Lista 4 - EDO

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Universidade Federal de Vic¸osa
Centro de Cieˆncias Exatas - CCE
Departamento de Matema´tica
4a Lista de MAT147 -Ca´lculo II
1. Determine a ordem de cada uma das equac¸o˜es diferenciais dadas e diga
tambe´m quando a equac¸a˜o e´ linear ou na˜o-linear:
(a) t3y′′′ + ty′ + y = cos(t)
(b) y′′ + tg(x+ y) = tg(x)
(c) (3 + x)y′ + 2xy = 3ex
2
(d) (2 + y2)
d2y
dx2
+ x
dy
dx
= senh(x)
(e) y′ + arctg2(x
√
y) = 3
2. Prove que y(x) = c1e
−3x + c2e3x +
1
6
xe3x − 1
10
sen(x) e´ uma soluc¸a˜o da
equac¸a˜o diferencial ordina´ria y′′ − 9y = e3x + sen(x).
3. Prove que y(t) = et
2
∫ t
0
e−s
2
ds+et
2
e´ uma soluc¸a˜o da EDO y′−2ty = 1.
4. Encontre a soluc¸a˜o geral das equac¸o˜es abaixo:
(a) x2dy − cossec(2y)dx = 0.
Resp: y(x) =
1
2
arccos
(
2 + cx
x
)
(b) (xy − 4x)dx+ (x2y + y) dy = 0.
Resp: (1 + x2)
−1
(y − 4)8e2y = c
(c) xtg(y)− y′sec(x) = 0.
Resp: xsen(x) + cos(x)− ln|sen(y)| = c
(d) y′ = cos2(x)cos2(2y).
Resp: 2x+ sen(2x)− 2tg(2y) = c
(e) y′ = x5ex
2
.
Resp: y(x) =
ex
2
2
(x4 − 2x2 + 2) + c
1
(f) y (1 + x3) y′ + x2 (1 + y) = 0.
Resp:
1
2
ln (1 + y2) +
1
3
ln (1 + x3) = c
(g) cotg(x)dy − (1 + y2) dx = 0.
Resp: arctg(y) + ln |cos(x)| = c
(h) cos(x)dy − (ysen(x) + e−x) dx = 0.
Resp: y(x) = − 1
excos(x)
+
c
cos(x)
(i) y2
√
1− x2y′ − arcsen(x) = 0.
Resp: y3 =
3 (arcsen(x))2
2
+ c
(j) y′ + ytg(x) = excos2(x).
Resp: y(x) = cos(x)
(
ex
2
(cos(x) + sen(x)) + c
)
.
(k) y′ − xy − x3y2 = 0.
Resp: y(x) = − e
x2
c+ ex2 (x2 − 2)
(l) xy′ + 2xy − y = 0, x > 0.
Resp: y2 =
5x
2 + 5cx5
(m) y′ + 2y = e2x.
Resp: y(x) =
1
4
e2x + ce−2x
(n) xy′ − 3y = x5.
Resp: y(x) =
1
2
x5 + cx3
(o) y′ + ycotg(x) = cossec(x).
Resp: y(x) = (x+ c)cossec(x)
(p) x2dy + (2xy − ex) dx = 0.
Resp: (1 + x2)
−1
(y − 4)8e2y = c
(q) (ysen(x)− 2)dx+ cos(x)dy = 0.
Resp: y(x) =
ex + c
x2
(r) (x+ 4)y′ + 5y = x2 + 8x+ 16.
Resp: y(x) =
(x+ 4)2
7
+
c
(x+ 4)5
2
(s) y′ + ytg(x) = sec(x) + 2xcos(x).
Resp: y(x) = sen(x) + (x+ c) cos(x).
5. Encontre a soluc¸a˜o particular da equac¸a˜o diferencial que satisfac¸a a
condic¸a˜o dada. Este tipo de problema e´ conhecido como Problema de
Valor Inicial (PVI) ou Problema com Valores de Contorno.
