Texas Instruments TI-89 Titanium Short User Manual page 181

Apply
to try to convert it to one or more equivalent
explicit solutions.
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.
1stOrderOde
deSolve(
independentVar
⇒
a particular solution
Returns a particular solution that satisfies
1stOrderOde
easier than determining a general solution,
substituting initial values, solving for the arbitrary
constant, and then substituting that value into
the general solution.
initialCondition
dependentVar
initialDependentValue
The
initialIndependentValue
initialDependentValue
and
y0
differentiation can help verify implicit solutions.
2ndOrderOde
deSolve(
initialCondition2
dependentVar
Returns a particular solution that satisfies
2ndOrderOde
dependent variable and its first derivative at one
point.
For
initialCondition1
dependentVar
initialDependentValue
For
initialCondition2
dependentVar
initial1stDerivativeValue
Appendix A: Functions and Instructions
to an implicit solution if you want
solve()
initialCondition
and
dependentVar
, )
and
initialCondition
is an equation of the form:
(
initialIndependentValue
and
can be variables such as
that have no stored values. Implicit
initialCondition1
and
independentVar
, ,
) ⇒
a particular solution
and has a specified value of the
, use the form:
( initialIndependentValue
, use the form:
' (
initialIndependentValue
deSolve(y'=(cos(y))^2ù x,x,y)
¸
solve(ans(1),y) ¸
Note: To type an @ symbol, press:
@
H
ans(1)|@3=cì 1 and @n1=0 ¸
sin(y)=(yù
,
¸
deSolve(ode and
y(0)=0,x,y)! soln ¸
. This is usually
soln|x=0 and y=0 ¸
d
) =
(
! impdif(eq,x,y) ¸
x0
ode|y'=impdif(soln,x,y) ¸
DelVar ode,soln ¸
deSolve(y''=y^(ë 1/2) and
and
y(0)=0 and y'(0)=0,t,y) ¸
solve(ans(1),y) ¸
) =
) =
xñ +2ø@3
(
y=tanê
¥
§
2
R
(
y=tanê
e
^(x)+cos(y))y'! ode
sin(y)=(
ë(2øsin(y)+yñ)
=ë(
2
(right(eq)ì left(eq),x)/
d
(left(eq)ì right(eq),y))
ø(3øt)
2
2/3
y=
xñ
tan(y)= +@3
2
)
+@n1øp
2
xñ +2ø(cì 1)
)
2
e øy+cos(y))øy'
x
e e
ì1)ø øsin(y)
ëx
x
true
Done
true
Done
2øy
3/4
=t
3
4/3
and t‚0
4
175