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author | Thomas Albers Raviola <thomas@thomaslabs.org> | 2025-01-04 23:28:34 +0100 |
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committer | Thomas Albers Raviola <thomas@thomaslabs.org> | 2025-01-05 17:11:20 +0100 |
commit | bc69faaa5bbd8a2d8afb9ab81882b1ba21c4bb8e (patch) | |
tree | 6ad76e58bb58f1ac00dcf4ec2dff8d4239f3592e /test/test4.chn | |
parent | 9412242d0aa9550860db364ddf715fd5c5855740 (diff) |
Reorganize and add documentation
Diffstat (limited to 'test/test4.chn')
-rw-r--r-- | test/test4.chn | 65 |
1 files changed, 0 insertions, 65 deletions
diff --git a/test/test4.chn b/test/test4.chn deleted file mode 100644 index e126091..0000000 --- a/test/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. |