 9r 2 + 12r + 4 = (3r + 2? = 0 => r =- ~and the general solution is y = c1e- 2 "'/ 3 + c 2 xe- 2 "'1 3 . Then y(O) = 1 =><br />

c1 = 1 and, since y' = - fr cle- 2 "' 13 + Cz (1 - ~x) e- 2 "'1 3 , y'(O) = 0 => - ~c1 + cz = 0, so c2 = t and the solution to<br />

the initial-value problem is y = e- 2 "'1 3 + ~xe- 2 "'/3.<br />

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