Appendix A: Functions and Instructions 175
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
(
xñ +2ø@3
2
)
+@n1øp
Note: To type an @ symbol, press:
@ ¥ §
H 2 R
ans(1)|@3=cì 1 and @n1=0 ¸
y=tanê
(
xñ +2ø(cì 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)+yñ)
2
=ë(
e
x
ì1)ø
e
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
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