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Some observations :
1) Not Bernoulli/homogeneous
2) Not exact

How to solve this and in general how to attempt equations that don't have a standard method ? Appreciate any help

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  • $\begingroup$ I'll try, thanks! $\endgroup$
    – rrr
    Oct 3, 2014 at 16:45

1 Answer 1

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Note that $y\equiv 1$ is a trivial solution, hereafter we assume that $y \neq 1$.

We have

\begin{eqnarray} y' &=& (1+x(1-y))(1-y)\\ \frac{d}{dx}\frac{1}{1-y}&=&x + \frac{1}{1-y} \end{eqnarray}

To simplify the notation, let $z := \frac{1}{1-y}$. We have

\begin{eqnarray} dz - z dx&=&x dx \\ d(\exp{(-x)}(1+x+z))&=&0\\ z&=& -x-1 + c\exp{x}\\ y&=& 1-\frac{1}{z} = 1- \frac{1}{ c\exp{x} -x-1}\\ \end{eqnarray}

There is no standard method to solve non standard differential equations, the only thing you can do is to play with it, and try to simplify it as much as you can, until you have an eureka moment. It is an art, sort of. Experience helps, too. There is no "royal" way to learn it quickly, unfortunately.

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