Lista_de_EDO

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E.D.O – Lista de Exercícios
Equações diferenciais de 1
a
 ordem 
Exercícios 
a) -2xy′ + 2y = 0 ; y = c ⋅e
b) y 0 ; y ax bx c2′ = = + +′
xx
′
− = = ⋅ + ⋅ −′ −
2
′ = = +
f) 2
2y
g) 
2-x
y
x
y 2 2′ = − + =
i) 2x x 2x
j) ( )
�
�
	
=
= ⋅ −
′
−
′ + =
4
x
y
2
2
2
1
2
y′ = = ⋅ ; y c x
x
y′ + 2xy = 0 ; y = c ⋅ e 
h) ; x y c
y′ − y = e ; y = c ⋅ e + e
1) Em cada caso, verificar que a função dada constitui uma solução da equação:
e) y 2x ; y x c
′
d) y y x ; y c e c e x
′
c) y ′ + y = 0 ; y = a ⋅ cos(x )+ b ⋅ sen(x )
y c x c
y xy y 0 ; 
′k) y ′ + y = 0 ; y = cos(x)
l) ( )
( )
( )
( )
�
�
	
= −
= +
=
′ =
5
4
3
2
1
m) 
�
�
	
= − ⋅
= ⋅
=
′
− =
x
3
x
2
x
1
e
5
y 2 e
y e
n) 
�
�
	
= ⋅ + ⋅
=
=
′
− ⋅ ′ + =⋅ ′
3
2
2
3 1
3
2
2
1
2
y c x c x
y x
y x
y cos x ; 
y sen x
y sen x 3
y sen x
6
y
y y 0 ; 
x y 4x y 6y 0 ; 
correspondente e determinar as constantes de modo que a solução particular 
satisfaça a condição dada: 
a) y y 0 ; y c e ; y(0) 3x′ + = = ⋅ =− R: xy 3 e−
x
′ + = = ⋅ + =− R: y e 51 x= +−
2x
′ + = = ⋅ = −− R: 
2xy 2 e−
2 2y = 3 ⋅ x
e) ( )��
	
′ =
= −y 1 8
0 ; y c x c ; 
dx
dy
dx
d y
x 2
2
12
2
 R: y 2x 102= −
( )
�
�
	
′ =
=
+ = = ⋅ +
3
2
3
a
2
3
2
2
pi
pi
 R: �
�
�
�
�
�
= ⋅ +2
y
2) Em cada caso, verificar que a função dada é solução da equação diferencial
− = = ⋅ +
y 1 4
dx
dy
x
2y
= = ⋅ = R: 
pi
y 2 cos x
6
y
y 0 ; y a cos x b ; f) ( )d y
dx
= − ⋅
d) ; y c x ; y(1) 3
c) y 2xy 0 ; y c e ; y(0) 2
= ⋅
b) y y 5 ; y c e 5 ; y(1) 6
1
Equações de Variáveis Separáveis
b) (1 x )xy1 ydxdy 2
2
+
+ kx
y
2
2
2
−=
sen(x )y =
2 2
⋅ + ⋅ = R: tg y = k ⋅cotg x
dx
dy
2y xy
dx
dy
a x � =
�
�
�
�
� 2a ay = ln kx ⋅ y
3) Determine, se possível, a solução geral das seguintes equações diferenciais:
 a) (x −1)dy − ydx = 0 R: y = k(x −1)
= R: 1
x 1
k
e
⋅ + R: ( )
d) sec (x) tg(y) dx sec (y) tg(x) dy 0
