From 6fb1cd644308ac46a71536cc7da3a16377107b4f Mon Sep 17 00:00:00 2001
From: Thomas Albers Raviola <thomas@thomaslabs.org>
Date: Tue, 21 May 2024 11:32:09 +0200
Subject: Rename test files

---
 t/test4.chn | 65 -------------------------------------------------------------
 1 file changed, 65 deletions(-)
 delete mode 100644 t/test4.chn

(limited to 't/test4.chn')

diff --git a/t/test4.chn b/t/test4.chn
deleted file mode 100644
index e126091..0000000
--- a/t/test4.chn
+++ /dev/null
@@ -1,65 +0,0 @@
-{@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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