HP 49g+, Руководство пользователя, Страница: 499 / 862

Page 16-25
OBJ
ƒ ƒ
Isolates right-hand side of last expression
ILAP Obtains the inverse Laplace transform
The result is ‘y1*SIN(X-1)+y0*COS(X-1)-(COS(X-3)-1)*Heaviside(X-3)’.
Thus, we write as the solution: y(t) = y
o
cos t + y
1
sin t + H(t-3)
⋅
(1+sin(t-3)).
Check what the solution to the ODE would be if you use the function LDEC:
‘H(X-3)’
`
[ENTER] ‘X^2+1’
`
LDEC
The result is:
Please notice that the variable X in this expression actually represents the
variable t in the original ODE, and that the variable
ttt in this expression is a
dummy variable. Thus, the translation of the solution in paper may be written
as:
Example 4 – Plot the solution to Example 3 using the same values of y
o
and y
1
used in the plot of Example 1, above. We now plot the function
y(t) = 0.5 cos t –0.25 sin t + (1+sin(t-3))
⋅
H(t-3).
In the range 0 < t < 20, and changing the vertical range to (-1,3), the graph
should look like this:
. ) 3 ( sin sin cos ) (
0
1
∫
∞
−
⋅ ⋅ − ⋅ + ⋅ + ⋅ =
du e u H t t C t Co t y
ut