lista 2 calc 4 EDO 1a ordem parte1

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IBMEC - ECONOMIA
Disciplina: Ca´lculo III Profa.: Cla´udia
SEGUNDA LISTA DE EXERCI´CIOS
1) Resolva as seguintes equac¸o˜es de 1a ordem:
(a) y′ + 3y = x+ e−2x
(b) (x2 + 9)
dy
dx
+ xy = 0
(c) (t2 − xt2)dx
dt
+ x2 + tx2 = 0
(d) 2xy =
dy
dx
(e) sen θ cosφ dθ − cos θ senφ dφ = 0
(f) y′ = 1 + x+ y2 + xy2
(g) y′ − 3y = sen 2x
(h) y′ =
2y4 + x4
xy3
(i) x y′ = 2
√
y − 1
(j) y′ = −2y + cos 2x
(k) x2y′ + 2xy − y3 = 0
(l) xdy − ydx = x2exdx
(m) x ln y
dy
dx
= y
(n)
dy
dx
=
ex
y2
(o)
dy
dx
=
x+ y
x
(p) eyy˙ − t− t3 = 0
(q)
dx
dt
= (x+ 2)x
(r)
dx
dy
=
1 + x2
1 + y2
(s) (x2 + 1)
√
yy′ = xe3x/2 + (1 + x2)y
√
y
(t) y′ − 3y = 6
(u)
dy
dx
+
2y
x
= x3
2) Resolva os seguintes problemas de valor inicial:
(a)
{
y′ − 2xy = x
y(0) = 1
(b)
{
(x2 + 1)y′ + y2 + 1 = 0
y(0) = 1
(c)
{
y˙ + 2y = te−2t
y(1) = 0
(d)
{
y′ + y tg x = 2senx cosx
y(0) = 1, 0 ≤ x ≤ pi/2
(e)
{
xy′ + 2y = 4x2
y(1) = 2
(f)
 (x2 + 1)
dy
dx
+ 3x(y − 1) = 0
y(
√
3) = 2
(g)
{
cotg x dy − (1 + y2)dx = 0
y(0) = 1
(h)
{
(x+ 4)y′ + 5y = x2 + 8x+ 16
y(−3) = 8/7
3) Encontre a soluc¸a˜o do problema de valor inicial dado, em forma expl´ıcita, e determine o intervalo
no qual a soluc¸a˜o e´ va´lida:
(a)

dy
dx
=
2x
y + x2y
y(0) = −2
(b)

dy
dx
=
2x
1 + 2y
y(2) = 0
(c)

dy
dx
=
x(x2 + 1)
4y3
y(0) = −1/√2
(d)
{
ex dx− y dy = 0
y(0) = 1
(e)
 y4e2x +
dy
dx
= 0, y < 0
y(0) = −1
4) Sabendo que a func¸a˜o y = y(t), t > 0, satisfaz a equac¸a˜o diferencial y˙ +
1
1 + t
y =
1
1 + t
e que
y(0) = 3, calcule y(1).
1
RESPOSTAS
1)
(a) y =
x
3
− 1
9
+ e−2x + ce−3x
(b) y =
c√
x2 + 9
(c)
t+ x
tx
+ ln |x
t
| = c
(d) y = cex
2
(e) cosφ = c cos θ
(f) y = tg(x+ x
2
2
+ c)
(g) y = ce3x − 3
13
sen 2x− 2
13
cos 2x
(h) y4 = cx8 − x4
(i) y = (ln |x|+ c)2 + 1
(j) y = ce−2x +
1
4
(cos 2x+ sen2x)
(k) y =
(
2
5x
+ cx4
)−1/2
, y ≡ 0
(l) y = cx+ xex
(m) y = e
√
2 ln |x|+c
(n) y = 3
√
3ex + c
(o) y = (c+ ln |x|)|x|
(p) y = ln
(
t2
2
+
t4
4
+ c
)
(q) x =
2ce2t
1− ce2t
(r) x =
y + c
1− cy
(s) y =
[
3
4
ln(1 + x2)e
3x
2 + ce
3x
2
]2/3
(t) y = ce3x − 2
(u) y =
1
6
x4 +
c
x2
2)
(a) y = −1
2
+
3
2
ex
2
(b) y =
1− x
1 + x
(c) y =
t2
2
e−2t − 1
2
e−2t
(d) y = 3 cos x− 2 cos2 x
(e) y = x2 +
1
x2
(f) y = 1 +
8
(x2 + 1)3/2
(g) arctg y − ln |secx| = pi
4
(h) y =
(x+ 4)2
7
+
1
(x+ 4)5
3)
(a) y = −[2 ln(1 + x2) + 4]1/2 ; (−∞,+∞)
(b) y = −1
2
+
1
2
√
4x2 − 15 ;
(
1
2
√
15,+∞
)
(c) y = −
√
x2 + 1
2
; (−∞,+∞)
(d) y =
√
2ex − 1 ; [ln 1/2,+∞)
(e) y =
(
2
3e2x − 5
)1/3
; (−∞, ln
√
5/3)
4) 6
2