Understanding Engineering Mathematics

Sincey( 0 )=0 this becomes

(s+ 1 )y(s) ̃ =

1

s

Solving fory(s) ̃ now gives

y(s) ̃ =

1

s(s+ 1 )

Now by separation of variables (454
➤
), or integrating factor (458
➤
), the solution of the
initial value problem is

y(t)= 1 −e−t

Taking the Laplace transform of this:

L[y(t)]=L[1−e−t]

=

1

s

−

1

s+ 1

=

1

s(s+ 1 )

= ̃y(s)

which is indeed the Laplace transform of the solution as obtained above. So the Laplace
transform of the equation did lead us to the Laplace transform of the solution. In fact,
in this case it is not too difficult to find theinverse Laplace transformofy(s) ̃ directly.
Denoting the inverse Laplace transform byL−^1

y(t)=L−^1 [y(s) ̃ ]=L−^1

[

1

s(s+ 1 )

]

=L−^1

[

1

s

−

1

s+ 1

]

on splitting into partial fractions (60

➤

)

=L−^1

[

1

s

]

−L−^1

[

1

s+ 1

]

Now read Table 17.1 backwards to give

L−^1

[

1

s

]

= 1 , L−^1

[

1

s+ 1

]

=e−t

so
y(t)= 1 −e−t

The above example contains all the essential elements of using the Laplace transform to
solve initial value problems. The main new feature is the occurrence of the inverse Laplace
transform. Also note that in this method the initial values are actually incorporated in the