Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 18 REVIEW ET CHAPTER 17 ¤ 319
7. r 2 − 2r +1=0 ⇒ r =1and y c(x) =c 1e x + c 2xe x .Tryy p(x) =(Ax + B)cosx +(Cx + D)sinx ⇒
y 0 p =(C − Ax − B)sinx +(A + Cx + D)cosx and y 00
p =(2C − B − Ax)cosx +(−2A − D − Cx)sinx. Substitution
gives (−2Cx +2C − 2A − 2D)cosx +(2Ax − 2A +2B − 2C)sinx = x cos x ⇒ A =0, B = C = D = − 1 2 .
The general solution is y(x) =c 1 e x + c 2 xe x − 1 cos x − 1 (x +1)sinx.
2 2
9. r 2 − r − 6=0 ⇒ r = −2, r =3and y c(x) =c 1e −2x + c 2e 3x .Fory 00 − y 0 − 6y =1,tryy p1 (x) =A. Then
y 0 p 1
(x) =y 00
p 1
(x) =0and substitution into the differential equation gives A = − 1 6 .Fory00 − y 0 − 6y = e −2x try
y p2 (x) =Bxe −2x
[since y = Be −2x satisfies the complementary equation]. Then y 0 p 2
=(B − 2Bx)e −2x and
yp 00
2
=(4Bx − 4B)e −2x , and substitution gives −5Be −2x = e −2x ⇒ B = − 1 5
. The general solution then is
y(x) =c 1 e −2x + c 2 e 3x + y p1 (x)+y p2 (x) =c 1 e −2x + c 2 e 3x − 1 6 − 1 5 xe−2x .
11. The auxiliary equation is r 2 +6r =0and the general solution is y(x) =c 1 + c 2 e −6x = k 1 + k 2 e −6(x−1) .But
3=y(1) = k 1 + k 2 and 12 = y 0 (1) = −6k 2 . Thus k 2 = −2, k 1 =5and the solution is y(x) =5− 2e −6(x−1) .
13. The auxiliary equation is r 2 − 5r +4=0and the general solution is y(x) =c 1e x + c 2e 4x .But0=y(0) = c 1 + c 2
and 1=y 0 (0) = c 1 +4c 2 , so the solution is y(x) = 1 3 (e4x − e x ).
15. Let y(x) = ∞ c n x n .Theny 00
(x) = ∞ n(n − 1)c n x n−2
= ∞ (n +2)(n +1)c n+2 x n and the differential equation
n=0
n=0
becomes ∞ [(n +2)(n +1)c n+2 +(n +1)c n ]x n =0. Thus the recursion relation is c n+2 = −c n /(n +2)
n=0
for n =0, 1, 2, ....Butc 0 = y(0) = 0,soc 2n =0for n =0, 1, 2, ....Alsoc 1 = y 0 (0) = 1,soc 3 = − 1 3 , c 5 = (−1)2
3 · 5 ,
c 7 = (−1)3
3 · 5 · 7 = (−1)3 2 3 3!
, ..., c 2n+1 = (−1)n 2 n n!
7!
(2n +1)!
is y(x) = ∞ c n x n
= ∞
n=0
n=0
(−1) n 2 n n!
(2n +1)!
x 2n+1 .
n=0
for n =0, 1, 2,.... Thus the solution to the initial-value problem
17. Here the initial-value problem is 2Q 00 +40Q 0 + 400Q =12, Q (0) = 0.01, Q 0 (0) = 0. Then
Q c(t) =e −10t (c 1 cos 10t + c 2 sin 10t) and we try Q p(t) =A. Thus the general solution is
Q(t) =e −10t (c 1 cos 10t + c 2 sin 10t)+ 3
100 .But0.01 = Q0 (0) = c 1 +0.03 and 0=Q 00 (0) = −10c 1 +10c 2 ,
so c 1 = −0.02 = c 2 . Hence the charge is given by Q(t) =−0.02e −10t (cos 10t +sin10t)+0.03.
19. (a) Since we are assuming that the earth is a solid sphere of uniform density, we can calculate the density ρ as follows:
ρ =
mass of earth
volume of earth =
M
4
3 πR3 .IfV r is the volume of the portion of the earth which lies within a distance r of the
center, then V r = 4 3 πr3 and M r = ρV r = Mr3
R .ThusF 3
r = − GM rm
= − GMm r.
r 2 R 3