Revisiting Laplace Transforms Applied To Differential Equations

The Laplace transform is perhaps one of the most useful devices for solving differential equations. In this post I look 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