HP 49g+ User Manual page 493

The result is:
h(t) = a/(k-1)⋅e
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:
L{d
Note: 'SIN(3*X)' ` LAP µ produces '3/(X^2+9)', i.e.,
With Y(s) = L{y(t)}, and L{d
h'(0), the transformed equation is
2 ⋅Y(s) – s⋅y
s
Use the calculator to solve for Y(s), by writing:
'X^2*Y-X*y0-y1+2*Y=3/(X^2+9)' ` 'Y' ISOL
The result is
'Y=((X^2+9)*y1+(y0*X^3+9*y0*X+3))/(X^4+11*X^2+18)'.
, i.e.,
-t
+((k-1)⋅cC -a)/(k-1)⋅e
o
2 2
d y/dt +2y = sin 3t.
2 2
L{d y/dt +2y} = L{sin 3t},
2 2
y/dt } + 2⋅L{y(t)} = L{sin 3t}.
2
L{sin 3t}=3/(s +9).
2 2 2
⋅Y(s) - s⋅y
y/dt } = s – y
o
– y + 2⋅Y(s) = 3/(s
o 1
-kt
.
(n-1)
(0).
, where y = h(0) and y
1 o
2
+9).
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1