(a)
√
xy′ −√y = x√y, y(9) = 4.
Resp:
√
y −√x
(x
3
+ 1
)
= −10
(b) 2y2y′ = 3y − y′, y(3) = 1.
Resp: y2 + ln|y| = 3x− 8.
(c) sec(2y)dx− cos2(x)dy = 0, y
(pi
4
)
=
pi
6
.
Resp: 2tg(x)− sen(2y) = 2−
√
3
2
.
(d) xdy −√1− y2dx = 0, y(1) = 1
2
.
Resp: y(x) = sen
(
ln |x|+ pi
6
)
.
(e) y′ − y = 2xe2x, y(0) = 1.
Resp: y(x) = xe2x − e2x + 2ex.
(f) t3y′ + 4t2y = e−t, y(−1) = 0.
Resp: y(x) = − 1
t3
e−t − 1
t4
e−t.
(g) ty′ + 2y = sen(t), y
(pi
2
)
= 1.
Resp: y(x) =
sen(t)
t2
− cos(t)
t
+
pi2
4t2
− 1
t2
.
(h) y′ =
e−x − ex
3 + 4y
, y(0) = 1.
Resp: 3y + y2 = −e−x − ex + 6
(i) sen(2x)dx+ cos(3y)dy = 0, y
(pi
2
)
=
pi
3
.
Resp:
sen(3y)
3
=
cos(2x)
2
+
1
2
.
6. Resolva a equac¸a˜o de Bernoulli dada:
3
(a) xy′ + y = y−2.
Resp: y3 = 1 + cx−3
(b) y′ − y = exy2.
Resp:
1
y
= −1
2
ex + ce−x
(c) y′ = y(xy3 − 1).
Resp:
1
y3
= x+
1
3
+ ce3x
(d) xy′ − (1 + x)y = xy2.
Resp:
1
y
= −1 + 1
x
+
ce−x
x
(e) x2y′ + y2 = xy.
Resp: y(x) =
x
c+ ln(x)
(f) 3 (1 + x2) y′ = 2xy (y3 − 1).
Resp: y3 =
1
1 + c(1 + x2)
.
(g) y′ − xy − x3y2 = 0
Resp: y(x) = − e
x2
2
c+ e
x2
2 (x2 − 2)
(h) x2y′ + 2xy − y3 = 0.
Resp: y2 =
5x
2 + cx5
.
7. Encontre a soluc¸a˜o geral da EDO:
(a) y′ = αy − βy3, onde α e β sa˜o constantes positivas;
Resp: y2 =
α
β + cαe−2αx
(b) y′ + cos(t+ 1)y − sen(t)y3 = 0.
Resp:
1
y2
= −e2sen(t+1)
(∫
2sen(t)e−2sen(t+1)dt+ c
)
8. Mostre que (x+ y)a+b · (x− y)a−b = c e´ soluc¸a˜o da equac¸a˜o diferencial
(ax− by)dx+ (bx− ay)dy = 0, onde a, b ∈ Z.
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9. Resolva as equac¸o˜es sem o termo em y ou sem o termo em x:
(a) y′′ + y′ = e−x.
Resp: y(x) = c1e
−x + c2 − xe−x
(b) yy′′ − (y′)3 = 0.
Resp: y = c ou yln|y| − y + c1y + x = c2.
(c) 2x2y′′ + (y′)3 = 2xy′, x > 0.
Resp: y = c ou y = ±2
3
(x− 2c1)
√
x+ c1 + c2.
10. Resolva as seguintes equac¸o˜es diferenciais:
(a) y(1 + 2xy)dx+ x(1− 2y)dy = 0.
Resp: x = 2cye
1
2xy
(b) 2xyy′ + 2y2 = 3x− 6.
Resp: x2y = x3 − 3x+ c
(c) y′′ = 2x(y′)2.
Resp: y + c2 = − 1
c1
arctg
(
x
c1
)
(d) xy′ − y = x
3
y
ey/x.
Resp: y + x = x(c− x)e yx
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