From d917f41beca176b8f2b682ac3a2c25b148752b71 Mon Sep 17 00:00:00 2001
From: Thomas Albers Raviola <thomas@thomaslabs.org>
Date: Wed, 15 May 2024 16:21:29 +0200
Subject: Add syntax form for writing macros

* src/core.lisp (syntax): Change evaluation rules to replace arguments inside
macro expansion.
* src/eval.lisp (tree-eval): Add Evaluation route for chains with a syntax
closure.
* t/test4.chn: New file.
---
 t/test3.chn |  4 ++--
 t/test4.chn | 65 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 67 insertions(+), 2 deletions(-)
 create mode 100644 t/test4.chn

(limited to 't')

diff --git a/t/test3.chn b/t/test3.chn
index d989187..42a0b97 100644
--- a/t/test3.chn
+++ b/t/test3.chn
@@ -1,3 +1,3 @@
 {@define}{@define-function}{{@syntax}{@name @args @body}{{@define}{@name}{{@lambda}{@args}{@body}}}}%
-{@define-function}{@foo}{@a}{hola @a}%
-%{@foo}{mundo}
+{@define-function}{@foo}{@a @b}{hola @b}%
+{@foo}{mundo}{casa}
diff --git a/t/test4.chn b/t/test4.chn
new file mode 100644
index 0000000..e126091
--- /dev/null
+++ b/t/test4.chn
@@ -0,0 +1,65 @@
+{@define}{@title}{Method for solving first order and Bernoulli's differential equations}
+{@define}{@author}{Thomas Albers Raviola}
+{@define}{@date}{2022-10-01}
+%
+{@template}
+{@section}{History}
+I came across the method concerning this article in an old math book from Doctor
+Granville (Elements of differential and integral calculus - ISBN-13:
+978-968-18-1178-5). It doesn't appear to be a popular technique as when using it
+for my assignments I always had to explain what I was doing. As of yet, I still
+haven't found another text referencing it, which is why I decided to include it
+in my website.
+
+In the original book this procedure is shown but never really explained, it is
+left as a sort of "it just works" thing. Here is my attempt to it clear.
+
+{@section}{Theory}
+Throughout this article we'll consider first order differential equations with
+function coefficients just as a special case of the Bernoulli's differential
+equation with {@eq*}{n = 0}.
+
+Consider now the following ODE:
+{@equation}{
+y' + P(x)y = Q(x)y^n
+}
+
+let {@eq*}{y} be the product of two arbitrary functions {@eq*}{w} and {@eq*}{z}
+such that
+
+{@equation}{
+y &= wz \\
+y' &= w'z + wz'
+}
+
+we now restrict {@eq*}{z} to be the solution of the ODE
+
+{@equation}{
+z' + P(x)z = 0
+}
+
+with this it is possible to solve for {@eq*}{z} by integrating
+
+{@equation}{
+\frac{z'}{z} = - P(x)
+}
+
+using {@eq}{z} we solve for {@eq}{w} by replacing {@eq}{y} inside the original
+ODE
+
+{@align}{
+w'z + wz' + P(x)wz &= Q(x)w^nz^n \\
+w'z + w\left(z' + P(x)z\right) &= Q(x)w^nz^n \\
+w'z &= Q(x)w^nz^n \\
+\frac{w'}{w^n} &= Q(x)z^{n-1}
+}
+
+the general solution to our original ODE can be simply obtained by multiplying
+{@eq*}{w} and {@eq*}{z}.
+
+{@section}{Comments}
+This method, while functional, may not always be the most practical. In some
+cases the differential equations for $w$ and $z$ may not have closed algebraic
+solutions. A more traditional substitution may in some situations also be easier
+than this method. Like always it is up to one to know which tool to apply for a
+given problem.
-- 
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