Linear differential equations on HP50g
vexx
conditions. Looked in manual, but there is only examples in RPN which I
don't use (jet to be exact). Tried to put this in my calculator (example
from manual I tried to convert from RPN to ALG mode):
> (d1d1y(t)+5*y(t)=2*cos(t/2),y(0)=6/5,d1y(0)=-1/2,y(t))
and I get:
> "Bad Argument Value"
Help?
Virgil
wrote:
Shouldn't the outermost pair of grouping symbols be either
braces ,"{ }", or brackets, "[ ]"?
--
Virgil
vexx
> In article <h25um1$78b$1...@news1.carnet.hr>, vexx <vexx...@gmail.com>
> wrote:
>
>> I need a little help with solving LDE of second order with initial
>> conditions. Looked in manual, but there is only examples in RPN which I
>> don't use (jet to be exact). Tried to put this in my calculator (example
>> from manual I tried to convert from RPN to ALG mode):
>>
>>> (d1d1y(t)+5*y(t)=2*cos(t/2),y(0)=6/5,d1y(0)=-1/2,y(t))
I forgot to write here the command... it was:
>> DESOLVE(d1d1y(t)+5*y(t)=2*cos(t/2),y(0)=6/5,d1y(0)=-1/2,y(t))
>> and I get:
>>
>>> "Bad Argument Value"
>> Help?
>
> Shouldn't the outermost pair of grouping symbols be either
> braces ,"{ }", or brackets, "[ ]"?
>
you mean something like this:
> DESOLVE({d1d1*y(t)+5*y(t)=2*cos(t/2),y(0)=6/5,d1y(0)=-1/2},y(t))
It doesnt work either...
Virgil
wrote:
From the AUR, DOWNLOADABLE FROM http://www.hpcalc.org/
DESOLVE
Type: Command
Description: Solves certain first-order ordinary differential equations
with respect to the current variable.
Access: Symbolic solve, !� or calculus, !�DIFFERENTIAL EQNS.
Input: Level 2/Argument 1: A first-order differential equation.
Level 1/Argument 2: The function to solve for.
Output: The solution to the equation, either y as a function of x or x
as a function of y, or x and y as
functions of a parameter.
Flags: Exact mode must be set (flag ?105 clear).
Numeric mode must not be set (flag ?3 clear).
Example: Solve the following differential equation:
y?(x) + 2y(x) = e3x
Command: DESOLVE(d1Y(X)+2*Y(X)=EXP(3*X),Y(X))
It appears from the manual that you cannot set initial conditions as you
have tried to do.
--
Virgil
Virgil
Virgil <vir...@nowhere.com> wrote:
In RPN mode3, I entered
'd1d1*y(t)+5*y(t)=2*cos(t/2)' and then 'y(t)' and then DESOLVE
and got
{'y(t)=COS(t)*(1/2))*(8/19)+COS(t*\sqrt5)*((95*cC0+-40)/95)+cC1+(\sqrt5*1
9)/95*SIN(t*\sqrt5))'}
where \sqrt stands for the square root symbol.
Your initial conditions
make cC0 = y0 = 6/5
and cC1 = d1y(0) = -1/2
--
Virgil
vexx
>> Level 1/Argument 2: The function to solve for.
>> Output: The solution to the equation, either y as a function of x or x
>> as a function of y, or x and y as
>> functions of a parameter.
>> Flags: Exact mode must be set (flag ?105 clear).
>> Numeric mode must not be set (flag ?3 clear).
>> Example: Solve the following differential equation:
>> y?(x) + 2y(x) = e3x
>> Command: DESOLVE(d1Y(X)+2*Y(X)=EXP(3*X),Y(X))
>>
>>
>> It appears from the manual that you cannot set initial conditions as you
>> have tried to do.
>
> In RPN mode3, I entered
> 'd1d1*y(t)+5*y(t)=2*cos(t/2)' and then 'y(t)' and then DESOLVE
> and got
> {'y(t)=COS(t)*(1/2))*(8/19)+COS(t*\sqrt5)*((95*cC0+-40)/95)+cC1+(\sqrt5*1
> 9)/95*SIN(t*\sqrt5))'}
>
> where \sqrt stands for the square root symbol.
>
> Your initial conditions
> make cC0 = y0 = 6/5
> and cC1 = d1y(0) = -1/2
>
in manual this is notation for initial conditions so I figured there
must be a way for initial conditions:
> ['d1d1y(t)+5*y(t)=2*COS(t/2)''y(0)=6/5''d1y(0)=-1/2']ENTER
> 'y(t)'ENTER
> DESOLVE
so I trided this
> DESOLVE['d1d1y(t)+5*y(t)=2*COS(t/2)''y(0)=6/5''d1y(0)=-1/2',y(t)]ENTER
and it works! I dont know what I was doing wrong last night, but this is
the right way in ALG mode... they say that morning is always smarter
than night he he he... tnx for the help!
renan.peli...@gmail.com
Any idea?
Thanks and sorry for my bad english.
Renan
Virgil
DESOLVE['d1d1y(t)+5*y(t)=2*COS(t/2) ,' y(0)=6/5','d1y(0)=-1/2',y(t)]
--