Working with Laplace Transforms

Working with Laplace Transforms


Using Laplace Transforms to solve various differential equations


Table of Laplace Transforms

Function:  y(t)
Transformed Function:  f(s)

f(s) = ℒ( y(t) )
= ∫ y(t) ∙ exp(-s ∙ t) dt from t = 0 to t = ∞,  s is a complex number

a1, a2, and a are constants

ℒ( a1 ∙ y1(t) + a2 ∙ y2(t) ) = a1 ∙ f1(s) + a2 ∙f2(s)

ℒ ( y'(t) ) = s ∙ f(s) - y(0)

ℒ( t ∙ y(t) ) = -f'(s)

ℒ( 1 ) = 1 ÷ s

ℒ( a ) = a ÷ s

ℒ( t^n ) = n! ÷ (s^(n+1))

ℒ( t^n ÷ n! ) = 1 ÷ (s^(n+1))

ℒ( e^(a ∙ t) ) = 1 ÷ (s - a)

ℒ( sin(a ∙ t) ) = a ÷ (s^2 + a^2)

ℒ( cos(a ∙ t) ) = s ÷ (s^2 + a^2)


All the problems presented today will have the initial condition y(0) = 0.  



Problem 1:  y'(t) + a ∙ y(t) = b with y(0) = 0

Note: y'(t) = dy/dt.   All of the problems this weekend will have an initial condition.

 y'(t) + a ∙ y(t) = b

Applying the Laplace Transform:

s ∙ f(s) - y(0) + a ∙ f(s) = b ÷ s

f(s) ∙ (s + a) = b ÷ s

f(s) = b ÷ (s ∙ (s + a))



Partial fractions:

b ÷ (s ∙ (s + a)) = G ÷ s + H ÷ (s + a)

Simplifying and considering the numerator:

b + 0 ∙ s = a ∙ G + s ∙ (G + H)

Then:

b = a ∙ G
0 = G + H

G = b/a,  H = -G, H = -b/a


f(s) = (b/a) ÷ s - (b/a) ÷ (s + a)

Reverse the Laplace Transform:

y(t) = b/a - b/a ∙ e^(a ∙ t)

y(t) = b/a ∙ (1 - e^(a ∙ t))



Problem 2:  y'(t) + a ∙ y(t) = e^(b ∙ t) with y(0) = 0

y'(t) + a ∙ y(t) = e^(b ∙ t)

Applying the Laplace Transform:

s ∙ (f(s) - y(0) + a ∙ f(s) = 1 ÷ (s - b)

(s + a) ∙ f(s) = 1 ÷ (s - b)

f(s) = 1 ÷ ((s - b) ∙ (s + a))


Partial fractions:

1 ÷ ((s - b) ∙ (s + a)) = E ÷ (s - a) + F ÷ (s + a)

Simplifying and considering the numerator:

1 = E ∙ (s + a) + F ∙ (s - b)

0 ∙ t + 1 = E ∙ (s + a) + F ∙ (s - b)

Solving for E and F:

1 = a ∙ E - b ∙ F
0 = E + F

E = -F;  F = -1 ÷ (a + b), E = 1 ÷ (a + b)


f(s) = (1/(a + b)) ÷ (s - b) - (1/(a - b)) ÷ (s + a)

Reverse the Laplace Transform:

y(t) = e^(b ∙ t) ÷ (a + b) - e^(-a ∙ t) ÷ (a + b)

y(t) = (e^(b ∙ t) - e^(-a ∙ t)) ÷ (a + b)



Problem 3:  y'(t) + a ∙ y(t) = b ∙ t + c with y(0) = 0

y'(t) + a ∙ y(t) = b ∙ t + c

Applying the Laplace Transform:

s ∙ f(s) - y(0) + a ∙ f(s) = (b ∙ 1!) ÷ (s^(1 + 1)) + c ÷ s

s ∙ f(s) + a ∙ f(s) = b ÷ s^2 + c ÷ s

f(s) = (b + c ∙ s) ÷ (s^2 ∙ (s + a))


Partial fractions:

(b + c ∙ s) ÷ (s^2 ∙ (s + a)) = G ÷ (s + a) + (s ∙ H + J) ÷ s^2

= G ÷ (s + a) + H ÷ s + J ÷ s^2

Simplifying and considering the numerator:
 
b + c ∙ s + 0 ∙ s^2  = G ∙ s^2 + H ∙ s^2 + H ∙ a ∙ s + J ∙ s + J ∙ a

b = a ∙ J
c = a ∙ H + J
0 = G + H

H = -G; J = b/a, H = (a ∙ c - b) ÷ a^2, G = (b - a ∙ c) ÷ a^2


f(s) = ((b - a ∙ c) ÷ a^2) ÷ (s + a) + ((a ∙ c - b) ÷ a^2) ÷ s + (b/a) ÷ s^2

Reverse the Laplace Transform:

y(t) = ((b - a ∙ c) ÷ a^2) ∙ e^(-a ∙ t) + ((a ∙ c - b) ÷ a^2)  + (b/a) ∙ t





Problem 4:  y'(t) + a ∙ y(t) = b ∙ t^2 with y(0) = 0

y'(t) + a ∙ y(t) = b ∙ t^2

Applying the Laplace Transform:

s ∙ f(s) - y(0) + a ∙ f(s) = 2 ∙ b ÷ s^3

f(s) = (2 ∙ b) ÷ (s^3 ∙ (s + a))


Partial fractions:

(2 ∙ b) ÷ (s^3 ∙ (s + a)) = (F ∙ s^2 + G ∙ s + H) ÷ s^3 + K ÷ (s + a)

(2 ∙ b) ÷ (s^3 ∙ (s + a)) = F ÷ s + G ÷ s^2 + H ÷ s^3 + K ÷ (s + a)

Simplifying and considering the numerator:

0 ∙ s^3 + 0 ∙ s^2 + 0 ∙ s + 2 ∙ b = F ∙ s^3 + G ∙ s^2 + H ∙ s + F ∙ a ∙ s^2 + G ∙ a ∙ s + H ∙ a + K ∙ s^3

2 ∙ b = a ∙ H
0 = H + a ∙ G
0 = G + a ∙ F
0 = F + K

F = -K; K = (-2 ∙ b) ÷ a^3 ; F = (2 ∙ b) ÷ a^3; H = (2 ∙ b) ÷ a; G = (-2 ∙ b) ÷ a


f(s) = (2 ∙ b)/a^3 ÷ s - (2 ∙ b)/a^2 ÷ s^2 + (2 ∙ b)/a ÷ s^3 - (2 ∙ b)/a^3 ÷ (s + a)

I use Wolfram Alpha and CAS calculators to help verify whether the work is done correctly.  Sometimes I have deal with problems in parts.  

Reverse the Laplace Transform:

y(t) = (2 ∙ b)/a^3 - (2 ∙ b)/a^2 ∙ t + (b/a) ∙ t^2 - (2 ∙ b)/a^3 ∙ e^(-a ∙ t)

Note:  ℒ( t^n ÷ n! ) = 1 ÷ (s^(n+1)), ℒ( t^2 ÷ 2) = 1 ÷ (s^2)


Eddie



All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.