Cálculo III (Lista VII)

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Lista VII (Cálculo III) 
 
 
1. Determine a solução geral da equação diferencial homogênea. 
 
a. 2 0y y′′′ ′′+ = b. 2 2 0y y y y′′′ ′′ ′− − + = 
resp: ( ) 21 2 3xy x c e c c x−= + + resp: ( ) 21 2 3x x xy x c e c e c e−= + + 
 
c. 2 2 0y y y′′′ ′′ ′+ + = d. 2 4 8 0y y y y′′′ ′′ ′− + − = 
resp: ( ) ( ) ( )( )1 2 3cos senxy x c e c x c x−= + + resp: ( ) ( ) ( )21 2 3cos 2 sen 2xy x c e c x c x= + + 
 
e. ( ) 2 0
iv
y y y′′+ + = 
resp: ( ) ( ) ( ) ( ) ( )( )1 2 3 4cos sen cos seny x c x c x x c x c x= + + + 
 
f. ( ) 5 6 4 8 0
iv
y y y y y′′′ ′′ ′− + + − = 
resp: ( ) 2 2 2 21 2 3 4x x x xy x c e c e c xe c x e−= + + + 
 
g. ( ) 16 0
iv
y y+ = 
resp: ( ) ( ) ( )( ) ( ) ( )( )2 21 2 3 4cos 2 sen 2 cos 2 sen 2x xy x e c x c x e c x c x−= + + + 
 
h. ( ) 2 2 2 0
iv
y y y y y′′′ ′′ ′− + + − = 
resp: ( ) ( ) ( )( )1 2 3 4cos senx x xy x c e c e e c x c x−= + + + 
 
i. ( ) 13 36 0
iv
y y y′′− + = j. 4 5 0y y y′′′ ′′ ′− + = 
resp: ( ) 3 2 2 31 2 3 4x x x xy x c e c e c e c e− −= + + + resp: ( ) ( ) ( )( )21 2 3cos senxy x c e c x c x= + + 
 
l. 0y y y y′′′ ′′ ′− − + = m. ( ) 0ivy y′′− = 
resp: ( ) 1 2 3x x xy x c e c e c xe−= + + resp: ( ) 1 2 3 4x xy x c e c c x c e−= + + + 
 
n. ( ) 8 0
iv
y y′− = 
resp: ( ) ( ) ( )( )21 2 3 4cos 3 sen 3x xy x c c e e c x c x−= + + + 
 
o. ( ) 0
vi
y y′′− = 
resp: ( ) ( ) ( )1 2 3 4 5 6cos senx xy x c e c c x c e c x c x−= + + + + + 
 
2. Determine a solução do problema de valor inicial dado em forma explícita. 
 
a. 0y y′′′ ′+ = , ( )0 0y = , ( )0 1y′ = , ( )0 2y′′ = 
resp: ( ) ( ) ( )2 2cos seny x x x= − + 
 
b. ( ) 4 4 0
iv
y y y′′′ ′′− + = , ( )1 1y = − , ( )1 2y′ = , ( )1 0y′′ = , ( )1 0y′′′ = 
resp: ( ) 2 3y x x= − 
c. 0y y y y′′′ ′′ ′− + − = , ( )0 2y = , ( )0 1y′ = − , ( )0 2y′′ = − 
resp: ( ) ( ) ( )2cos seny x x x= − 
 
d. 6 5 0y y y′′′ ′′ ′+ + = , ( )0 2y = − , ( )0 2y′ = , ( )0 0y′′ = 
resp: ( ) 3 28 18 8
x x
y x e e
− −= − + 
 
e. ( ) 0
iv
y y+ = , ( )0 0y = , ( )0 0y′ = , ( )0 1y′′ = − , ( )0 0y′′′ = 
resp: ( )
2 2
2 2
1 2 1 2
sen sen
2 2 2 2
x x
y x e x e x
−    
= −      
   
 
 
f. 4 5 0y y y′′′ ′+ + = , ( )0 2y = , ( )0 1y′ = , ( )0 1y′′ = − 
resp: ( ) ( ) ( )2 22 24 3cos sen
13 13 13
x x
x
y x e e x e x
−= + + 
 
