HP 49g+, Руководство пользователя

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Example 1 – To solve the first order equation,
dh/dt + k
⋅
h(t) = a
⋅
e
–t
,
by using Laplace transforms, we can write:
L{dh/dt + k
⋅
h(t)} = L{a
⋅
e
–t
},
L{dh/dt} + k
⋅
L{h(t)} = a
⋅
L{e
–t
}.
Note
: ‘EXP(-X)’
`
LAP , produces ‘1/(X+1)’, i.e., L{e
–t
}=1/(s+1).
With H(s) = L{h(t)}, and L{dh/dt} = s
⋅
H(s) - h
o
, where h
o
= h(0), the transformed
equation is s
⋅
H(s)-h
o
+k
⋅
H(s) = a/(s+1).
Use the calculator to solve for H(s), by writing:
‘X*H-h0+k*H=a/(X+1)’
`
‘H’ ISOL
The result is ‘H=((X+1)*h0+a)/(X^2+(k+1)*X+k)’.
To find the solution to the ODE, h(t), we need to use the inverse Laplace
transform, as follows:
OBJ
ƒ ƒµ
Isolates right-hand side of last expression
ILAP Obtains the inverse Laplace transform
The result is . Replacing X with t in this expression
and simplifying, results in h(t) = a/(k-1)
⋅
e
-t
+((k-1)
⋅
h
o
-a)/(k-1)
⋅
e
-kt
.
Check what the solution to the ODE would be if you use the function LDEC:
‘a*EXP(-X)’
`
‘X+k’
`
LDEC
µ