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‘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),
Check what the solution to the ODE would be if you use the function LDEC:
‘Delta(X-3)’
`
‘X^2+1’
`
LDEC
µ
The result is:
‘SIN(X-3)*Heaviside(X-3) + cC1*SIN(X) + cC0*COS(X)+’.
Please notice that the variable X in this expression actually represents the
variable t in the original ODE. Thus, the translation of the solution in paper
may be written as:
When comparing this result with the previous result for y(t), we conclude that
cC
o
= y
o
, cC
1
= y
1
.
)
3
(
)
3
sin(
sin
cos
)
(
1
−
⋅
−
+
⋅
+
⋅
=
t
H
t
t
C
t
Co
t
y