584 CHAPTER 30. LAPLACE TRANSFORM METHODS

Thus we find the Laplace transform of the function desired.

Y (s) = y′ (0)1

s2−2s+1+ y(0)

s−2s2−2s+1

+1

s+1

(s2−2s+1)

= y′ (0)1

s2−2s+1+ y(0)

s−2s2−2s+1

+1

s+1

(s2−2s+1)

Now you go backwards in the table. This typically involves doing partial fractions to getsomething which is in the table. It may be tedious, but is completely routine. You can alsoget this from a computer algebra system. More on this later. Thus we need

y′ (0)L −1(

1s2−2s+1

)+ y(0)L −1

(s−2

s2−2s+1

)+L −1

(1

(s+2)(s2−2s+1)

)1

s+1

(s2−2s+1)=

14(s+1)

+1

2(s−1)2 −1

4(s−1)

Now you go backwards in the table to find that this comes from

14

e−t +12

tet − 14

et .

Next consider the other two terms.

s−2s2−2s+1

=− 1

(s−1)2 +1

s−1

These are in the table.

L −1(

s−2s2−2s+1

)=−tet + et

L −1(

1s2−2s+1

)= tet

Therefore, our solution is

y′ (0) tet + y(0)(−tet + et)+ 1

4e−t +

12

tet − 14

et

If you specify y′ (0) = y(0) = 1, then you will find the unique solution to the differentialequation with initial conditions. It is

y(t) =12

tet +34

et +14

e−t

You can check that this satisfies the initial conditions and the equation.Another important formula mentioned in the above table is Formula 19. In this formula,

f ∗g(t)≡∫ t

0f (t−u)g(u)du =

∫ t

0g(t−u) f (u)du

You can use change of variables to observe that the last equation is true so f ∗g = g∗ f .Why is this formula so? It follows from the definition and interchanging the order of

integration.