Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 9 REVIEW ¤ 403
3. (a) We estimate that when x =0.3, y =0.8,soy(0.3) ≈ 0.8.
(b) h =0.1, x 0 =0, y 0 =1and F (x, y) =x 2 − y 2 .Soy n = y n−1 +0.1 x 2 n−1 − y 2 n−1
.Thus,
y 1 =1+0.1 0 2 − 1 2 =0.9, y 2 =0.9+0.1 0.1 2 − 0.9 2 =0.82, y 3 =0.82 + 0.1 0.2 2 − 0.82 2 =0.75676.
This is close to our graphical estimate of y(0.3) ≈ 0.8.
(c) The centers of the horizontal line segments of the direction field are located on the lines y = x and y = −x.
When a solution curve crosses one of these lines, it has a local maximum or minimum.
5. y 0 = xe − sin x − y cos x ⇒ y 0 +(cosx) y = xe − sin x (). This is a linear equation and the integrating factor is
I(x) =e cos xdx = e sin x . Multiplying ()bye sin x gives e sin x y 0 + e sin x (cos x) y = x ⇒ (e sin x y) 0 = x ⇒
e sin x y = 1 2 x2 + C ⇒ y = 1
2 x2 + C e − sin x .
7. 2ye y2 y 0 =2x +3 √ y2 dy
x ⇒ 2ye
dx =2x +3√ x ⇒ 2ye y2 dy =
9.
2x +3 √
x dx
2ye
y 2 dy = 2x +3 √ x
dx ⇒ e y2 = x 2 +2x 3/2 + C ⇒ y 2 =ln(x 2 +2x 3/2 + C) ⇒
y = ± ln(x 2 +2x 3/2 + C)
dr
dr
dr
+2tr = r ⇒
dt dt = r − 2tr = r(1 − 2t) ⇒ r = (1 − 2t) dt ⇒ ln |r| = t − t 2 + C ⇒
|r| = e t−t2 +C = ke t−t2 .Sincer(0) = 5, 5=ke 0 = k. Thus,r(t) =5e t−t2 .
11. xy 0 − y = x ln x ⇒ y 0 − 1 x y =lnx. I(x) =e −1
(−1/x) dx = e − ln|x| = e ln|x| = |x| −1 =1/x since the condition
y(1) = 2 implies that we want a solution with x>0. Multiplying the last differential equation by I(x) gives
1
x y0 − 1 x 2 y = 1 x ln x ⇒ 1
x y 0
= 1 x ln x ⇒ 1 x y = ln x
x dx ⇒ 1 x y = 1 2 (ln x)2 + C ⇒
y = 1 2 x(ln x)2 + Cx. Nowy(1) = 2 ⇒ 2=0+C ⇒ C =2,soy = 1 2 x(ln x)2 +2x.
⇒
13.
d
dx (y) = d
dx (kex ) ⇒ y 0 = ke x = y, so the orthogonal trajectories must have y 0 = − 1 y
⇒
dy
dx = − 1 y
⇒
ydy= −dx ⇒ ydy= − dx ⇒ 1 2 y2 = −x + C ⇒ x = C − 1 2 y2 , which are parabolas with a horizontal axis.