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lista4 Calc III EDO 2a ordem, Exercícios de Engenharia Civil

Exercícios com resposta de EDO 2ª ordem

Tipologia: Exercícios

2011

Compartilhado em 20/11/2011

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UGF
Disciplina: C´
alculo III Profa.: Cl´
audia
QUARTA LISTA DE EXERC´
ICIOS
1) Resolva as equa¸oes:
(a) y00 8y0+ 7y= 14
(b) y00 +y=x3+ 6
(c) y00 + 2y0+y= 4ex
(d) y00 + 3y05y= 0
(e) y00 7y0+ 12y=e4x
(f) y00 +y06y=xe2x
(g) y00 +y= cos x
(h) y00 y=x2ex
(i) y00 + 4y=xsen 2x
(j) y00 2y0+ 10y= sen 3x+ex
(k) y00 y= 2xsen x
(l) y00 3y0=x+ cos x
(m) y00 3y0+ 2y=et+e2t
(n) y00 +y0+ 4y=exex
(o) y00 y02y= (e2x+e2x)/2
(p) y00 4y0+ 4y= 2e2x
(q) 12y00 5y02y= 0
R:y=c1e7x+c2ex+ 2
R:y=c1cos x+c2senx+x36x+ 6
R:y=c1ex+c2xex+ex
R:y=c1e(3+29)x/2+c2e(329)x/2
R:y=c1e3x+c2e4xxe4x
R:y=c1e2x+c2e3x+xe2xx
10
1
25
R:y=c1cos x+c2senx+x
2senx
R:y=c1ex+c2ex+1
4
1
4x+1
6x2xex
R:y=c1cos 2x+c2sen2x+1
16x(sen2x2xcos 2x)
R:y=ex(c1cos 3x+c2sen3x) + 1
9ex+1
37(sen3x+ 6 cos 3x)
R:y=c1ex+c2exxsenxcos x
R:y=c1+c2e3x
1
10 cos x
3
10senx
1
6x2
1
9x
R:y=c1et+c2e2ttet+te2t
R:y=ex/2(c1cos 15
2x+c2sen15
2x) + 1
6ex
1
4ex
R:y=c1e2x+c2ex+1
8e2x+1
6xe2x
R:y=c1e2x+c2xe2x+x2e2x
R:y=c1e2x/3+c2ex/4
2) Resolva os problemas de valor inicial:
(a) x00 + 6x0+ 9x=t+ 2
x(0) = 0, x0(0) = 0 R: x(t) =
4
27
5
9te3t+t
9+4
27
(b) y00 3y0+ 2y= 0
y(1) = 0, y0(1) = 1 R: y(t) = e2t2et1
(c) x00 + 2x0+ 2x= sen2t+ cos 2t
x(0) = 0, x0(0) = 1 R: x(t) = et3
10 cos t+11
10sen t+1
10sen 2t
3
10 cos 2t
(d) y00 + 16y= 0
y(0) = 2, y0(0) = 2R: y(t) = 2 cos 4t
sen 4t
2
(e) 2x00 4x0+ 2x= 0
x(0) = 1, x0(0) = 0 R: x(t) = ettet
(f) 5y00 +y0=6t
y(0) = 0, y0(0) = 10 R: y(t) = 200 + 200et/53t2+ 30t
1
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UGF Disciplina: C´alculo III Profa.: Cl´audia

QUARTA LISTA DE EXERC´ICIOS

  1. Resolva as equa¸c˜oes:

(a) y′′^ − 8 y′^ + 7y = 14 (b) y′′^ + y = x^3 + 6 (c) y′′^ + 2y′^ + y = 4ex (d) y′′^ + 3y′^ − 5 y = 0 (e) y′′^ − 7 y′^ + 12y = −e^4 x (f) y′′^ + y′^ − 6 y = xe^2 x (g) y′′^ + y = cos x (h) y′′^ − y = x^2 ex (i) y′′^ + 4y = xsen 2x (j) y′′^ − 2 y′^ + 10y = sen 3x + ex (k) y′′^ − y = 2xsen x (l) y′′^ − 3 y′^ = x + cos x (m) y′′^ − 3 y′^ + 2y = et^ + e^2 t (n) y′′^ + y′^ + 4y = ex^ − e−x (o) y′′^ − y′^ − 2 y = (e^2 x^ + e−^2 x)/ 2 (p) y′′^ − 4 y′^ + 4y = 2e^2 x (q) 12y′′^ − 5 y′^ − 2 y = 0

