Solução_Calculo_Stewart_6e

F.
SECTION TX.10 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ET SECTION 17.2 ¤ 311
18.2 Nonhomogeneous Linear Equations ET 17.2
1. The auxiliary equation is r 2 +3r +2=(r +2)(r +1)=0, so the complementary solution is y c (x) =c 1 e −2x + c 2 e −x .
We try the particular solution y p (x) =Ax 2 + Bx + C,soy 0 p =2Ax + B and y 00
p =2A. Substituting into the differential
equation, we have (2A)+3(2Ax + B)+2(Ax 2 + Bx + C) =x 2 or 2Ax 2 +(6A +2B)x +(2A +3B +2C) =x 2 .
Comparing coefficients gives 2A =1, 6A +2B =0,and2A +3B +2C =0,soA = 1 2 , B = − 3 2 ,andC = 7 4 .Thusthe
general solution is y(x) =y c(x)+y p(x) =c 1e −2x + c 2e −x + 1 2 x2 − 3 2 x + 7 4 .
3. The auxiliary equation is r 2 − 2r = r(r − 2) = 0, so the complementary solution is y c (x) =c 1 + c 2 e 2x . Try the particular
solution y p (x) =A cos 4x + B sin 4x,soy 0 p = −4A sin 4x +4B cos 4x and y 00
p = −16A cos 4x − 16B sin 4x. Substitution
into the differential equation gives (−16A cos 4x − 16B sin 4x) − 2(−4A sin 4x +4B cos 4x) =sin4x
⇒
(−16A − 8B)cos4x +(8A − 16B)sin4x =sin4x. Then−16A − 8B =0and 8A − 16B =1 ⇒ A = 1 40 and
B = − 1 . Thus the general solution is y(x) =y 20 c(x)+y p (x) =c 1 + c 2 e 2x + 1 cos 4x − 1 sin 4x.
40 20
5. The auxiliary equation is r 2 − 4r +5=0with roots r =2± i, so the complementary solution is
y c (x) =e 2x (c 1 cos x + c 2 sin x). Tryy p (x) =Ae −x ,soy 0 p = −Ae −x and y 00
p = Ae −x . Substitution gives
Ae −x − 4(−Ae −x )+5(Ae −x )=e −x ⇒ 10Ae −x = e −x ⇒ A = 1 10
. Thus the general solution is
y(x) =e 2x (c 1 cos x + c 2 sin x)+ 1
10 e−x .
7. The auxiliary equation is r 2 +1=0with roots r = ±i, so the complementary solution is y c(x) =c 1 cos x + c 2 sin x.
For y 00 + y = e x try y p1 (x) =Ae x .Theny 0 p 1
= y 00
p 1
= Ae x and substitution gives Ae x + Ae x = e x ⇒ A = 1 2 ,
so y p1 (x) = 1 2 ex .Fory 00 + y = x 3 try y p2 (x) =Ax 3 + Bx 2 + Cx + D. Theny 0 p 2
=3Ax 2 +2Bx + C and
y 00
p 2
=6Ax +2B. Substituting, we have 6Ax +2B + Ax 3 + Bx 2 + Cx + D = x 3 ,soA =1, B =0,
6A + C =0 ⇒ C = −6, and2B + D =0 ⇒ D =0. Thus y p2 (x) =x 3 − 6x and the general solution is
y(x) =y c(x)+y p1 (x)+y p2 (x) =c 1 cos x + c 2 sin x + 1 2 ex + x 3 − 6x. But2=y(0) = c 1 + 1 2
⇒
c 1 = 3 and 2 0=y0 (0) = c 2 + 1 − 6 ⇒ c 2 2 = 11 . Thus the solution to the initial-value problem is
2
y(x) = 3 2
cos x +
11
2 sin x + 1 2 ex + x 3 − 6x.
9. The auxiliary equation is r 2 − r =0with roots r =0, r =1so the complementary solution is y c(x) =c 1 + c 2e x .
Try y p (x) =x(Ax + B)e x so that no term in y p is a solution of the complementary equation. Then
y 0 p =(Ax 2 +(2A + B)x + B)e x and y 00
p =(Ax 2 +(4A + B)x +(2A +2B))e x . Substitution into the differential equation
gives (Ax 2 +(4A + B)x +(2A +2B))e x − (Ax 2 +(2A + B)x + B)e x = xe x ⇒ (2Ax +(2A + B))e x = xe x ⇒
A = 1 2 , B = −1. Thus yp(x) = 1
2 x2 − x e x and the general solution is y(x) =c 1 + c 2e x + 1
2 x2 − x e x .But
2=y(0) = c 1 + c 2 and 1=y 0 (0) = c 2 − 1, soc 2 =2and c 1 =0. The solution to the initial-value problem is
y(x) =2e x + 1
2 x2 − x e x = e x 1
2 x2 − x +2 .