HP 49g+ - Manual (Page 492)

Page 16-18
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