f(s) = ò ¥o exp(-st) F(t) dt
For all s such that the integral exists, then f is called the Laplace transform of F and is written as:
f = £ {F} or f(s) = £ {F(t) }
Example: Compute £ {t} where F(t) = t
£ {t} = ò ¥o exp(-st) t dt = lim R ® 0 ò Ro exp(-st) F(t) dt
= lim R ® 0 [ - t/s exp (-st)] Ro + 1/s ò Ro t exp(-st) dt
= lim R ® 0 - R/s exp (-Rs) + (-1/ s 2 exp(-Rs) + 1/ s 2 )
General properties of Laplace Transforms:
1)Let F1 and F2 both have Laplace transforms on some common
interval. Let c1 and c2 be constants. Then:
£ {c1
F1 + c2 F2} = c1 £ {F1}
+ c2 £ {F2}
Let F1 = 1, and F2 = cos t
Then: £ {1 – cos t} = £ {1} - £ {cos t} = 1/s - s /1 + s 2
2)Let F be continuous for t > 0 and of exponential
order exp (a t). Assume F’ is piecewise continuous on every
interval of the form [0, b], and 0 <
b < ¥ .
Then,
£ {F’}exists
and:
£ {F’(t)
} =
s £ {F(t) }
- F (0)
Solution
example:
Solve: dY / dt +
2Y = cos t
Using Laplace
transforms:
Then:
£ {Y’(t)
} + 2 £ {Y (t) }
= £ {cos (t) }
And:
£ {Y’(t)
} + 2 £ {Y (t) }
= s / s 2 + 1
Further: £ {Y’(t) }
=
y(s)
s £ {Y’(t) }
- Y(0) + 2 £ {Y (t) }
= s / s 2 + 1
s y(s) + 1 + 2 y(s)
= s / s 2 + 1
y(s) [s + 2}
= s - s 2 + 1 / s 2 + 1
Whence: y(s) = - s 2 +
s - 1 / ( s 2 + 2) ( s + 2)
Separate using partial fractions:
As + B/ s 2 + 1
+ C/ s + 2 =
(As + B) (s + 2) + Cs2 + C/ ( s 2 + 2) ( s + 1)
SO:
(C
+ A) s2 + (2A +
B) s + 2B + C = = - s 2 +
s - 1
From which we see by inspection:
A + C = -1, 2A +
B = 1, 2B +
C = -1
Add:
-2A – 2C = 2
2A + B
= 1
-----------------
B – 2C = 3
Add:
B
- 2C = 3
4B
+ 2C = -2
----------------
5B =
1 Therefore: B = 1/5
2A + B = 1 and 2A = 1 - 1/5
= 4/5
A = ½ (4/5)
= 2/5 so: C
= -1 – 2B = -1 – 2(1/5) = -7/5
The inverse transform is therefore:
£ -1 {y(s)}
= 2 cos t/ 5 - sin t/5 – 7/5 exp (-2t) = Y(1)
Problem for Math Mavens:
Solve , using the Laplace transform:
d
3 Y/ dt 3 - d Y /dt =
0
Using the conditions:
Y(0) = 1, Y’(0) =
0 and Y’ (0) = 1
----------------------------------
Some common Laplace transforms:
Some common Laplace transforms:
