Lista_Revisão

Prévia do material em texto

Lista de Revisa˜o para a Avaliac¸a˜o Regimental - Prof. Renato Cruz
1) Resolva as seguintes equac¸o˜es separa´veis:
a) 4xy2dx + (x2 + 1)dy = 0 Resp: 2 ln (x2 + 1)− 1y = K
b) xydx− 3(y − 2)dy = 0 Resp: 6y − x2 = 12 ln (Ky)
c) xdx + y.e−x2dy = 0 Resp: y2 + ex2 = K2
d) (2 + y)dx− (3− x)dy = 0 Resp: (2 + y).(3− x) = K
e) xydx− (1 + x2)dy = 0 Resp: Ky2 = 1 + x2
f) (1− x)dy − y2dx = 0 Resp: y lnK(1− x) = 1
g)
dy
dx
=
e−2y
x2 + 4
Resp: e2y − arctan
(x
2
)
= K
2) Resolva as seguintes equac¸oes homogeˆneas:
a) 2xydx + (y2 − x2)dy = 0 Resp: y = K(y2 + x2)
b) (x2 − 3y2)dx + 2xydy = 0 Resp: y = x√Kx + 1
c) 2x(x + y)dx + (x2 + y2)dy = 0 Resp: 2x3 + 3x2y + y3
d) xdy − ydx =
√
x2 + y2 Resp: y = Kx2 −
√
x2 − y2
e) (x2 − xy + y2)dx− xydy = 0 Resp: ln (y − x) + yx = K
f) ydx + (2
√
xy − x)dy = 0 Resp:
√
x
y
+ ln y = K
3) Resolva as seguintes equac¸o˜es pelo me´todo do fator integrante:
a)
dy
dx
− 5y = e3x Resp: −2y = e3x + Ke5x
b)
dy
dx
+
3y
x
= x3 − 2 Resp: 14x3y = 2x7 − 7x4 + K
c)
dy
dx
+ 2xy = e3x(3 + 2x) Resp: ex
2y = e3x+x
2
+ K
d) x2
dy
dx
+ 2xy = x4 − 7 Resp: 5x2y = x5 − 35x + K
e)
dy
dx
+ 2yx = 5x Resp: yex
2
= 52e
x2 + K
4) Resolva as seguintes equac¸o˜es de 2a ordem com coeficientes na˜o constantes:
a) (1otipo)
d2y
dx2
= e2x + cos 2x Resp: y = 14e
2x − 14 cos 2x + K1x + K2
b) (2otipo) (−3− 4x)d
2y
dx2
+
dy
dx
= 0 Resp: y = K15
4
√
(3 + 4x)5 + K2
c) (3otipo)
d2y
dx2
− 4 =
(
dy
dx
)2
Resp: y = ln [sec (2x + 2K1)] + K2
c) (4otipo) y2
d2y
dx2
= y
dy
dx
− y
(
dy
dx
)2
Resp: y −K1 ln (y + K1) = x + K2
1