• Author / Uploaded
• Jose Antonio Huacpe

Citation preview

EJERCICIO 4 a) y ′′ (t) + 4 y(t) = 9t ; y(0) = 0 y′(0) = 7 ℒ{y ′′ } + 4ℒ{y} = 9ℒ{t} 9 s 2 Y(s) − 7 + 4Y(s) = 2 s 2 9 + 7s Y(s)(s 2 + 4) = s2 2 9 + 7s Y(s) = 2 2 s (s + 4) Aplicamos fracciones parciales: 9 + 7s 2 As + B C = 2 + 2 2 s (s + 4) s + 4 s 2 9 + 7s 2 As 3 + Bs 2 + Cs 2 + 4C Y(s) = 2 2 = s (s + 4) s 2 (s2 + 4) 2 3 2 Y(s) = 9 + 7s = As + s (B + C) + 4C Y(s) =

Obtenemos: A=0

B+C=7

A=0

B+C=7

A=0

B+

A=0

9 =7 4 19 B= 4

4C = 9 9 C= 4 9 C= 4 9 C= 4

Reemplazamos: As + B C + s2 + 4 s2 19 1 9 1 Y(s) = ( 2 ) + ( 2) 4 s +4 4 s 19 1 9 1 ℒ −1 {Y(s)} = ℒ −1 { 2 } + ℒ −1 { 2 } 2 4 s +2 4 s 19 1 9 y(t) = ∗ sin 2t + t 4 2 4 9 19 y(t) = t + sin 2t 4 8 Y(s) =

EJERCICIO 7 b) Si ℒ{F(t)} = H(s) hallar x(t) en: x ′′ (t) + 6x ′ (t) + 7x(t) = F(t)

; x(0) = 0 x ′ (0) = 0

ℒ{x ′′ } + 6ℒ{x ′ } + 7ℒ{x} = ℒ{F(t)} s 2 X(s) + 6sX(s) + 7X(s) = H(s) X(s)(s2 + 6s + 7) = H(s) X(s) 1 = 2 H(s) s + 6s + 7 X(s) 1 = 2 H(s) (s + 3) + 7 − 9

X(s) 1 = (s H(s) + 3)2 − 2 X(s) 1 = H(s) (s + 3)2 − (√2)2 ℒ −1 {

X(s) 1 −1 √2 ℒ { }= 2} H(s) √2 (s + 3)2 − (√2)

x(t) 1 −3t = e sinh √2t F(t) √2 F(t) −3t x(t) = e sinh √2t √2 EJERCICIO 26 x ′′ (t) + 2x ′ (t) + x(t) = F(t) ; x(0) = 0 ℒ{x ′′ } + 2ℒ{x ′ } + ℒ{x} = ℒ{F(t)} s 2 X(s) + 2sX(s) + X(s) = f(s) X(s)(s 2 + 2s + 1) = f(s) X(s) 1 = 2 f(s) s + 2s + 1 X(s) 1 = f(s) (s + 1)2 X(s) 1 ℒ −1 { } = ℒ −1 { } (s + 1)2 f(s) x(t) = te−t F(t) x(t) = F(t)te−t

x ′ (0) = 0

EJERCICIO 41 y ′ + y + 2z ′ + 3z = e−t h) { ′ ; y(0) = 0 z(0) = 1 3y − y + 4z ′ + z = 0 y ′ + y + 2z ′ + 3z = e−t ℒ{ ′ 3y − y + 4z ′ + z = 0 1 sY(s) + Y(s) + 2sZ(s) − 2 + 3Z(s) = { s+1 3sY(s) − Y(s) + 4sZ(s) − 4 + Z(s) = 0 2s + 3 Y(s)(s + 1) + Z(s)(2s + 3) = { s+1 Y(s)(3s − 1) + Z(s)(4s + 1) = 4 Aplicaremos la regla de Cramer: 2s + 3 s + 1 2s + 3 [ ]=[s+1] 3s − 1 4s + 1 4 Y(s) =

