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Lista de Exercı́cios 06 - EDO - 2015.2
1. Resolva a equação dada pelo método dos coeficientes indeterminados:
(a) y′′ + 3y′ + 2y = 6
(b) y′′ − 10y′ + 25y = 30x + 3
(c)
1
4
y′′ + y′ + y = x2 − 2x
(d) y′′ + 3y = −48x2e3x
(e) y′′ − y′ = −3
(f) y′′ − y′ +
1
4
y = 3 + ex/2
(g) y′′ + 4y = 3sen2x
(h) y′′ + y = 2xsenx
(i) y′′ − 2y′ + 5y = ex cos 2x
(j) y′′ + 2y′ + y = senx + 3 cos 2x
2. Resolva cada problema de valor inicial a seguir:
(a) y′′ + 4y = −2, y(π/8) = 1/2, y′(π/8) = 2
(b) 5y′′ + y′ = −6x, y(0) = 0, y′(0) = −10
(c) y′′ + 4y′ + 5y = 35e−4x, y(0) = −3, y′(0) = 1
(d) y′′ + y = cos x − sen2x, y(π/2) = 0, y′(π/2) = 0
3. Resolva a equação dada pelo método da variação de parâmetros, indicando um inter-
valo no qual a solução geral seja válida.
(a) y′′ + y = sec x
(b) y′′ + y = senx
(c) y′′ + y = cos2 x
(d) y′′ + 3y′ + 2y =
1
1 + ex
(e) y′′ + 3y′ + 2y = sen(ex)
(f) y′′ − 2y′ + y =
e−x
1 + x2
(g) y′′ − 2y′ + y = e−x ln x
(h) 3y′′ − 6y′ + 30y = extg3x
4. Resolva a equação 4y′′ − y = xex/2 sujeita às condições y(0) = 1, y′(0) = 0.
5. Considere a equação x2y′′ − xy′ + y = 4x ln x
(a) Verifique que y1(x) = x e y2(x) = x ln x formam um conjunto fundamental de
soluções para a equação homogênea associada, no intervalo (0,∞).
(b) Encontre a sua solução geral.
1
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Gabarito
1. (a) y = c1e−x + c2e−2x + 3
(b) y = c1e5x + c2xe5x +
6
5
x +
3
5
(c) y = c1e−2x + c2xe−2x + x2 − 4x +
7
2
(d) y = c1 cos
√
3x + c2sen
√
3x + (−4x2 + 4x − 43 )e
3x
(e) y = c1 + c2ex + 3x
(f) y = c1ex/2 + c2xex/2 + 12 +
1
2
x2ex/2
(g) y = c1 cos 2x + c2sen2x − 34x. cos 2x
(h) y = c1 cos x + c2senx − 12x
2 cos x + 12xsenx
(i) y = c1ex cos 2x + c2exsen2x + 14xe
xsen2x
(j) y = c1e−x + c2xe−x − 12 cos x +
12
25sen2x −
9
25 cos 2x
2. (a) y =
√
2sen2x −
1
2
(b) y = −200 + 200e−x/5 − 3x2 + 30x
(c) y = −10e−2x cos x + 9e−2xsenx + 7e−4x
(d) y = − 16 cos x −
π
4 senx +
1
2x.senx +
1
3sen2x
3. (a) y = c1 cos x + c2senx + x.senx + (cos x). ln | cos x|, I = (−π/2, π/2)
(b) y = c1 cos x + c2senx − 12x cos x, I = (−∞,∞)
(c) y = c1 cos x + c2senx + 12 −
1
6 cos 2x, I = (−∞,∞)
(d) y = c1e−x + c2e−2x + (e−x + e−2x). ln(1 + ex), I = (−∞,∞)
(e) y = c1e−2x + c2e−x − e−2xsenex , I = (−∞,∞)
(f) y = c1ex + c2xex − 12e
x ln(1 + x2) + xextg−1x, I = (−∞,∞)
(g) y = c1e−x + c2xe−x − 12x
2e−x ln x − 34x
2e−x, I = (0,∞)
(h) y = c1ex cos 3x + c2exsenx − 127e
x(cos 3x) ln | sec 3x + tg3x|, I = (−π/6, π/6)
4. y = 14e
−x/2 + 34e
x/2 + 18x
2ex/2 − 14xe
x/2
5. (b) y = c1x + c2x ln x + 23x(ln x)
3
2