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+#+title: Method for solving first order and Bernoulli's differential equations
+#+author: Thomas Albers Raviola
+#+date: 2022-10-01
+#+setupfile: ../../math_options.org
+
+* Disclaimer
+This site as of now just a technology demonstration and its claims
+should not be taken as true (even though I myself am pretty confident
+they are)
+
+* 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.
+
+* 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 $n = 0$.
+
+Consider now the following ODE:
+\begin{equation*}
+y' + P(x)y = Q(x)y^n
+\end{equation*}
+
+let $y$ be the product of two arbitrary functions $w$ and $z$ such that
+
+\begin{align*}
+y &= wz \\
+y' &= w'z + wz'
+\end{align*}
+
+we now restrict $z$ to be the solution of the ODE
+
+\begin{equation*}
+z' + P(x)z = 0
+\end{equation*}
+
+with this it is possible to solve for $z$ by integrating
+
+\begin{equation*}
+\frac{z'}{z} = - P(x) \\
+\end{equation*}
+
+using $z$ we solve for $w$ by replacing $y$ inside the original ODE
+
+\begin{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} \\
+\end{align*}
+
+the general solution to our original ODE can be simply obtained by multiplying
+$w$ and $z$.
+
+* 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.
+
+* Example
+Let's solve an example to show the method in practice
+
+\begin{align*}
+y' - 3x^2y &= -x^2y^3 \\
+w'z + wz' - 3x^2wz &= -x^2w^3z^3 \\
+w'z + w\left(z' - 3x^2z\right) &= -x^2w^3z^3
+\end{align*}
+
+solve now for $z$
+
+\begin{align*}
+z' - 3x^2z &= 0 \\
+\frac{z'}{z} &= 3x^2 \\
+z &= c e^{x^3}
+\end{align*}
+
+replace $z$ in the equation
+
+\begin{align*}
+w' &= -x^2w^3z^2 \\
+- \frac{w'}{w^3} &= c^2 e^(2x^3)x^2 \\
+\frac{1}{2 w^2} &= c^2\left(\frac{e^{2x^3}}{6} + k\right) \\
+w &= \pm \frac{1}{c} \sqrt{\frac{3}{e^{2x^3} + k}}
+\end{align*}
+
+finally with $w$ and $z$ get $y$
+
+\begin{equation*}
+y = \pm e^{x^3} \sqrt{\frac{3}{e^{2x^3} + k}}
+\end{equation*}