Texas Instruments PLUS TI-89 - Manual (Page 452)

Appendix A: Functions and Instructions 435
8992APPA.DOC TI-89 / TI-92 Plus: Appendix A (US English) Susan Gullord Revised: 02/23/01 1:48 PM Printed: 02/23/01 2:21 PM Page 435 of 132
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))^2ù x,x,y)
¸
tan(y)=
xñ
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=tanê
(
xс +2ш@3
2
)
+@n1ø p
Note: To type an @ symbol, press:
TI-89: ¥ §
TI-92 Plus: 2 R
ans(1)|@3=cì 1 and @n1=0 ¸
y=tanê
(
xс +2ш(cм 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)=(yù e^(x)+cos(y))y'! ode
¸
sin(y)=(e x øy+cos(y))øy'
deSolve(ode and
y(0)=0,x,y)! soln ¸
ë(2øsin(y)+yс)
2
=л(e x м1)шe ëx øsin(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) ¸
2øy 3/4
3
=t
solve(ans(1),y) ¸
y=
2 2/3 ø(3øt) 4/3
4
and t ‚0
For initialCondition1 , use the form:
dependentVar ( initialIndependentValue ) =
initialDependentValue
For initialCondition2 , use the form:
dependentVar ' ( initialIndependentValue ) =
initial1stDerivativeValue