HP F2226A - 48GII Graphic Calculator User Manual page 499

' ` LAP , the calculator produces EXP(-3*X), i.e., L{δ(t-3)}
With '
Delta(X-3)
–3s
= e . With Y(s) = L{y(t)}, and L{d
and y = h'(0), the transformed equation is s
1
the calculator to solve for Y(s), by writing:
'X^2*Y-X*y0-y1+Y=EXP(-3*X)' ` 'Y' ISOL
The result is 'Y=(X*y0+(y1+EXP(-(3*X))))/(X^2+1)'.
To find the solution to the ODE, y(t), we need to use the inverse Laplace
transform, as follows:
ƒ ƒ
OBJ
µ
ILAP
The result is 'y1*SIN(X)+y0*COS(X)+SIN(X-3)*Heaviside(X-3)'.
Notes:
[1]. An alternative way to obtain the inverse Laplace transform of the
expression '(X*y0+(y1+EXP(-(3*X))))/(X^2+1)' is by separating the
expression into partial fractions, i.e.,
'y0*X/(X^2+1) + y1/(X^2+1) + EXP(-3*X)/(X^2+1)',
and use the linearity theorem of the inverse Laplace transform
-1
L {a⋅F(s)+b⋅G(s)} = a⋅L
to write,
-1 ⋅s/(s
L {y
o
-1
⋅L
y {s/(s
o
Then, we use the calculator to obtain the following:
2 2
L{d y/dt } + L{y(t)} = L{δ(t-3)}.
2 2 2
⋅Y(s) - s⋅y
y/dt } = s
2
⋅Y(s) – s⋅y
Isolates right-hand side of last expression
Obtains the inverse Laplace transform
-1
{F(s)} + b⋅L
2 2
+1)+y /(s +1)) + e
1
2 -1 2
⋅L
+1)}+ y {1/(s +1)}+ L
1
– y , where y
o 1
–3s
– y + Y(s) = e
o 1
-1
{G(s)},
–3s 2
/(s +1)) } =
-1 –3s 2
{e /(s +1))},
Page 16-21
= h(0)
o
. Use