Page 16-22
‘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()3sin(sincos)(
1
−⋅−+⋅+⋅= tHttCtCoty
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