Looking Again At Laplace Transforms Applied To Differential Equations

 As seen in earlier blog posts, the Laplace transform is one of the most useful methods for solving differential equations. In this post I look again at some of the basics underlying its use, especially in solving  DEs.

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

f(s) =  ò ^¥_o    exp(-st) 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 ^R_o exp(-st) F(t) dt   

=     lim _R ® 0    [ - t/s  exp (-st)]  ^R_o   + 1/s   ò ^R_o  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)

Problem example:

Solve:   dY / dt  +   2Y = cos t

Using Laplace transforms:

Write:

£ {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)

                                            Table of common Laplace transforms:

Suggested Problems:

1)  Solve:    

d ³ Y/ dt ³ -    d Y /dt   =   0 

Using the Laplace transform and the conditions:

Y(0) =  1,  Y’(0) =  0  and  Y’ (0) = 1

2)   Solve:    

  d² y/ dt ²  +  4y    =   3 sin t 

 Using the appropriate Laplace transform, given: 

y= F(t),   F(0)= 1,  F'(t) = 0