EULER (Numerical Version) - TI 83 Plus

Introduction
This program gives coordinates for an approximate solution for the differential equation ${\hbox{dy}\over \hbox{dx}}=\hbox{f(x,y)}$ using Euler's method.
 
If you have not used one of the programs posted on this website before, you should read through the information in the Intro to Programming section first.


If you have a TI Connectivity Cable, you can download the program EULNUM.8xp


The Program
:FnOff {FnOff is in VARS under Y-Vars}
:Disp "INITIAL X"  {Disp is in PRGM under I/O}
:Input X  {Input is in PRGM under I/O} {X is the X,T button}
:Disp "INITIAL Y" 
:Input Y 
:Disp "STEP SIZE"    
:Input H 
:0$\to$ {The 0 is a zero}{The arrow is STO }
:Lbl 1  {Lbl is in PRGM under CTL}
:C+1$\to$
:X+H$\to$
:Y+Y1*H$\to$ {Y1 is in VARS under Y-VARS} 
:ClrHome  {ClrHome is in PRGM under I/O}
:Disp "STEP",C 
:Disp "X=", U  {The = is in TEST }
:Disp "Y=", V   
:Pause  {Pause is in PRGM under CTL}
:ClrHome  {ClrHome is in PRGM under I/O}
:U$\to$  
:V$\to$  
:Goto 1  {Goto is in PRGM under CTL}


Running the Program
 
You will need to enter a function  f(x,y) into Y1 before running the program. The program will ask for an initial set of coordinates for X and Y. You will also need to provide a stepsize. This can be either positive or negative. Hit ENTER to see the consecutive coordinates.
 
To test the program try the following:
  f(x,y)  =  x + y,   Initial X  =  1,  Initial Y  =  1,  Stepsize  =  0.1
 
Your answer will be
X = 1.1,  Y = 1.2
X = 1.2,  Y = 1.43
X = 1.3,  Y = 1.693
X = 1.4,  Y = 1.9923
and so on