HP 50g User Manual page 498

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
to write,
y L
o
Then, we use the calculator to obtain the following:
'X/(X^2+1)' ` ILAP
'1/(X^2+1)' ` ILAP
'EXP(-3*X)/(X^2+1)' ` ILAP Result, SIN(X-3)*Heaviside(X-3)'.
[2]. The very last result, i.e., the inverse Laplace transform of the expression
'(EXP(-3*X)/(X^2+1))', can also be calculated by using the second shifting
theorem for a shift to the right
if we can find an inverse Laplace transform for 1/(s
try '1/(X^2+1)' ` ILAP. The result is 'SIN(X)'. Thus, L
sin(t-3) H(t-3),
Check what the solution to the ODE would be if you use the function LDEC:
-1
L {a F(s)+b G(s)} = a L
-1 2
L {y s/(s +1)+y
o
-1 2
{s/(s +1)}+ y
1
-1 –as
L {e
'Delta(X-3)' ` 'X^2+1' ` LDEC μ
-1
{F(s)} + b L
2 –3s
/(s +1)) + e
1
-1 2
L {1/(s +1)}+ L
Result, 'COS(X)', i.e., L
Result, 'SIN(X)', i.e., L
F(s)}=f(t-a) H(t-a),
-1
{G(s)},
2
/(s +1)) } =
-1 –3s 2
{e /(s +1))},
-1 2
{s/(s +1)}= cos t.
-1 2
{1/(s +1)}= sin t.
2
+1). With the calculator,
-1 –3s
{e /(s
2
+1)}} =
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