1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

552 CHAPTER 12 • FOURIER SERIES AND THE LAPLACE TRANSFORM

T he Laplace transform Y (s) of the solution y (t) is easily found to be

y (

8

) = F ( s) + sy (0) + y' (0) + ay (0).

s^2 +as+ b

(12-31)

For many physical problems involving mechanical systems and electrical cir-

cuits, the transform F (s) is known, and the inverse of Y (s) can easily be com-

puted. This process is referred to as operational calculus and has the advantage

of changing problems in differential equations into problems in algebra. Then

the solution obtained will satisfy the specific initial conditions.

• EXAMPLE 12.15 Solve the initial value problem

y" (t) +y(t) = 0, with y (0) = 2 and y' (0) = 3.

Solution The right side of the differential equation is f (t) = 0, so we have

F (s) = 0. The initial conditions yield£ (y'' (t)) = s^2 Y (s)- 2s- 3 and Equation

. 2s + 3
(12-30) becomes s^2 Y (s) + Y (s) = 2s + 3. Solvmg we get Y (s) = ~
1
. We
s +
then solve y (t) with the help of Table 12.2 to compute

y(t)=c-^1 (~~:~) =2£-^1 ( 82 :

1

) +3£-^1 (s 2 ~

1

) =2cost+3sint.

• EXAMPLE 12 .16 Solve the initial value problem

y" (t) + y' (t) - 2y (t) = 0, with y (0) = 1 and y' (O) = 4.

Solution As in Example 12 .15, we use the initial conditions and Equation

(12-31) becomes

Y() 8 s + 4 + 1 s+5

= s^2 + s - 2 = (s - 1) (s + 2}'

The partial fraction expansion Y (s) = -

2

• 
  
  
  
  • __!
    2
    gives the solution
    s - 1 s +
  
  
  
  

y (t) = c -^1 (Y (s)) = 2c-^1 ( -

1

• ) - c.-^1 (-

1

-) = 2et - e-^2 t.

s- 1 s+2

-------~EXERCISES FOR SECTION 12.6

1. Derive C (sin t) from C (cost).
   2. Derive C (cosh t) from C (sinh t).