## Saturday, March 2, 2019

### Revisiting Basic Laplace Transforms In Solving Differential Equations

Definition : Let  F be a function defied for t > 0. Then define a new function f by:

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 ® 0    [ - t/s  exp (-st)]  Ro        + 1/s   ò Ro     t exp(-st) dt

=    lim ® 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) s +  (2A + B) s  + 2B + C  =  =    -  s 2 + s  -  1

From which we see by inspection:

A + C = -1,   2A +  B  = 1,  2B  + C  = -1

-2A – 2C  = 2
2A   + B = 1
-----------------
B – 2C   =   3

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

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Some  common Laplace transforms: