HP 49g+ - Manual (Page 495)

Page 16-21
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: