HP 50G, User Manual

Page: 498 / 887

Page 16-21
Check what the solution to the ODE would be if you use the function LDEC:
‘Delta(X-3)’
`
‘X^2+1’
`
LDEC
μ
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
L
-1
{a
⋅
F(s)+b
⋅
G(s)} = a
⋅
L
-1
{F(s)} + b
⋅
L
-1
{G(s)},
to write,
L
-1
{y
o
⋅
s/(s
2
+1)+y
1
/(s
2
+1)) + e
–3s
/(s
2
+1)) } =
y
o
⋅
L
-1
{s/(s
2
+1)}+ y
1
⋅
L
-1
{1/(s
2
+1)}+ L
-1
{e
–3s
/(s
2
+1))},
Then, we use the calculator to obtain the following:
‘X/(X^2+1)’
`
ILAP Result, ‘COS(X)’, i.e., L
-1
{s/(s
2
+1)}= cos t.
‘1/(X^2+1)’
`
ILAP Result, ‘SIN(X)’, i.e., L
-1
{1/(s
2
+1)}= sin t.
‘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
L
-1
{e
–as
⋅
F(s)}=f(t-a)
⋅
H(t-a),
if we can find an inverse Laplace transform for 1/(s
2
+1). With the calculator,
try ‘1/(X^2+1)’
`
ILAP. The result is ‘SIN(X)’. Thus, L
-1
{e
–3s
/(s
2
+1)}} =
sin(t-3)
⋅
H(t-3),