MM1 19 Hoja2 sol - Problemas propuestos y resueltos de EDOs Y Variable compleja PDF

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Problemas propuestos y resueltos de EDOs Y Variable compleja...

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´ ´ I. GRUPOS D,E METODOS MATEMATICOS CURSO 2018-2019. HOJA 2. SOLUCIONES

1. Encontrar la soluci´on general (real) de las siguientes ecuaciones: a) x′′ + 2x′ + x = 0; b) x′′ + x′ = 0; c) x′′ + 2x′ + 2x = 0; d) x′′′ + 3x′′ + 3x′ + x = 0; e) x′′′ − x = 0; f ) x′′ − x = e−t ; g) x′′′ − x = e−t ; h) x′′′ + x = sen t; i) x′′′ + x = et . Soluci´ on: a) p(λ) = λ2 + 2λ + 1 = 0,

x(t) = (c1 + c2 t)e−t

λ = −1,

b) p(λ) = λ2 + λ = 0,

x(t) = c1 + c2 e−t

λ = 0, −1,

c) p(λ) = λ2 + 2λ + 2 = 0,

λ = −1 ± i,

x(t) = e−t (c1 cos t + c2 sen t)

d) p(λ) = λ3 + 3λ2 + 3λ + 1 = 0, e) p(λ) = λ3 − 1 = 0,

x(t) = (c1 + c2 t + c3 t2 )e−t

λ = −1,

√ 1 3 λ = 1, − ± , 2 2

x(t) = c1 et + e−t/2

√ √ ! 3 3 t c2 cos t + c3 sen 2 2

f) p(λ) = λ2 − 1 = 0,

λ = ±1,

xh (t) = c1 et + c2 e−t ,

1 A=− , 2

1 xp (t) = − te−t , 2

p(λ) = λ − 1 = 0, −Ae

−t

− Ae

−t

=e

−t

,

t

xh (t) = c1 e + e

1 A=− , 2

h) p(λ) = λ3 + 1 = 0,

√ 3 1 , λ = −1, ± 2 2

√ √ ! 3 3 c2 cos t + c3 sen t , 2 2

−t/2

1 xp (t) = − e−t , 2

A(t − 2)e−t − Ate−t = e−t

1 x(t) = c1 et + c2 e−t − te−t 2

g) 3

xp (t) = Ate−t ,

t

x(t) = c1 e + e

xh (t) = c1 e

−t

+e

−t/2

t/2

xp (t) = Ae−t

√ ! √ 1 3 3 t + c3 sen c2 cos t − e−t 2 2 2 √ √ ! 3 3 c2 cos t + c3 sen t 2 2

xp (t) = A cos t + B sen t, 1 A=B = , 2

A sen t − B cos t + A cos t + B sen t = sen t, A + B = 1, A − B = 0 √ ! √ 1 1 3 3 −t t/2 xp (t) = (cos t + sen t), x(t) = c1 e + e t + c3 sen c2 cos t + (cos t + sen t) 2 2 2 2

i) p(λ) = λ3 + 1 = 0,

2A = 1,

xh (t) = c1 e−t + et/2

A=

1 , 2

1 xp = e t , 2

√ √ ! 3 3 t , c2 cos t + c3 sen 2 2 x(t) = c1 e

−t

1

+e

t/2

xp (t) = Aet ,

Aet + Aet = et

√ √ ! 1 t 3 3 c2 cos t + c3 sen t + e 2 2 2

´ ´ I. GRUPOS D,E CURSO 2018-2019. HOJA 2. SOLUCIONES METODOS MATEMATICOS

2

2. Hallar la soluci´on general de las ecuaciones: a) x′′ − x = e2t ; b) x′′ + x = tet cos t; c) x′′′ + 2x′′ + 5x′ = 5t; d) x(4) + 4x = tet cos t; e) x′′ + x = cos3 t; f ) t2 x′′ − 3tx′ + 3x = 9 log t; g) (t + 1)x′′ − x′ = (t + 1)2 ; h) t2 x′′ − t(t + 2)x′ + (t + 2)x = t3 . Soluci´ on: a) p(λ) = λ2 − 1 = 0,

λ = ±1, xh (t) = c1 et + c2 e−t , xp (t) = Ae2t , 4Ae2t − Ae2t = e2t 1 1 1 A = , xp (t) = e2t , x(t) = c1 et + c2 e−t + e2t 3 3 3

b) p(λ) = λ2 + 1 = 0,

λ = ±i,

xh (t) = c1 cos t + c2 sen t,

xp (t) = et ((At + B) cos t + (Ct + D) sen t)

et [((A + 2C)t + 2A + B + 2C + 2D) cos t + ((−2A + C)t − 2A − 2B + 2C + D) sen t] = tet cos t A + 2C = 1, 2A + B + 2C + 2D = 0, −2A + C = 0, −2A − 2B + 2C + D = 0 A=

