Solução_Calculo_Stewart_6e

F.
388 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
TX.10
21. u = x + y ⇒ d
dx (u) = d
du
(x + y) ⇒
dx
du
1+u = dx [u 6=−1] ⇒
du
1+u =
dx =1+dy dx
dy
du
,but = x + y = u, so
dx dx =1+u
dx ⇒ ln |1+u| = x + C ⇒ |1+u| = e x+C ⇒
1+u = ±e C e x ⇒ u = ±e C e x − 1 ⇒ x + y = ±e C e x − 1 ⇒ y = Ke x − x − 1, whereK = ±e C 6=0.
If u = −1,then−1 =x + y ⇒ y = −x − 1,whichisjusty = Ke x − x − 1 with K =0. Thus, the general solution
⇒
is y = Ke x − x − 1,whereK ∈ R.
23. (a) y 0 =2x 1 − y 2 ⇒ dy
dx =2x 1 − y 2
⇒
dy
=2xdx ⇒ 1 − y
2
dy
= 1 − y
2
2xdx
⇒
sin −1 y = x 2 + C for − π 2 ≤ x2 + C ≤ π 2 .
(b) y(0) = 0 ⇒ sin −1 0=0 2 + C ⇒ C =0,
so sin −1 y = x 2 and y =sin x 2 for
− π/2 ≤ x ≤ π/2.
25.
(c) For 1 − y 2 to be a real number, we must have −1 ≤ y ≤ 1;thatis,−1 ≤ y(0) ≤ 1. Thus, the initial-value problem
y 0 =2x 1 − y 2 , y(0) = 2 does not have a solution.
dy
dx = sin x
sin y , y(0) = π 2 . So sin ydy= sin xdx ⇔
− cos y = − cos x + C ⇔ cos y =cosx − C. From the initial condition,
we need cos π 2
=cos0− C ⇒ 0=1− C ⇒ C =1, so the solution is
cos y =cosx − 1. Note that we cannot take cos −1 of both sides, since that would
unnecessarily restrict the solution to the case where −1 ≤ cos x − 1 ⇔ 0 ≤ cos x,
as cos −1 is defined only on [−1, 1]. Instead we plot the graph using Maple’s
plots[implicitplot] or Mathematica’s Plot[Evaluate[···]].
27. (a)
x y y 0 =1/y
0 0.5 2
0 −0.5 −2
0 1 1
0 −1 −1
0 2 0.5
x y y 0 =1/y
0 −2 −0.5
0 4 0.25
0 3 0.3
0 0.25 4
0 0.3 3