Let Solve the differential equation using Laplace transforms. y(t) = g(t) = { t if t≤2π 2π if t>2π y" + 16y= g(t), y(0) = 5, y(0) = 5 ift ≤2x ift > 2π function f(t) 1 tn eat sin at cos at sinh at cosh at e-at f(t) U(ta) or Ua(t) (a ≥ 0) (ta) (a> 0) U(t − a)f(t − a) or Ua(t)f(t − a) f(n)(t) (−t)" f (t) - (fg)(t) = f(r)g(t − T) dr Laplace transform F(s) 1/s (s > 0) n!/sn+1 (s > 0) 1/(s – a) (s> a) a/(s²+a²) (s > 0) s/(s² + a²) (s > 0) a/(s² - a²) (s> |a|) s/(s² – a²) F(s+a (s>|a|) -as e S S 0) as e as e S F(s F(n)(s) F(s)G(s) sn ¯¹ƒ (0) — sn−² ƒ'(0) … … … — ƒ (n−¹) (0)

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.1: Solutions Of Elementary And Separable Differential Equations
Problem 54E: Plant Growth Researchers have found that the probability P that a plant will grow to radius R can be...
Question
Let
Solve the differential equation
using Laplace transforms.
y(t) =
g(t) = {
t
if t≤2π
2π if t>2π
y" + 16y= g(t),
y(0) = 5, y(0) = 5
ift ≤2x
ift > 2π
Transcribed Image Text:Let Solve the differential equation using Laplace transforms. y(t) = g(t) = { t if t≤2π 2π if t>2π y" + 16y= g(t), y(0) = 5, y(0) = 5 ift ≤2x ift > 2π
function f(t)
1
tn
eat
sin at
cos at
sinh at
cosh at
e-at f(t)
U(ta) or Ua(t) (a ≥ 0)
(ta) (a> 0)
U(t − a)f(t − a) or Ua(t)f(t − a)
f(n)(t)
(−t)" f (t)
-
(fg)(t) = f(r)g(t − T) dr
Laplace transform F(s)
1/s
(s > 0)
n!/sn+1 (s > 0)
1/(s – a)
(s> a)
a/(s²+a²) (s > 0)
s/(s² + a²)
(s > 0)
a/(s² - a²)
(s> |a|)
s/(s² – a²)
F(s+a
(s>|a|)
-as
e
S S
0)
as
e
as
e
S
F(s
F(n)(s)
F(s)G(s)
sn
¯¹ƒ (0) — sn−² ƒ'(0) … … … — ƒ (n−¹) (0)
Transcribed Image Text:function f(t) 1 tn eat sin at cos at sinh at cosh at e-at f(t) U(ta) or Ua(t) (a ≥ 0) (ta) (a> 0) U(t − a)f(t − a) or Ua(t)f(t − a) f(n)(t) (−t)" f (t) - (fg)(t) = f(r)g(t − T) dr Laplace transform F(s) 1/s (s > 0) n!/sn+1 (s > 0) 1/(s – a) (s> a) a/(s²+a²) (s > 0) s/(s² + a²) (s > 0) a/(s² - a²) (s> |a|) s/(s² – a²) F(s+a (s>|a|) -as e S S 0) as e as e S F(s F(n)(s) F(s)G(s) sn ¯¹ƒ (0) — sn−² ƒ'(0) … … … — ƒ (n−¹) (0)
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