Understanding Engineering Mathematics

(やまだぃちぅ) #1

Note how differentiation with respect totin ‘t-space’ becomes multiplication bysin
‘s-space’. Also, note the appearance of the initial value,f( 0 ),off(t). This makes the
Laplace transform particularly suitable for solving initial value problems. We transform a
differential equation fromt-space tos-space, thereby obtaining an algebraic equation for
the Laplace transform of the solution. We solve this algebraic equation to get the Laplace
transform of the solution of the differential equation, which we then invert, going back
tot-space to get the solution as a function oft. This methodology is characteristic of
‘transform methods’ in mathematics, whatever the particular type of transform.
We can now derive the Laplace transform of the second order derivative.


Problem 17.5
Show that

L[f′′.t/]=s^2 L[f.t/]−sf. 0 /−f′. 0 /

We only have to apply the derivative rule twice in succession:


L[f′′(t)]=sL[f′(t)]−f′( 0 )
=s(sL[f(t)]−f( 0 ))−f′( 0 )
=s^2 L[f(t)]−sf ( 0 )−f′( 0 )

The general result for the Laplace transform of thenth derivative is


L[f(n)(t)]=snL[f(t)]−sn−^1 f( 0 )−sn−^2 f′( 0 )−···−sf(n−^2 )( 0 )−f(n−^1 )( 0 )

Note again the occurrence of the initial values.
To see how we can use the Laplace transform to solve initial value problems, consider
the following problem.


Problem 17.6
Solve the initial value problem

y′Yy= 1 y. 0 /= 0

as follows: take the Laplace transform of the equation and obtain an equa-
tion for the Laplace transformy ̃.s/ofy. Solve this equation. By taking the
Laplace transform of the solution found by, say, separation of variables,
confirm thaty ̃.s/is the Laplace transform of the solutiony.t/.

Taking the Laplace transform of the equation gives, using linearity and the derivative rule


L[y′+y]=L[y′]+L[y]=L[1]

or


sy(s) ̃ −y( 0 )+ ̃y(s)=

1
s
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