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L{d
2
y/dt
2
} + L{y(t)} = L{
δ
(t-3)}.
With ‘
Delta(X-3)
’
`
LAP , the calculator produces EXP(-3*X), i.e., L{
δ
(t-3)}
= e
–3s
. With Y(s) = L{y(t)}, and L{d
2
y/dt
2
} = s
2
⋅
Y(s) - s
⋅
y
o
– y
1
, where y
o
= h(0)
and y
1
= h’(0), the transformed equation is s
2
⋅
Y(s) – s
⋅
y
o
– y
1
+ Y(s) = e
–3s
. Use
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:
OBJ
ƒ ƒ
Isolates right-hand side of last expression
ILAP
µ
Obtains the inverse Laplace transform
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
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: