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{@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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