EULER (Numerical Version) - TI 89 or TI 92

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 main.eulnum.89p


The Program
:eulnum( ) {This will already appear if you named the program eulnum}
:Prgm {This will already appear}
:ClrIO {ClrIO is in the CATALOG menu}
:FnOff {FnOff is in the CATALOG menu}
:Input "initial x", x  {Input is in the F3 menu}
:Input "initial y", y 
:Input "step size", s             
:0$\to$ {The 0 is a zero}{The arrow is STO }
:Lbl p  {Lbl is in the CATALOG menu}
:c+1$\to$c
:x+s$\to$
:y+y1(x)*s$\to$ {y1 is the Y followed by a 1}
:Disp "this is step number", c          {Disp is in the F3 menu}
:Disp "the x coord is", u
:Disp "the y coord is", v
:Pause  {Pause is in the CATALOG menu}
:ClrIO
:u$\to$
:v$\to$
:Goto p  {Goto is in the CATALOG menu}
:EndPrgm {This will already appear}


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