**Introduction**

This program gives coordinates for an approximate solution for the differential
equation
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.

**The Program **

:FnOff | {FnOff is in Y-VARS under ON/OFF} |

: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 | |

:0C | {The 0 is a zero}{The arrow is STO } |

:Lbl 1 | {Lbl is in PRGM under CTL} |

:C+1C | |

:X+HU | |

:Y+Y1*HV | {Y1 is in Y-VARS under FUNCTION} |

: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} |

:UX | |

:VY | |

: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