R : y = c 1 e^7 x^ + c 2 ex^ + 2 R : y = c 1 cos x + c 2 senx + x^3 − 6 x + 6 R : y = c 1 e−x^ + c 2 xe−x^ + ex R : y = c 1 e(−3+

√29)x/ 2

  • c 2 e(−^3 −

√29)x/ 2

R : y = c 1 e^3 x^ + c 2 e^4 x^ − xe^4 x

R : y = c 1 e^2 x^ + c 2 e−^3 x^ + xe^2 x

( (^) x 10

)

R : y = c 1 cos x + c 2 senx + x 2 senx

R : y = c 1 e−x^ + c 2 ex^ +

x +

x^2

) xex

R : y = c 1 cos 2x + c 2 sen2x +

x(sen2x − 2 x cos 2x)

R : y = ex(c 1 cos 3x + c 2 sen3x) +

ex^ +

(sen3x + 6 cos 3x) R : y = c 1 ex^ + c 2 e−x^ − xsenx − cos x R : y = c 1 + c 2 e^3 x^ −

cos x −

senx −

x^2 −

x R : y = c 1 et^ + c 2 e^2 t^ − tet^ + te^2 t

R : y = e−x/^2 (c 1 cos

√ 15 2 x^ +^ c^2 sen

√ 15 2 x) +

ex^ −

e−x

R : y = c 1 e^2 x^ + c 2 e−x^ +

e−^2 x^ +

xe^2 x R : y = c 1 e^2 x^ + c 2 xe^2 x^ + x^2 e^2 x R : y = c 1 e^2 x/^3 + c 2 e−x/^4

  1. Resolva os problemas de valor inicial:

(a)

{ (^) x′′ (^) + 6x′ (^) + 9x = t + 2 x(0) = 0, x′(0) = 0 R:^ x(t) =

( −

t

) e−^3 t^ + t 9

(b)

{ (^) y′′ (^) − 3 y′ (^) + 2y = 0 y(1) = 0, y′(1) = 1 R:^ y(t) =^ e

2 t− (^2) − et− 1

(c)

{ (^) x′′ (^) + 2x′ (^) + 2x = sen2t + cos 2t x(0) = 0, x′(0) = 1 R: x(t) = e−t

cos t +

sen t

)

sen 2t −

cos 2t

(d)

{ (^) y′′ (^) + 16y = 0 y(0) = 2, y′(0) = − 2 R:^ y(t) = 2 cos 4t^ −^

sen 4t 2

(e)

{ (^2) x′′ (^) − 4 x′ (^) + 2x = 0 x(0) = 1, x′(0) = 0 R:^ x(t) =^ e

t (^) − tet

(f)

{ 5 y′′^ + y′^ = − 6 t y(0) = 0, y′(0) = − 10 R:^ y(t) =^ −200 + 200e

−t/ (^5) − 3 t (^2) + 30t

  1. Resolva as equa¸c˜oes usando o m´etodo de varia¸c˜ao dos parˆametros:

(a) y′′^ − 2 y′^ + y = ex x^5 R: y = c 1 ex^ + c 2 xex^ + 121 x−^3 ex

(b) y′′^ + 6y′^ + 9y = e−^3 x x^3

R: y =

( (^1) 2 x +^ c^1 +^ c^2 x

) e−^3 x

(c) y′′^ + y = tg x , 0 < x < π 2 R: y = c 1 cos x + c 2 senx − cos x ln(tgx + sec x)

(d) y′′^ − 3 y′^ + 2y = xex^ R: y = c 1 ex^ + c 2 e^2 x^ − 1 − x − x^2 2 e^2 x

(e) y′′^ = ln x, x > 0 R: y = c 1 + c 2 x +

x^2 ln x −

x^2

  1. Se o Wronskiano de f (x) e g(x) for 3e^4 x^ e se f (x) = e^2 x, ache g(x). R: g(x) = 3xe^2 x^ + ce^2 x