Solução_Calculo_Stewart_6e

F.
310 ¤ CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ETX.10
CHAPTER 17
21. r 2 +16=0 ⇒ r = ±4i and the general solution is y = e 0x (c 1 cos 4x + c 2 sin 4x) =c 1 cos 4x + c 2 sin 4x. Then
y
π
4 = −3 ⇒ −c1 = −3 ⇒ c 1 =3and y 0
π
4 =4 ⇒ −4c2 =4 ⇒ c 2 = −1, so the solution to the
initial-value problem is y =3cos4x − sin 4x.
23. r 2 +2r +2=0 ⇒ r = −1 ± i and the general solution is y = e −x (c 1 cos x + c 2 sin x). Then2=y(0) = c 1 and
1=y 0 (0) = c 2 − c 1 ⇒ c 2 =3and the solution to the initial-value problem is y = e −x (2 cos x +3sinx).
25. 4r 2 +1=0 ⇒ r = ± 1 2 i and the general solution is y = c1 cos 1
2 x + c 2 sin 1
2 x .Then3=y(0) = c 1 and
−4 =y(π) =c 2 , so the solution of the boundary-value problem is y =3cos 1
2 x − 4sin 1
2 x .
27. r 2 − 3r +2=(r − 2)(r − 1) = 0 ⇒ r =1, r =2and the general solution is y = c 1 e x + c 2 e 2x .Then
1=y(0) = c 1 + c 2 and 0=y(3) = c 1 e 3 + c 2 e 6 so c 2 =1/(1 − e 3 ) and c 1 = e 3 /(e 3 − 1). The solution of the
boundary-value problem is y = ex+3
e 3 − 1 +
e2x
1 − e . 3
29. r 2 − 6r +25=0 ⇒ r =3± 4i and the general solution is y = e 3x (c 1 cos 4x + c 2 sin 4x). But1=y(0) = c 1 and
2=y(π) =c 1 e 3π ⇒ c 1 =2/e 3π , so there is no solution.
31. r 2 +4r +13=0 ⇒ r = −2 ± 3i and the general solution is y = e −2x (c 1 cos 3x + c 2 sin 3x). But2=y(0) = c 1
and 1=y π
2
= e −π (−c 2 ), so the solution to the boundary-value problem is y = e −2x (2 cos 3x − e π sin 3x).
33. (a) Case 1 (λ =0): y 00 + λy =0 ⇒ y 00 =0which has an auxiliary equation r 2 =0 ⇒ r =0 ⇒ y = c 1 + c 2 x
where y(0) = 0 and y(L) =0.Thus,0=y(0) = c 1 and 0=y(L) =c 2L ⇒ c 1 = c 2 =0. Thus y =0.
Case 2 (λ