Solução_Calculo_Stewart_6e

F.
386 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
TX.10
(c) The exact value of y(1) is 2+e −13 =2+e −1 .
(i) For h =1: (exact value) − (approximate value) =2+e −1 − 3 ≈−0.6321
(ii) For h =0.1: (exact value) − (approximate value) =2+e −1 − 2.3928 ≈−0.0249
(iii) For h =0.01: (exact value) − (approximate value) =2+e −1 − 2.3701 ≈−0.0022
(iv) For h =0.001: (exact value) − (approximate value) =2+e −1 − 2.3681 ≈−0.0002
In (ii)–(iv), it seems that when the step size is divided by 10, the error estimate is also divided by 10 (approximately).
27. (a) R dQ
dt + 1 C Q = E(t) becomes 5Q0 + 1
0.05 Q =60
or Q 0 +4Q =12.
(b) From the graph, it appears that the limiting value of the
charge Q is about 3.
(c) If Q 0 =0,then4Q =12 ⇒ Q =3is an
equilibrium solution.
(d)
(e) Q 0 +4Q =12 ⇒ Q 0 =12− 4Q. NowQ(0) = 0,sot 0 =0and Q 0 =0.
Q 1 = Q 0 + hF (t 0 ,Q 0 )=0+0.1(12 − 4 · 0) = 1.2
Q 2 = Q 1 + hF (t 1,Q 1)=1.2+0.1(12 − 4 · 1.2) = 1.92
Q 3 = Q 2 + hF (t 2 ,Q 2 )=1.92 + 0.1(12 − 4 · 1.92) = 2.352
Q 4 = Q 3 + hF (t 3 ,Q 3 )=2.352 + 0.1(12 − 4 · 2.352) = 2.6112
Q 5 = Q 4 + hF (t 4,Q 4)=2.6112 + 0.1(12 − 4 · 2.6112) = 2.76672
Thus, Q 5 = Q(0.5) ≈ 2.77 C.
9.3 Separable Equations
1.
dy
dx = y x ⇒ dy
y = dx dy dx
x [y 6= 0] ⇒ y
= x
⇒ ln |y| =ln|x| + C ⇒
|y| = e ln|x|+C = e ln|x| e C = e C |x| ⇒ y = Kx,whereK = ±e C is a constant. (In our derivation, K was nonzero,
but we can restore the excluded case y =0by allowing K to be zero.)
3. (x 2 +1)y 0 = xy ⇒ dy
dx = xy
x 2 +1
⇒
dy
y = xdx
x 2 +1
[y 6= 0] ⇒
dy
y =
xdx
x 2 +1
ln |y| = 1 2 ln(x2 +1)+C [u = x 2 +1, du =2xdx] =ln(x 2 +1) 1/2 +lne C =ln e C√ x 2 +1 ⇒
|y| = e C√ x 2 +1 ⇒ y = K √ x 2 +1,whereK = ±e C is a constant. (In our derivation, K was nonzero, but we can
restore the excluded case y =0by allowing K to be zero.)
⇒