14 2 2 1 , B=− , C = , D=− , 25 5 25 5 x(t) = c1 cos t + c2 sen t +

xp (t) =

1 t e [(5t − 2) cos t + 2 (5t − 7) sen t] 25

1 t e [(5t − 2) cos t + 2 (5t − 7) sen t] 25

c) p(λ) = λ3 + 2λ2 + 5λ = 0, λ = 0, −1 ± 2i, xh (t) = c1 + e−t (c2 cos 2t + c3 sen 2t), xp (t) = t(At + B) 1 1 1 2 4A + 10At + 5B = 5t, A = , B = − , xp (t) = t(5t − 4), x(t) = c1 + e−t (c2 cos 2t + c3 sen 2t) + t(5t − 4) 10 5 2 10 d) p(λ) = λ4 + 4 = 0,

λ = ±1 ± i,

xh (t) = c1 e(1+i)t + c2 e(1−i)t + c3 e(−1+i)t + c4 e(−1−i)t

xp (t) = t(At + B)e(1+i)t ,

8ie(1+i)t (2(1 + i)At + 3A + (1 + i)B) = te(1+i)t 3 1+i , B= 16(−1 + i)A = 1, 3A + (1 + i)B = 0, A = − 32 32 tet t (t sen t + (3 − t) cos t − i((t − 3) sen t + t cos t)) (−(1 + i)t + 3)e(1+i)t = xp (t) = 32 32 tet x(t) = et (c1 cos t + c2 sen t) + e−t (c3 cos t + c4 sen t) + (t sen t + (3 − t) cos t) 32 e) 1 cos3 t = (cos 3t + 3 cos t) 4 xp (t) = A cos 3t + B sen 3t + Ct cos t + Dt sen t 1 3 1 −8A cos 3t − 8B sen 3t − 2C sen t + 2D cos t = (cos 3t + 3 cos t), A = − , B = 0, C = 0, D = 8 32 4 3 1 3 1 cos 3t + t sen t cos 3t + t sen t, x(t) = c1 cos t + c1 sen t − xp (t) = − 32 8 32 8 p(λ) = λ2 + 1 = 0,

λ = ±2i,

xh (t) = c1 cos t + c1 sen t,

f) p(λ) = r(r − 1) − 3r + 3 = 0,

r = 1, 3,

xh (t) = c1 t + c2 t3 ,

xp (t) = A log t + B

−A − 3A + 3A log t + 3B = 9 log t, A = 3, B = 4, xp (t) = 3 log t + 4, x(t) = c1 t + c2 t3 + 3 log t + 4 1 1 1 ˙ x′′ = − 2 x˙ + 2 x ¨, x ¨ − 4x˙ + 3x = 9s, λ2 − 4λ + 3 = 0, λ = 1, 3, xh (s) = c1 es + c2 e3s t = es , x′ = x, t t t xp (s) = As + B, −4A + 3(As + B) = 9s, A = 3, B = 4, xp (s) = 3s + 4, x(s) = c1 es + c2 e3s + 3s + 4 x(t) = c1 t + c2 t3 + 3 log t + 4

g) ′

x = v,

2

R

dt t+1

 Z R c1 + (t + 1)e−

dt t+1



dt = (t + 1)(c1 + t) (t + 1)v − v = (t + 1) , v = e Z Z 1 1 x = vdt + c2 = (t + 1)(c1 + t)dt + c2 = t2 (2t + 3) + t(t + 2)c1 + c2 6 2 ′

h) Una soluci´on particular de la homog´enea es x1 (t) = t. x(t) = tu(t),

t2 (tu′′ + 2u′ ) − t(t + 2)(tu′ + u) + (t + 2)tu = t3 ,

u′′ − u′ = 1,

up (t) = −t

´ ´ I. GRUPOS D,E METODOS MATEMATICOS

CURSO 2018-2019. HOJA 2. SOLUCIONES

u(t) = c1 + c2 et − t,

3

x(t) = c1 t + c2 tet − t2

3. Estudiar existencia, unicidad y hallar la soluci´on general de: a) x′′ − x = 2(1 + et )−1 ; b) t2 x′′ − 2x = t2 ; c) t2 x′′ + 5tx′ + 4x = t−2 ; d) (1 + t2 )x′′ + 2tx′ = 2t−3 Soluci´ on: a) Z x1 (s)x2 (t) − x1 (t)x2 (s) b(s)ds x(t) = c1 et + c2 e−t + xp (t), xp (t) = W [x1 (s), x2 (s)]  Z  t Z Z es−t − et−s es e−s e e−t t −t = −2, x (t) = − ds ds + e ds = −e W [x1 , x2 ] = det et −e−t p s s 1 + es 1+e 1+e  Z Z Z  Z 1 es 1 e−s t ds = − dy = − 1 − ds = log(1 + e ), dy = log(1 + e−t ) − e−t 1 + es 1 + 1/y 1 + es 1+y xp (t) = −e−t log(1 + et ) + et (log(1 + e−t ) − e−t ) = et log(1 + e−t ) − e−t log(1 + et ) − 1 x(t) = c1 et + c2 e−t + et log(1 + e−t ) − e−t log(1 + et ) − 1

