Summary Differential Equations with Boundary-Value Problems, Exercise 7.1, 11 to 18

Dennis G. Zill - Differential Equations with Boundary-Value
Problems, Exercise 7.1. Form 11 to 18
Leonardo J. Perez Gomez
7 de agosto de 2022


1. f (t) = et+7

Z ∞
L {f (t)} = e−st (et+7 )dt
0
et(−s+1)+7
Z Z
∴ e−st et+7 dt = et(−s+1)+7 =
1−s
 (−s+1) ∞
e +7 e7
∴ L {f (t)} = =−
1−s 0 s


2. f (t) = e−2t−5

Z ∞
L {f (t)} = e−st e−2t−5 dt
0
et(−s−2)−5
Z
∴ et(−s−2)−5 dt =
−s − 2
 t(−s−s)−5 ∞
e e−5
∴ L {f (t)} = =−
−s − 2 0 −s − 2




1

, 3. f (t) = te4t

Z ∞
L {f (t)} = e−st (te4t )
Z Z0
∴ e−st+4t = et(4−s)

dv = et(4−s) dt et(4−s) t
Z t(4−s)
u=t
Z
t(4−s) e dt
et(4−s) ∴ e tdt = −
du = dt v = 4−s 4−s 4−s
t(4−s)
et(4−s) t
Z  t(4−s) 
e t 1 t(4−s) 1 e
= − e dt = −
4−s 4−s 4−s 4−s 4−s
et(4−s) t et(4−s)
 
t(4−s) t 1
= − =e −
4−s (4 − s)2 4 − s (4 − s)2
  b
t 1 1 1
∴ L {f (t)} = lı́m e t(4−s)
− = − =
b→∞ 4 − s (4 − s)2 0 (4 − s)2 (s − 4)2


4. f (t) = t2e−2t

Z ∞
L {f (t)} = e−st (t2 e−2t )dt
Z Z0
∴ e−st−2t t2 dt = et(−s−2) t2 dt

u = t2 dv = et(−s−2) dt et(−s−2) t2
Z Z t(−s−2)
t(−s−2) 2 e
t(−s−2) ∴ e t dt = − 2tdt
du = 2tdt v = e −s−2 −s − 2 −s − 2
et(−s−2) t2
Z
2
= + et(−s−2) tdt
−s − 2 −s − 2
R
1. et(−s−2) tdt

dv = et(−s−2) dt et (−s − 2)
Z t(−s−2)
u=t
Z
t(−s−2) e
t(−s−2) ∴ e tdt = − dt
du = dt v = e −s−2 −s − 2 −s − 2
et (−s − 2)
 t(−s−2) 
1 e
= −
−s − 2 −s − 2 −s − 2




2