dy
+ ⋅ = R: c) y cos(x ) 0
dx
e) 
2 3 2 3 1
x
1
2
1
y
x
ln
2 2
=�
�
�
�
�
�
�
�
− +��
�
�
��
�
�
2 2 2 2 2 2 2 2+ + + − − = 
 R: c
b
y
y 2b arctg
x a
x a
x ln
a
� =
�
�
�
�
�
� + − ⋅
�
�
�
�
�
+
−
h) ( ) 0dytg y1
1
ln x 1
22 + − =
− = R: x ⋅cos(y) = k 
g) (x a )(y b )dx (x a )(y b )dy 0
y
f) (1 x )y dx (1 y )x dy 0+ + − = R: k
+
2 2+ + = R: ( ) c
y
i) 4xy dx (x 1)dy 0
2
− − = = R: 
x11
1
y
−
x
− = = R: y 2e 12 x
2
y = x +1 
2 + − = = R: y
1 y
1 x e
−
3
�
�
�
�
�
�
�
�
−
=y ln
2
1 y dx 12 2
9(x 1)
x 1
−
=
+
1 2 3− + = = R: 
�
�
�
�
�
�
�
�
−
= ⋅
−
+ 2
2
x
1 x
3 e
y 1
h) 1 y dx 1 x dy 0 ; y(1) 12 2
1 y dx 12 2+ + + = = R: ( ) arctg(x)arctg y = −
2 y
3
2
4) Resolva os seguintes problemas de valor inicial (PVI): 
a) (y y )dx dy 0 ; y(0) 2 =
− e
2
= −
− =
3x
2 x 1
+ − = = R: e) dx (x x)dy 0 ; y(2) ln( 3)
d) y dx (x 1)dy 0 ; y(0) 1
c) ydx − xdy = 0 ; y(1) = 4 R: ( )
b) e dx ydy 0 ; y(0) 1
−y 1
y 1
− + − = = R: f) ( ) ( x )dy 0 ; y(2) 2
g) ( y )dx x dy 0 ; y(1) 2 y 1
−
=
7 x 6
x
+ + − = = R: j) (x 3)ydx (6x x )dy 0 ; y(7) 1
− + − = = R: arccos(x)+ arccos(y) = 0 
i) ( ) ( x )dy 0 ; y(1) 1
2
Equações Diferenciais Exatas 
2 2
− + − + =
b) ( ) ( ) dy 0
y
1
dx x cos xy 2 x
x
y
y cos xy =
�
�
�
�
�
+ ⋅ + +
�
�
�
�
R: sen xy + 2y x + ln y = C
c) dy 0
y
y 3x
dx
4
2 2
3
=
− 1
y
x
3
2
− =
2 2 2 3 3 2 2 4
2 2
22 2+ − =
2t sen y y e t cos y 3y e3 t 2 2 t⋅ + ⋅ + ⋅ + ⋅ = 
2 3 t
y sec t sec t tg t 2y tg t2 2
2 2 2 2y x y
dx
⋅ +
+ = 2 2+ + =
k) 
y x y
x xy
dx
dy
2
2
+
+ 2 2 2 2
R: x 2y sen x k2
2 2
5) Determine, se possível, a solução geral das seguintes equações diferenciais:
 a) (2x − y +1)dx − (x + 3y − 2)dy = 0 R: 2x 2xy 2x 3y 4y k
�
⋅ + 
2x
y
+ R: C
y
+ + =
e) 
j) 
dy
+
+ + =
d) (3x 6xy )dx (6x y 4y )dy 0
f) (1+ y ⋅sen(x))dx + (1− cos(x))dy = 0 R: x − y ⋅cos(x)+ y = C 
g) (sec(t)⋅ tg(t)− w )dt + (sec(w )⋅ tg(w )− t + 2)dw = 0 
dy
dt
 