∆Y(s) ∆

Z(s) =

Hallamos los determinantes: ∆= −2s2 − 2s − 4

∆Z(s) ∆

−3(2s + 3) s+1 −2s2 + s + 7 ∆Z(s) = s+1 ∆Y(s) =

Hallamos Y(s): −3(2s + 3) (s + 1)(−2s2 − 2s − 4) 3(2s + 3) Y(s) = 2(s + 1)(s − 1)(s + 2) 3 2s + 3 Y(s) = ( 2 ) 2 (s − 1)(s + 2) Y(s) =

Aplicamos fracciones parciales: 2s + 3 As + B C = + (s 2 − 1)(s + 2) s 2 − 1 s + 2 2s + 3 s 2 (A + C) + s(2A + B) + 2B − C = (s 2 − 1)(s + 2) (s 2 − 1)(s + 2) 2 (A 2s + 3 = s + C) + s(2A + B) + 2B − C Obtenemos: A+C=0 A = −C A=

1 3

2A + B = 2 C+3 − 2C + =2 2 1 C=− 3

2B − C = 3 C+3 B= 2 4 B= 3

Reemplazamos: 3 As + B C Y(s) = ( 2 + ) 2 s −1 s+2 3 1 s 4 1 1 1 Y(s) = [ ( 2 )+ ( 2 )− ( )] 2 3 s −1 3 s −1 3 s+2 1 s 1 1 1 Y(s) = ( 2 ) + 2( 2 )− ( ) 2 s −1 s −1 2 s+2 1 s 1 1 1 ℒ −1 {Y(s)} = ℒ −1 { 2 } + 2 ℒ −1 { 2 } − ℒ −1 { } 2 s −1 s −1 2 s+2 1 1 y(t) = cosh t + 2 sinh t − e−2t 2 2 1 𝑒 𝑡 + 𝑒 −𝑡 𝑒 𝑡 − 𝑒 −𝑡 1 y(t) = ( )+ 2( ) − e−2t 2 2 2 2 t −t e e 1 y(t) = + + et − e−t − e−2t 4 4 2 5 t 3 −t e−2t y(t) = e − e − 4 4 2 5𝑒 𝑡 − 3𝑒 −𝑡 − 2𝑒 −2𝑡 y(t) = 4

Hallamos Z(s): −2s 2 + s + 7 Z(s) = (s + 1)(−2s2 − 2s − 4) 2s 2 − 𝑠 − 7 Z(s) = 2(s + 1)(s − 1)(s + 2) 1 2s 2 − 𝑠 − 7 Z(s) = ( 2 ) 2 (s − 1)(s + 2) Aplicamos fracciones parciales: 2s 2 − 𝑠 − 7 As + B C = + (s 2 − 1)(s + 2) s 2 − 1 s + 2 2s 2 − 𝑠 − 7 s 2 (A + C) + s(2A + B) + 2B − C = (s 2 − 1)(s + 2) (s 2 − 1)(s + 2) 2 2 (A 2s − 𝑠 − 7 = s + C) + s(2A + B) + 2B − C Obtenemos: A+C=2 A=2−C A=1

2A + B = −1 C−7 4 − 2C + = −1 2 C=1

2B − C = −7 C−7 B= 2 B = −3

Reemplazamos: 1 As + B C Z(s) = ( 2 + ) 2 s −1 s+2 1 s 3 1 1 1 Z(s) = ( 2 )− ( 2 )+ ( ) 2 s −1 2 s −1 2 s+2 1 s 1 1 Z(s) = [( 2 )−3( 2 )+( )] 2 s −1 s −1 s+2 1 s 3 1 1 1 ℒ −1 {Z(s)} = ℒ −1 { 2 } − ℒ −1 { 2 } + ℒ −1 { } 2 s −1 2 s −1 2 s+2 1 3 1 z(t) = cosh t − sinh t + e−2t 2 2 2 1 𝑒 𝑡 + 𝑒 −𝑡 3 𝑒 𝑡 − 𝑒 −𝑡 1 z(t) = ( )− ( ) + e−2t 2 2 2 2 2 t −t e e 3 3 1 z(t) = + + et + e−t + e−2t 4 4 4 4 2 −2t 1 t e z(t) = − e + e−t + 2 2 𝑒 −2𝑡 + 2𝑒 −𝑡 − 𝑒 𝑡 z(t) = 2