b) r(r − 1) − 2 = 0,

r = −1, 2,

At2 (2 log t + 4 − 1) − 2At2 log t = t2 ,

x(t) = A=

1 , 3

c1 + c2 t2 + xp (t), xp (t) = At2 log t t 1 c1 1 + c2 t2 + t2 log t xp (t) = t2 log t, x(t) = 3 t 3

c) r(r − 1) + 5r + 4 = 0, 2At−2 = t−2 ,

A=

1 , 2

r = −2,

xp = At−2 (log t)2 1 x(t) = c1 t−2 + c2 t−2 log t + t−2 (log t)2 2

x(t) = c1 t−2 + c2 t−2 log t + xp (t),

1 xp = t−2 (log t)2 , 2

d) x(t) =

Z

˜c1 − t dt + c2 = 1 + t2 −2

Z

((1 + t2 )x′ )′ = 2t−3 , (1 + t2 )x′ = −t−2 + ˜c1 Z 1 ˜c1 + 1 dt dt − + c2 = (˜c1 + 1) arc tg t + + c2 , t 1 + t2 t2

x(t) = c1 arc tg t + c2 +

1 t

4. Hallar la soluci´on general sabiendo que la x1 que se indica es soluci´on de la homog´enea: ( ( ( tx′′ − (t + 1)x′ + x = t2 et t2 x′′ − 2tx′ + (2 − t2 )x = 2t3 cosh t (t2 − 1)x′′ − 2x = 0 a) : ; c) ; b) : 2 t x1 = t senh t x1 = t − 1 x1 = e Soluci´ on: a) x(t) = et u(t), et (t(u + 2u′ + u′′) − (t + 1)(u + u′ ) + u) = t2 et , u′ = v, tv ′ + (t − 1)v = t2     Z R Z R t−1 t−1 1 c1 (t + 1)e−t + c2 + t2 v(t) = e− t dt ˜c1 + te t dt dt = te−t ˜c1 + et dt = c˜1 te−t + t, u(t) = −˜ 2 1 x(t) = c1 (t + 1) + c2 et + t2 et 2 b) x(t) = (t2 − 1)u(t), v(t) = ˜c1 e

c)

−

R

(t2 − 1)(2u + 4tu′ + (t2 − 1)u′′)) − 2(t2 − 1)u = 0, u′ = v, (t2 − 1)v′ + 4tv = 0   4t 2 c˜1 2t 1 t+1 dx t2 −1 = c˜1 e−2 log(t −1) = 2 − , u(t) = c ˜ log + c2 1 t−1 (t − 1)2 4 t2 − 1     t + 1   − 2t + c2 x(t) = c1 (t2 − 1) log  t − 1

t3 (sinh(t)u′′(t) + 2 cosh(t)u′ (t)) = 2t3 cosh t, u′ = v, v′ sinh t + 2v cosh t = 2 cosh t     Z Z R t 1 cosh t 2 R cosh t dt dt −2 cosh sinh t sinh t c˜1 + 2 senh t cosh t dt e dt = c˜1 + 2 v(t) = e senh t senh2 t

x(t) = tu(t) senh t,

´ ´ I. GRUPOS D,E CURSO 2018-2019. HOJA 2. SOLUCIONES METODOS MATEMATICOS

4

v(t) =

c˜1 2

senh t

+ 1,

c1 + c2 + t, u(t) = tanh t

x(t) = c1 t cosh t + c2 t senh t + t2 senh t

5. Calcular la soluci´on particular que se indica: ( ( x′′ − 2x′ + 2x = et cos t x′′′ + x′ − 10x = 36tet ; ′ x(0) = 0, x′ (0) = −1, x′′(0) = 2 x(0) = x (0) = 0

;