 R: sec(t)− wt + sec(w )+ 2w = k 
h) ( ) ( ( ) ) 0
+ ⋅ + =⋅ + ⋅ + + = R: y y tg(t) sec(t) C
= − R: x x y y k
− ⋅ =
n) sec (x) tg(y) dx sec (y) tg(x) dy 0
l) (x − 2y)dx − 2xdy = 0 R: x ⋅ (x − 4y) = C 
m) (x − y ⋅cos(x))dx − sen(x)dy = 0 
⋅ + ⋅ = R: tg(x)⋅ tg(y) = C 
⋅ + ⋅ =
i) ( ) ( ) ( ) ( ( )) 0
 R: x x y k
x y
xdy
y
 R: t sen(y) y e C
dy
dt
+ + + = R: x 3x y y C
+ = R: (x y ) 4xy kxdx ydy
+
xdy+ydx 
 x y
3 2
3
y 2
y
ee
y
2
3 2
− + = R: 
4
λ =
6) Determine os fatores integrantes para as seguintes equações:
− + = R: 
+ − = R: 
1
x
1λ = ⋅a) (x y x )dx xdy 0
b) ydx (ye x y )dy 0
x
3e
x
λ =
d) (x 2xy)dx x dy 0
c) (y ⋅cos(x)− tg(x))dx − sen(x)dy = 0 R: cossec (x)
1λ = ⋅
3
2 2
2 2
=
+
b) y dy ydx xdy2 2
y 3 2 3+ =
2 2
2 3 2 2+ + + + = R: ye y 3x C3x 2 2
2x 2x x
− = ⋅ +
��
�
�
��
�
7) Determine, se possível, a solução geral das seguintes equações diferenciais: 
dy
dx
x
dx =
y
2 2 2
x
+ =
+ − =
e) (3x y 2xy y )dx (x y )dy 0
2y+ − =− 2y
x x+ + ⋅ = R: e sen y y Kx 2
4+ − = R: 9y x 1 ln x Cx4 3
2 2+ − = R: 2 2 3
4 3 4+ + + − = R: 3 2
3 x 2
3
+ =�
�
� + R: x y e K
3
2 3 x+ =
y y x y x y x+ =+
�
�
�
g) sen(y) dy 0
x y
− + = R: C
+ − = R: ln(x ) y ky
f) e y 1
+ − =
k) 2xydx (2y 3x )dy 0 x − 2y = Ky
l) (y 2y)dx (xy 2y 4x)dy 0
a) (x y )dx 2xydy 0
+ = R: y x Cy
c) dx (y ln(x))dy 0
d) (x x y )dx xydy 0
n) (e xe tg(e ))dx xe dy 0
h) ydx (2xy e )dy 0
i) e dx (e cotg(y) 2y cossec(y))dy 0
− = R: xe ln(y) c
+ =
j) (y x ln(x ))dx xdy 0
�
+ − R: xy + y ⋅ cos(y)− sen(y) = k 
= + − R: y e k e 1
+ − − = R: x (3x 4x 6y ) C
+ =
y x + y + 2x = cy
m) 2y 3xe dx 3xy dy 0
+ + + = R: xe ln(sec(e )) C
xy
1 1
x 2
2
2
( ) ( ) xxx = =
�
�
�
�
�
� +
+��
�
�
�
λ
x + ⋅ =
x 2 x 2
− =
multiplicadas pelo fator integrante dado ao lado. Portanto, resolva as 
equações: 
8) Mostre que as equações abaixo não são exatas mas tornam-se exatas quando 
�
−
 R: e sen(y) 2y cos(x) k
�
�
dy 0 ; x, y ye
y
−
−
− + =
y
x y dx + x 1+ y dy = 0 ; x, y =2 3 2
y
�
�
cos y 2e cos x
2e sen x dx
a) ( ) ( )
3
λ R: ln(y ) C
(2x cos(y) e )dx x sen(y)dy 0⋅ − − = R: e x cos(y) 19) Achar a solução particular para x = 0 na equação: 
sen(y)
b) 
dy
2ty 3t y3 2 2 y t
−
dy 3 2 2+ + =
c) ; y(1) 1
3x 4y 2
2 2
− + + − =
− +
=
− +
dy
dx
2x 3y 5
2 2
= R: x 3xy 5x 2y 2y 3
=
2
3+ = = R: a) 0 ; y(1) 1
dt
+ + + = = R: t 2t y y 13t 4ty 2y 2tb) ( ) 0 ; y(0) 1
10) Resolver os seguintes problemas de valor inicial (PVI):
4
Equações Lineares 
dy
− ⋅ = R: ( )
�
�
�
�
�
�
= ⋅ + Cy sec x
2
1 x2 y arctg x 1 k e−
y = ⋅ +
�
�
�
�
�
= ⋅ + sen 2x + C
4
1x
2
f) 2 2
g) 3
−
2 2
= −
xy x 1 k e−
j) y sen(x) sen x cos xy −+ ⋅
k) 
4x3 2e
1
4y
dx
dy
�
�
= ⋅ + +−y e ln 3 2e C8
1
4x 4x
l) ( ) yx ln y y
dy
dx y yx y 1 k e−= ⋅ + ⋅
2 2 xdy y x 2
2
=
2y 2+ = ⋅ ⋅− 2y= ⋅ +−
dx
dy
y sec x 
x Cy
+ = R: 
h) y dx (2xy 3)dy 0
dy
x − = R: 
dy
i) y x
dx
+ = R: 
+
dy
dx
2xy x e
dx
1
x 1
x +1 + = ⋅ R: [e (x 2x 2) C]
+
− ⋅ = R: ( )
�
+ = R: ( ) ��
= k e
2
( ) ( ) x
dy
= − + ⋅
+ = R: 
y = Cx + x
e) y tg(x) cos(x)
dx
y x
dx
dy
dx
2y
x
x
�
�
sen (x)
2
dy
dx
+ − = 0 R: [ln(sen(x)) C]
1
= ⋅ + R: ( ) ( ) �
1
x
x
y
4
2
− + = R: 
y
= +
1
Cx
6
= − + ⋅
b) (x + sen(y)−1)dy − cos(y)dx = 0 
 R: x = [sec(y)+ tg(y)]⋅[2sec(y)− 2tg(y)+ y + C]
c) ( ) y arctg(x)+ + = R: ( ) arctg (x )
d) 
dx
y
x
cotg x
x
( )
m) ( )
n) dx 2xdy e sec (y) dy R: x e [tg(y) C]
⋅ ⋅ − + +
11)Determine, se possível, a solução geral das seguintes equações diferenciais: 
a) y tg(x) sen(x)
dy
dy
x x x a
1
y = ⋅ + −
2) Achar a solução particular para y = 0 e x = 0 na equação: 
3) Achar a solução particular para y = b e x = a na equação: 
y tg(x) sec(x)
dx
− ⋅ = R: y = x ⋅sec(x)
e ab e
x
⋅ + − = R: ( )y e 0
dx
5