HP 48g Graphing Calculator - Manual (Page 231)

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K ®
1000 £3 Q SKSD © Y 0
(I n } (0 T ® (E) (T} (C^ 0
T (ENTER)
1000
(^ (ENTER)
1000 0 ( c ^ © T ® Q dD
0 T
(ENTER)
0 T
(ENTER)
0
(ENTER)1 (ENTER)
( E N T E R ) s o l v e : e d i t
SDLVE V'CTJ=F(T.V)
F: '-1„. *F*V:-10„, *FiT: ' 10„,
INDER: J INIT: 0 FIHftL: 1______
SDLN: Y INIT: 1 FINflL:PE3Ell
TDL:.0001 STEP: Df It ¿STIFF
.841569099036
liBiBiiiBmmiiMiBBHMMiw
The problem takes about a minute to solve. (If you had used the
standard method, it would have taken over five minutes.)
How accurate is the answer? With the giveir initial conditions the
solution equation is:
y = e + sin(t)
Solving for y(i) gives + sin(l) = 0.841470984808. Comparing
the results, you can see there is an error of approximately 0.000098,
which is within the specified error tolerance of 0.0001.
19
Solving a Vector-Value Differential Equation
You can use vector-valued equations to solve second-order (or higher)
differential equations given two or more initial values.
Another way to write the second-order equation
y" = a i ( t ) y ' + a o ( i ) y + g i t )
V
/
_y\
0 1
ao(t) ai(t)
y ' o '
+
[ y J
1
You can then substitute w for­
gives
fw for
w' = f w * u; -b c * g(t)
which is a first-order differential equation.
g i t )
0 1
ao(i) a i { t )
, and c for
Differential Equations 19-5