EDOS Lineales No Homogéneas: Metodo del Anulador
Alumno:
Andres Garcia Cruz
Numero de control: 181M0284
Grupo: 4.1 M
Materia: Ecuaciones Diferenciales
Ing. Hugo Alberto Carillo Serrano
ˆ
1
1. 9y 00 − 4y = Senx
Solucionar la parte Homogenea
9y 00 − 4y = 0
9D2 − 4 = 0
9m2 − 4 = 0
9m2 = 4
√
p
m2 = 4/9
m1 = +2/3 m2 = −2/3
y(h) = C1 e2/3x + C2 e−2/3x
V olvemos a la ecuacion original
[D2 + 1][9D2 − 4] = [Senx][D2 + 1]
[D2 + 1][9D2 − 4] = 0
m2 + 1 = 0
m2 = 1
m1 = 1i
m2 = −1i
y = ACosx + BSenx + C1 e2/3x + C2 e−2/3x
y(p) = ACosx + BSenx
y 0 (p) = ASenx − BCosx
y 00 (p) = −ACosx − BSenx
Sustituyendo los valores
9(−ACosx − BSenx) − 4(ACosx + BSenx) = Senx
2
−9ACosx − 9BSenx − 4ACosx − 4BSenx = Senx
Sumo terminos semejantes
−13ACosx − 13BSenx = Senx
−13ACosx = 0 − 13BSenx = Senx
A = 0 − 13B = 1
B = −1/13
Respuesta
y = −1/13Senx + +C3 e2/3x + C4 e−2/3x
3
3. y 00 − 4y 0 − 12y = x − 6
Comenzamos haciendo la parte homogenea
y 00 − 4y 0 − 12y = 0
D2 − 4D − 12 = 0
m2 − 4m − 12 = 0
(m + 2)(m − 6) = 0
m+2=0 m−6=0
m1 = −2 m2 = 6
V olvemos a la ecuacion original
y(h) = C1 e−2x + C2 e6x
y 00 − 4y 0 − 12y = x − 6
[D2 ][D2 − 2D − 12] = [X − 6][D2 ]
D4 − 4D3 − 12D2 = 0
m2 (m2 − 4m − 12) = 0
m1 = 0 m2 = 0
y = C1 + C2 xC3 e−2x + C4 e6x
y(p) = C1 + C2 x
y 0 (p)C2
y 00 = 0
Sustituimos
−4(C2 − 12(C1 + C2 X) = X − 6
−4C2 − 12C1 − 12C2 X = X − 6
−4C2 − 12C1 = −6 − 12C2 = X
4
−4(−1/12) − 12C1 = −6 C2 = −1/12
19/3
−12
C1 = 19/36
−12C1 = −
Respuesta
Y = 19/36 − 1/12x + C3 e−2x + C4 e6x
5
5. y 000 + 10y 00 + 25y 0 = ex
Comenzamos haciendo la parte homogenea
D3 + 10D2 + 25D = 0
m3 + 10m2 + 25m = 0
m(m + 5)(m + 5) = 0
m1 = 0, m2 = −5, m3 = −5
y = C1 + C2 e−5x + C3 xe−5x
[D − 1][D3 + 10D2 + 25D] = ex [D − 1]
(m − 1)(m3 + 10m2 + 25m) = 0
(m − 1) = 0, m1 = 1
(m3 + 10m2 + 25m) = 0
m(m + 5)(m + 5) = 0
m2 = 0, m3 = −5, m4 = −5
y = Aex + B + Ce−5x + De−5x
yp = Aex
yp0 = Aex
yp00 = Aex
yp000 = Aex
Sustituimos
x
Ae + 10Aex + 25Aex = ex
36Aex = ex
36A = 1, A = 1/36
Resultado
y = C1 + C2 e−5x + C3 xe−5 + 1/36ex
6
7. y 000 + 2y 00 − 13y 0 + 10y = xe−x
Comenzamos haciendo la parte homogenea
y(h) = D3 + 2D2 − 13D + 10 = 0
m3 + 2m2 − 13m + 10 = 0
m1 = 1, m2 = 2, m3 = −5
y(h) = C1 ex + C2 e2x + C3 e−5x
[D + 1]2 [D3 + 2D2 − 13D + 10] = xe−x [D + 1]2
[D2 + 2D + 1]2 [D3 + 2D2 − 13D + 10] = 0
(m2 + 2m + 1) = 0, (m + 1)(m + 1) = 0
m1 = −1, m2 = −1
(m3 + 2m2 − 13m + 10) = 0
m1 , m4 = 2, m5 = −5
y(p) = Ae−x + Bxe−x
y 0 (p) = Ae−x − Bxe−x − Bxe−x
y 00 (p) = Ae−x + Bxe−x − Be−x − De−x
y 000 (p) = −Ae−x − Bxe−x + Be−x + Be−x
Sustituimos
−Ae−x − Bxe−x + Be−x + Be−x
+2(Ae−x + Bxe−x − Be−x − De−x ) − 13(Ae−x − Bxe−x − Bxe−x )
+10(Ae−x + Bxe−x ) = xe−x
24Ae−x + 24Bxe−x − 14Be−x = xe−x
24Bxe−x = xe−x
24B = 1, B = 1/24
7
24A − 14B = 0, 24A − 14(1/24) = 0, A = 7/288
Resultado
y = C1 ex + C2 e2x + C3 e−5x + 1/24xe−x + 7/288e−x
8
9. y IV + 8y 0 = 4
Comenzamos haciendo la parte homogenea
D4 + 8D = 0
m4 + 8m = 0
m(m3 + 8) = 0, m1 = 0
(m3 + m2 + m + 8) = 0
m2 = −2
m2 − 2m + 4m =
aplicamos f ormula general
√
∝= 1, β 3
y(h) = C1 C2 e−2x + ex (ACossqrt3 + BSensqrt3)
V olvemos a la ecuacion original
[D][D4 + 8D] = 4[D]
[D5 + 8D2 ] = 0
m5 + 8m2 = 0
m2 (m3 + 8) = 0
(m3 + m2 + m + 8) = 0
m2 = −2
m2 − 2m + 4 = 0
√
∝= 1, beta = 3
√
√
y = A + Bx + Ce−2x + ex (DCos 3 + ESen 3)
y(p) = Bx
y 0 (p) = B
y 00 (p) = 0
9
y 000 (p) = 0
y IV (p) = 0
Resultado
√
√
y = C1 + C2 e−2x + ex (DCos 3 + ESen 3) + 1/2x
10