3. Ache a solução geral da equação diferencial proposta. 
 
a. 3 24 2 0x y x y y′′′ ′′+ − = , 0x > b. 3 2 2 2 0x y x y xy y′′′ ′′ ′+ − + = , 0x > 
resp: ( ) 2 21 2 3
c
y x c x c x
x
−= + + resp: ( ) 21 2 3
c
y x c x c x
x
= + + 
 
c. 3 26 4 4 0x y x y xy y′′′ ′′ ′+ + − = , 0x > 
resp: ( ) ( )( ) 21 2 3lny x c c x x c x−= + + 
 
d. 3 6 0x y y′′′ − = , 0x > 
resp: ( ) ( )( ) ( )( )31 2 3cos 2 ln sen 2 lny x c x c x c x= + + 
 
4. Use o método da variação de parâmetros para determinar a solução geral da equação 
diferencial proposta. 
 
a. ( )tgy y x′′′ ′+ = , 0 x π< < 
resp: ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )1 2 3cos sen ln cos sen ln sec tgy x c c x c x x x x x= + + − − + 
 
b. y y x′′′ ′− = c. 42 2 xy y y y e′′′ ′′ ′− − + = 
resp: ( ) 21 2 3
1
2
x x
y x c c e c e x
−= + + − resp: ( ) 2 41 2 3
1
30
x x x x
y x c e c e c e e
−= + + + 
 
d. ( )senxy y y y e x−′′′ ′′ ′− + − = 
resp: ( ) ( ) ( ) ( )1 2 3
1
cos sen cos
5
x x
y x c e c x c x e x
−= + + − 
 
e. 
( ) ( )2 senivy y y x′′+ + = 
resp: ( ) ( ) ( ) ( ) ( ) ( )21 2 3 4
1
cos sen cos sen sen
8
y x c x c x c x x c x x x x= + + + − 
5. Use o método dos coeficientes indeterminados para determinar a solução geral da equação 
diferencial proposta. 
 
a. 2 3xy y y y e−′′′ ′′ ′− − + = + 
 resp: ( ) ( )1 2 3
1
3
2
x x x
y x c e e c c x xe
− −= + + + + 
 
b. 
( ) ( )3 cosivy y x x− = + 
resp: ( ) ( ) ( ) ( )1 2 3 4
1
cos sen 3 sen
4
x x
y x c e c e c x c x x x x
−= + + + − − 
 
c. 4xy y y y e x−′′′ ′′ ′+ + + = + 
resp: ( ) ( ) ( ) ( )1 2 3
1
cos sen 4 1
2
x x
y x c e c x c x xe x
− −= + + + + − 
 
d. ( )2seny y x′′′ ′− = 
resp: ( ) ( )1 2 3 cosx xy x c c e c e x−= + + + 
 
e. 
( ) ( )2 3 cos 2ivy y y x′′+ + = + 
resp: ( ) ( ) ( ) ( ) ( ) ( )1 2 3 4
1
cos sen cos sen 3 cos 2
9
y x c x c x c x x c x x x= + + + + + 
 
f. 
( ) ( )sen 2ivy y x′′′+ = 
resp: ( ) ( ) ( )21 2 3 4
1 1
cos 2 sen 2
40 20
x
y x c c x c x c e x x
−= + + + + + 
 
6. Determine a solução do problema de valor inicial dado em forma explícita. 
 
a. 4y y t′′′ ′+ = , ( )0 0y = , ( )0 0y′ = , ( )0 1y′′ = 
resp: ( ) ( )( ) 23 11 cos 2
16 8
y t t t= − + 
 
b. ( ) 2 3 4
iv
y y y t′′+ + = + , ( )0 0y = , ( )0 0y′ = , ( )0 1y′′ = , ( )0 1y′′′ = 
resp: ( ) ( ) ( ) ( )34 cos 4 sen 3 4
2
y t t t t t t
 = − − + + + 
 
 
 
c. 3 2 ty y y t e′′′ ′′ ′− + = + , ( )0 1y = , ( ) 10
4
y′ = − , ( ) 30
2
y′′ = − 
resp: ( ) ( )211 3
4
t
y t t t te= + + −