Texas Instruments TI-89 Titanium User Manual page 788

Apply
solve()
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
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
can be variables such as
stored values. Implicit differentiation can help verify
implicit solutions.
deSolve( 2ndOrderOde
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 to
initialCondition
and ,
dependentVar
, )
and . This is usually easier
initialCondition
is an equation of the form:
( ) =
initialIndependentValue
and
initialDependentValue
and that have no
x0 y0
and 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) ¸
xñ +2ø@3
(
y=tanê
Note: To type an @ symbol, press:
@
¥
§
H 2
R
ans(1)|@3=cì 1 and @n1=0 ¸
(
y=tanê
sin(y)=(yù e ^(x)+cos(y))y'! ode
¸
e
sin(y)=(
deSolve(ode and
y(0)=0,x,y)! soln ¸
ë(2øsin(y)+yñ)
=ë( e
2
soln|x=0 and y=0 ¸
d
(right(eq)ì left(eq),x)/
( d (left(eq)ì right(eq),y))
! impdif(eq,x,y) ¸
ode|y'=impdif(soln,x,y) ¸
DelVar ode,soln ¸
deSolve(y''=y^(ë 1/2) and
y(0)=0 and y'(0)=0,t,y) ¸
solve(ans(1),y) ¸
ø(3øt)
2
2/3
y=
xñ
tan(y)= +@3
2
)
+@n1øp
2
xñ +2ø(cì 1)
)
2
øy+cos(y))øy'
x
ì1)ø øsin(y)
e
x ëx
true
Done
true
Done
2øy
3/4
=t
3
4/3
and t‚0
4
785