Texas Instruments TI-89, TI-92

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))^2ùx,x,y)

tan(y)=

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(x

+2ø@3

2)+@n1øp

Note: To type an @ symbol, press:

TI-89: ¥ §

TI-92 Plus: 2R

ans(1)|@3=cì1 and @n1=0 ¸

y=tanê(x

+2ø(cì1)

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)=(yùe^(x)+cos(y))y'!ode

sin(y)=(exøy+cos(y))øy'

deSolve(ode and

y(0)=0,x,y)!soln ¸

ë(2øsin(y)+yñ)

2 =ë(exì1)øeëxøsin(y)

soln|x=0 and y=0 ¸true

d(right(eq)ìleft(eq),x)/

(d(left(eq)ìright(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^(ë1/2) and

y(0)=0 and y'(0)=0,t,y) ¸

2øy3/4

3 =t

solve(ans(1),y) ¸

22/3ø(3øt)4/3

4 and t‚0

For initialCondition1, use the form:

dependentVar (initialIndependentValue) =

initialDependentValue

For initialCondition2, use the form:

dependentVar' (initialIndependentValue) =

initial1stDerivativeValue