Solução_Calculo_Stewart_6e

F.
5. (1 + tan y) y 0 = x 2 +1 ⇒ (1 + tan y) dy
dx = x2 +1
1 − − sin y
cos y
7.
9.
⇒
1+ sin y
cos y
dy = (x 2 +1)dx ⇒ y − ln |cos y| = 1 3 x3 + x + C.
Note: The left side is equivalent to y +ln|sec y|.
dy
dt =
te t
y 1+y 2 ⇒ y 1+y 2 dy = te t dt ⇒ y 1+y 2 dy = te t dt ⇒ 1 3
SECTION 9.3 SEPARABLE EQUATIONS ¤ 387
dy =(x 2 +1)dx ⇒
1+y
2 3/2 = te t − e t + C
[where the first integral is evaluated by substitution and the second by parts] ⇒ 1+y 2 =[3(te t − e t + C)] 2/3 ⇒
y = ± [3(te t − e t + C)] 2/3 − 1
du
dt
TX.10
=2+2u + t + tu ⇒
du
dt =(1+u)(2 + t) ⇒
du
1+u = (2 + t)dt [u 6=−1] ⇒
ln |1+u| = 1 2 t2 +2t + C ⇒ |1+u| = e t2 /2+2t + C = Ke t2 /2+2t ,whereK = e C ⇒ 1+u = ±Ke t2 /2+2t
⇒
u = −1 ± Ke t2 /2+2t where K>0. u = −1 is also a solution, so u = −1+Ae t2 /2+2t ,whereA is an arbitrary constant.
11.
dy
dx = x y
⇒ ydy= xdx ⇒ ydy= xdx ⇒ 1 2 y2 = 1 2 x2 + C. y(0) = −3 ⇒
1
2 (−3)2 = 1 2 (0)2 + C ⇒ C = 9 ,so 1 2 2 y2 = 1 2 x2 + 9 ⇒ y 2 = x 2 +9 ⇒ y = − √ x
2 2 +9since y(0) = −3 < 0.
13. x cos x =(2y + e 3y ) y 0 ⇒ x cos xdx =(2y + e 3y ) dy ⇒ (2y + e 3y ) dy = x cos xdx ⇒
y 2 + 1 3 e3y = x sin x +cosx + C
[where the second integral is evaluated using integration by parts].
15.
Now y(0) = 0 ⇒ 0+ 1 3 =0+1+C ⇒ C = − 2 3 . Thus, a solution is y2 + 1 3 e3y = x sin x +cosx − 2 3 .
We cannot solve explicitly for y.
du
dt = 2t +sec2 t
, u(0) = −5.
2u
2udu=
2t +sec 2 t dt ⇒ u 2 = t 2 +tant + C,
where [u(0)] 2 =0 2 +tan0+C ⇒ C =(−5) 2 =25. Therefore, u 2 = t 2 +tant +25,sou = ± √ t 2 +tant +25.
Since u(0) = −5,wemusthaveu = − √ t 2 +tant +25.
17. y 0 tan x = a + y, 0