HP 50g User Manual page 495

The result is
To find the solution to the ODE, h(t), we need to use the inverse Laplace
transform, as follows:
ƒ ƒ
OBJ
ILAP μ
The result is
simplifying, results in
Check what the solution to the ODE would be if you use the function LDEC:
The result is:
Thus, cC0 in the results from LDEC represents the initial condition h(0).
Note: When using the function LDEC to solve a linear ODE of order n in f(X),
the result will be given in terms of n constants cC0, cC1, cC2, ..., cC(n-1),
representing the initial conditions f(0), f'(0), f"(0), ..., f
Example 2 – Use Laplace transforms to solve the second-order linear equation,
Using Laplace transforms, we can write:
'H=((X+1)*h0+a)/(X^2+(k+1)*X+k)'.
Isolates right-hand side of last expression
Obtains the inverse Laplace transform
'a*EXP(-X)' ` 'X+k' ` LDEC μ
h(t) = a/(k-1) e
2
d y/dt
2
L{d y/dt
2 2
L{d y/dt } + 2 L{y(t)} = L{sin 3t}.
. Replacing X with t in this expression and
h(t) = a/(k-1) e
, i.e.,
-t
+((k-1) cC -a)/(k-1) e
o
2
+2y = sin 3t.
2
+2y} = L{sin 3t},
-t
+((k-1) h -a)/(k-1) e
o
-kt
.
(n-1)
(0).
-kt
.
Page 16-18