Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 9.5 LINEAR EQUATIONS ¤ 397
9. Since P (x) is the derivative of the coefficient of y 0 [P (x) =1and the coefficient is x], we can write the differential equation
xy 0 + y = √ x in the easily integrable form (xy) 0 = √ x ⇒ xy = 2 3 x3/2 + C ⇒ y = 2 3
√ x + C/x.
11. sin x dy
dx +(cosx) y =sin(x2 ) ⇒ [(sin x) y] 0 =sin(x 2 ) ⇒ (sin x) y =
sin(x 2
sin(x 2 ) dx + C
) dx ⇒ y =
.
sin x
13. (1 + t) du
du
+ u =1+t, t>0 [divide by 1+t] ⇒
dt dt + 1 u =1 (), which has the
1+t
form u 0 + P (t) u = Q(t). The integrating factor is I(t) =e P (t) dt = e [1/(1+t)] dt = e ln(1+t) =1+t.
Multiplying () byI(t) gives us our original equation back. We rewrite it as [(1 + t)u] 0 =1+t. Thus,
(1 + t)u = (1 + t) dt = t + 1 2 t2 + C ⇒ u = t + 1 2 t2 + C
1+t
or u = t2 +2t +2C
.
2(t +1)
15. y 0 = x + y ⇒ y 0 +(−1)y = x. I(x) =e (−1) dx = e −x . Multiplying by e −x gives e −x y 0 − e −x y = xe −x ⇒
(e −x y) 0 = xe −x ⇒ e −x y = xe −x dx = −xe −x − e −x + C [integration by parts with u = x, dv = e −x dx] ⇒
y = −x − 1+Ce x . y(0) = 2 ⇒ −1+C =2 ⇒ C =3,soy = −x − 1+3e x .
17.
dv
dt − 2tv =3t2 e t2 , v (0) = 5.
−t2 dv
e
dt − 2te−t2 v =3t 2
⇒
I(t) =e (−2t) dt = e −t2 . Multiply the differential equation by I(t) to get
0
e −t2 v =3t
2
⇒ e −t2 v = 3t 2 dt = t 3 + C ⇒ v = t 3 e t2 + Ce t2 .
5=v(0) = 0 · 1+C · 1=C,sov = t 3 e t2 +5e t2 .
19. xy 0 = y + x 2 sin x ⇒ y 0 − 1 x y = x sin x. I(x) =e (−1/x) dx = e − ln x = e ln x−1 = 1 x .
Multiplying by 1 x gives 1 x y0 − 1 0 1
x y =sinx ⇒ 2 x y =sinx ⇒ 1 y = − cos x + C ⇒ y = −x cos x + Cx.
x
y(π) =0 ⇒ −π · (−1) + Cπ =0 ⇒ C = −1,soy = −x cos x − x.
21. xy 0 +2y = e x ⇒ y 0 + 2 x y = ex x .
I(x) =e (2/x) dx = e 2ln|x| =
e ln|x| 2
= |x| 2 = x 2 .
Multiplying by I(x) gives x 2 y 0 +2xy = xe x ⇒ (x 2 y) 0 = xe x ⇒
x 2 y = xe x dx =(x − 1)e x + C [by parts] ⇒
y =[(x − 1)e x + C]/x 2 . The graphs for C = −5, −3, −1, 1, 3, 5,and7 are
shown. C =1is a transitional value. For C1, there is a local minimum. As |C| gets larger, the “branches” get
further from the origin.