HP 50g User Manual page 501

where H(t) is Heaviside's step function. Using Laplace transforms, we can
2 2
write: L{d y/dt
this expression is: L{H(t-3)} = (1/s) e
2
s Y(s) - s y – y
o
2
s Y(s) – s y – y
o
necessary. Use the calculator to solve for Y(s), by writing:
'X^2*Y-X*y0-y1+Y=(1/X)*EXP(-3*X)' ` 'Y' ISOL
The result is
To find the solution to the ODE, y(t), we need to use the inverse Laplace
transform, as follows:
ƒ ƒ
OBJ
ILAP
The result is
Thus, we write as the solution: y(t) = y
Check what the solution to the ODE would be if you use the function LDEC:
The result is:
Please notice that the variable X in this expression actually represents the
variable t in the original ODE, and that the variable ttt in this expression is a
dummy variable. Thus, the translation of the solution in paper may be written
as:
) ( =
y t Co
+y} = L{H(t-3)}, L{d
, where y = h(0) and y
1 o
+ Y(s) = (1/s) e
1
'Y=(X^2*y0+X*y1+EXP(-3*X))/(X^3+X)'.
Isolates right-hand side of last expression
Obtains the inverse Laplace transform
'y1*SIN(X-1)+y0*COS(X-1)-(COS(X-3)-1)*Heaviside(X-3)'.
'H(X-3)' `[ENTER] 'X^2+1' ` LDEC
⋅ cos + ⋅
t C
1
2 2
y/dt } + L{y(t)} = L{H(t-3)}. The last term in
–3s
. With Y(s) = L{y(t)}, and L{d
= h'(0), the transformed equation is
1
–3s
. Change CAS mode to Exact, if
cos t + y
o
∞
∫
sin + sin ⋅
t t
0
sin t + H(t-3) (1+sin(t-3)).
1
− ut
( − ) 3 ⋅ ⋅
H u e
2 2
y/dt } =
.
du
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