( x(4) − 16x = 8t2 , x(0) = −1, x′ (0) = −2, x′′(0) = 3, x′′′ (0) = −8

Soluci´ on: a) λ2 − 2λ + 2 = 0,

x(t) = et (c1 cos t + c2 sen t) + xp (t) 1 1 xp (t) = tet (A cos t + B sen t), A = 0, 2B = 1, B = , xp (t) = tet sen t 2 2 1 1 x(t) = et (c1 cos t + c2 sen t) + tet sen t, c1 = 0, c1 + c2 = 0, x(t) = tet sen t 2 2 λ = 1 ± i,

b) λ3 + λ − 10 = 0,

x(t) = c1 e2t + e−t (c2 cos 2t + c3 sen 2t) + xp (t) 9 9 9 xp (t) = (At + B)et , −8A = 36, 4A − 8B = 0, A = − , B = − , xp = − (2t + 1)et 4 2 4 9 45 27 9 x(t) = c1 e2t + e−t (c2 cos 2t + c3 sen 2t) − (2t + 1)et , c1 + c2 = , 2c1 − c2 + 2c3 = , 4c1 − 3c2 − 4c3 = 4 4 4 4 9 1 −t 36 2t 2 27 36 t e − e (27 cos 2t + 8 sen 2t) − (2t + 1)e c1 = , c2 = − , c3 = − , x(t) = 52 13 13 52 13 4 λ = 2, −1 ± 2i,

c) λ4 − 16 = 0,

λ = ±2, ±2i, x(t) = c1 e2t + c2 e−2t + c3 cos 2t + c4 sen 2t + xp (t) 1 1 1 xp (t) = At2 , −16A = 8, A = − , xp (t) = − t2 , x(t) = c1 e2t + c2 e−2t + c3 cos 2t + c4 sen 2t − t2 2 2 2 c1 + c2 + c3 = 0, 2c1 − 2c2 + 2c4 = 0, 4c1 + 4c2 − 4c3 = 1, 8c1 − 8c2 − 8c4 = 0 1 1 1 1 1 c1 = − , c2 = , c3 = −1, c4 = 0, x(t) = − e2t + e−2t − cos 2t − t2 2 2 2 2 2 6. Resolver: ( x′′ + x = θ(t − π) x(0) = 1, x′ (0) = 0

;

( x′′ − x = 2δ(t − 1) x(0) = 1, x′ (0) = 0

;

( x′′′ − 3x′′ + 2x′ = f (t) x(0) = x′ (0) = x′′(0) = 0

( 0, t 0 2 4 0 0 2 0 0 2

son positivos:

´ ´ I. GRUPOS D,E CURSO 2018-2019. HOJA 2. SOLUCIONES METODOS MATEMATICOS

6

Para a = 0: p(λ) = λ2 (λ + 5) que tiene una ra´ız igual a cero doble, luego no es estable. La soluci´on del problema homog´eneo es: x(t) = c1 + c2 t + c3 e−5t b) Para a = 4, las ra´ıces son: p(λ) = (λ + 1)(λ2 + 2)2 que son reales y negativas, luego la soluci´on es estable asint´oticamente: x(t) = c1 e−t + (c2 + c3 t)e−2t + xp (t), x(t) = c1 e

−t

+ (c2 + c3 t)e

−2t

2 −2t

+ 4t e

,

xp (t) = At2 e−2t ,

−2A = −8,

A=4

c1 + c2 = 1, −c1 − 2c2 + c3 = −1, c1 + 4c2 − 4c3 = 1

c1 = 1, c2 = 0, c3 = 0,

x(t) = e−t + 4t2 e−2t

( x′′ + 2cx′ + x = δ(t − 1) distinguiendo los casos 0 ≤ c < 1, c = 1 y c > 1. 9. a) Resolver el problema: x(0) = x′ (0) = 0 b) Escribir el resultado para c = 0, dibujar la soluci´on. Soluci´ on: a) En 0 < x < 1, c 6= 1: λ2 + 2cλ + 1 = 0,

λ = −c ±

En x > 1: c 6= 1,

p

c2 − 1,

λ2 + 2cλ + 1 = 0,

Por continuidad en t = 1:

  √ √ 2 2 x(t) = e−ct a1 et c −1 + a2 e−t c −1 ,

λ = −c ±

p

c2 − 1,

a1 = a2 = 0,

  √ √ 2 2 x(t) = e−ct ˜a1 et c −1 + a˜2 e−t c −1

  √2 √ 2 0 = e−c ˜a1 e c −1 + a˜2 e− c −1

Salto en la derivada:   p √ √ √ √ 2 2 2 2 e−c −c ˜a1 e c −1 − c a˜2 e− c −1 + c2 − 1(˜a1 e c −1 − a˜2 e− c −1 ) = 1 Tenemos un sistema para calcular ˜a1 y ˜a2 :  √ √ ˜a1 e c2 −1 + a˜2 e− c2 −1 = 0 √ √ ˜a1 √c2 − 1 − c e c2 −1 − a˜2 √c2 − 1 + c e− c2 −1 = ec

con soluci´on:

√

ec− c −1 , ˜a1 = √ 2 c2 − 1

Por tanto: x(t) = Si c > 1:

Si c < 1:

2

√

ec+ c −1 ˜a2 = − √ 2 c2 − 1 2

 0,   

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