12.7 • SHIITING THEOREMS AND THE STEP FUNCTION 55 3- Find .C(sin^2 t).
4. Show that .C (te') =
1
(s -1)^2. Hint: Let f (t) = te' and f' (t) = te' + e'. - Find .c-^1 C(s~ 4 )).
- Find.c-
1
Cc)+ 4 )).
7. Show that .c-^1 (s 2 (s\i)) = t-1 +e•. - Show that .c-^1 ( 82 (s~ + l)) = t - sin t.
For Exercises 9-18, solve the initial value problem.- y" (t) + 9y (t) = 0, with y (0) = 2 and y' (0) = 9.
- y" (t) + y (t) = 1, with y (0) = 0 and y' (0) = 2.
- y" (t) + 4y (t) = -8, with y (0) = 0 and y' (0) = 2.
1 2. y' (t) + y (t) = 1, with y (0) = 2. - y' (t) -y (t) = -2, with y (0) = 3.
- y" (t) - 4y (t) = 0, with y (0) = 1 and y' (0) = 2.
- y" (t) - y (t) = 1, with y (0) = 0 and y' (0) = 2.
- y' (t) + 2y (t) = 3e^1 , with y (0) = 2.
1 7. y" (t) + y' (t) - 2y(t) = 0, with y(O) = 2 and y' (0) = -1. - y" (t) - y' (t) - 2y (t) = 0, with y (0) = 2 and y' (0) = 1.
12. 7 Shifting Theorems and the Step Function
We have shown how to use the Laplace transform to solve linear differential
equations. Familiar functions that arise in solutions to differential equations are
e'" cosbt and eat sinbt. Theorem 12. 15 shows how their transforms are related to
those of cos bt and sin bt by shifting the variable s in F ( s) and is called the first
shifting theorem. A companion result, called the second shifting theorem, Theo-
rem 12.16, shows how the transform off (t - a) can be obtained by multiplying
F (s) bye-a•. Loosely speaking, these results show that multiplication of f (t)
by eat corresponds to shifting F (s - a) and that shifting f (t - a) corresponds
to multiplication of the transform F ( s) by ea•.