Appendix A: Functions and Instructions
785
Apply
solve()
to an implicit solution if you want to
try to convert it to one or more equivalent explicit
solutions.
deSolve(y'=(cos(y))^2
ù
x,x,y)
¸
tan(y)=
x
ñ
2
+@3
When comparing your results with textbook or
manual solutions, be aware that different methods
introduce arbitrary constants at different points in
the calculation, which may produce different general
solutions.
solve(ans(1),y)
¸
y=tan
ê
(
x
ñ
+2
ø
@3
2
)
+@n1
ø
p
Note:
To type an @ symbol, press:
@
¥
§
H
2
R
ans(1)|@3=c
ì
1 and @n1=0
¸
y=tan
ê
(
x
ñ
+2
ø
(c
ì
1)
2
)
deSolve(
1stOrderOde
and
initialCondition
,
independentVar
,
dependentVar
)
⇒
a particular solution
Returns a particular solution that satisfies
1stOrderOde
and
initialCondition
. This is usually easier
than determining a general solution, substituting
initial values, solving for the arbitrary constant, and
then substituting that value into the general solution.
initialCondition
is an equation of the form:
dependentVar
(
initialIndependentValue
) =
initialDependentValue
The
initialIndependentValue
and
initialDependentValue
can be variables such as
x0
and
y0
that have no
stored values. Implicit differentiation can help verify
implicit solutions.
sin(y)=(y
ù
e
^(x)+cos(y))y'
!
ode
¸
sin(y)=(
e
x
ø
y+cos(y))
ø
y'
deSolve(ode and
y(0)=0,x,y)
!
soln
¸
ë
(2
ø
sin(y)+y
ñ
)
2
=
ë
(
e
x
ì
1)
ø
e
ë
x
ø
sin(y)
soln|x=0 and y=0
¸
true
d
(right(eq)
ì
left(eq),x)/
(
d
(left(eq)
ì
right(eq),y))
!
impdif(eq,x,y)
¸
Done
ode|y'=impdif(soln,x,y)
¸
true
DelVar ode,soln
¸
Done
deSolve(
2ndOrderOde
and
initialCondition1
and
initialCondition2
,
independentVar
,
dependentVar
)
⇒
a particular solution
Returns a particular solution that satisfies
2ndOrderOde
and has a specified value of the
dependent variable and its first derivative at one
point.
deSolve(y''=y^(
ë
1/2) and
y(0)=0 and y'(0)=0,t,y)
¸
2
ø
y
3/4
3
=t
solve(ans(1),y)
¸
y=
2
2/3
ø
(3
ø
t)
4/3
4
and t
‚
0
For
initialCondition1
, use the form:
dependentVar
(
initialIndependentValue
) =
initialDependentValue
For
initialCondition2
, use the form:
dependentVar
' (
initialIndependentValue
) =
initial1stDerivativeValue
Summary of Contents for Titanium TI-89
Page 9: ...Getting Started 6 TI 89 Titanium keys Ë Ì Í Ê ...
Page 34: ...Getting Started 31 2 or D 2 B u s i n e s s D B D B Press Result ...
Page 43: ...Getting Started 40 3 0 D B D D B D Press Result ...
Page 44: ...Getting Started 41 D 2 0 0 2 D B Scroll down to October and press Press Result ...
Page 58: ...Getting Started 55 Example Set split screen mode to TOP BOTTOM Press Result 3 B D ...
Page 70: ...Getting Started 67 ...
Page 175: ...Operating the Calculator 172 From the Keyboard ...
Page 456: ...Tables 453 ...
Page 527: ...Data Matrix Editor 524 ...
