Solução_Calculo_Stewart_6e

F.
312 ¤ CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ETTX.10
CHAPTER 17
11. The auxiliary equation is r 2 +3r +2=(r +1)(r +2)=0,sor = −1, r = −2 and y c(x) =c 1e −x + c 2e −2x .
Try y p = A cos x + B sin x ⇒ y 0 p = −A sin x + B cos x, y 00
p = −A cos x − B sin x. Substituting into the differential
equation gives (−A cos x − B sin x)+3(−A sin x + B cos x)+2(A cos x + B sin x) =cosx or
(A +3B)cosx +(−3A + B)sinx =cosx. Then solving the equations
A +3B =1, −3A + B =0gives A = 1 , B = 3 and the general
10 10
solution is y(x) =c 1 e −x + c 2 e −2x + 1 cos x + 3 sin x. Thegraph
10 10
shows y p and several other solutions. Notice that all solutions are
asymptotic to y p as x →∞.Exceptfory p , all solutions approach either ∞
or −∞ as x →−∞.
13. Here y c (x) =c 1 cos 3x + c 2 sin 3x. Fory 00 +9y = e 2x try y p1 (x) =Ae 2x and for y 00 +9y = x 2 sin x
try y p2 (x) =(Bx 2 + Cx + D)cosx +(Ex 2 + Fx+ G)sinx. Thus a trial solution is
y p (x) =y p1 (x)+y p2 (x) =Ae 2x +(Bx 2 + Cx + D)cosx +(Ex 2 + Fx+ G)sinx.
15. Here y c (x) =c 1 + c 2 e −9x .Fory 00 +9y 0 =1try y p1 (x) =Ax (since y = A is a solution to the complementary equation)
and for y 00 +9y 0 = xe 9x try y p2 (x) =(Bx + C)e 9x .
17. Since y c (x) =e −x (c 1 cos 3x + c 2 sin 3x) we try y p (x) =x(Ax 2 + Bx + C)e −x cos 3x + x(Dx 2 + Ex + F )e −x sin 3x
(sothatnotermofy p is a solution of the complementary equation).
Note: Solving Equations (7) and (9) in The Method of Variation of Parameters gives
u 0 Gy 2
1 = −
a (y 1 y2 0 − y 2y1 0 ) and u 0 Gy 1
2 =
a (y 1 y2 0 − y 2y1 0 )
We will use these equations rather than resolving the system in each of the remaining exercises in this section.
19. (a) Here 4r 2 +1=0 ⇒ r = ± 1 2 i and y c(x) =c 1 cos 1
2 x + c 2 sin 1
2 x . We try a particular solution of the form
y p(x) =A cos x + B sin x ⇒ y 0 p = −A sin x + B cos x and y 00
p = −A cos x − B sin x. Then the equation
4y 00 + y =cosx becomes 4(−A cos x − B sin x)+(A cos x + B sin x) =cosx or
−3A cos x − 3B sin x =cosx ⇒ A = − 1 , B =0.Thus,yp(x) =− 1 cos x and the general solution is
3 3
y(x) =y c (x)+y p (x) =c 1 cos 1
x + c
2 2 sin 1
x − 1 cos x.
2 3
(b) From (a) we know that y c(x) =c 1 cos x 2 + c2 sin x 2 .Settingy1 =cosx 2 , y2 =sinx 2 ,wehave
y 1 y 0 2 − y 2 y 0 1 = 1 2 cos2 x 2 + 1 2 sin2 x 2 = 1 2 .Thusu0 1 = − cos x sin x 2
4 · 1
2
= − 1 cos
2 · x
2 2 sin
x
= − 1
2 2 2cos
2 x
− 1 sin x 2 2
and u 0 2 = cos x cos x 2
4 · 1
2
= 1 cos
2
2 · x
2 cos
x
= 1
2 2 1 − 2sin
2 x
2 cos
x
.Then 2
u 1(x) = 1
2 sin x 2 − cos2 x 2 sin x 2
dx = − cos
x
2 + 2 3 cos3 x 2 and