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$y'=-\frac{2x+3y+1}{3x+4y+1}$

Basically, i have solution to this equation and the whole solving process isn't very difficult, i just have to follow well-known pattern for these kind of equations, but there are few thing that are not completely clear to me and i need to understand them better in order to be able to solve other differential equations.

Anyway, there it is:

As i stated before, this is the standard form of differential equation and these kind of equations are solvable by substitution, anyway, in order to find good substitution i need to look the system of equations where one of the equations is equation (equation that matches the function) in the numerator on the right side of DE and the second one is the one in the denominator. Now, when i solve that system i get:

$x=1, y=-1$

Which means that i have to introduce following substitutions:

$x=u+1, y=v-1, y'(x)=v'(u)$

And his is the first part that i don't understand. Why is function $v$ function of $u$ i mean $y$ is the function of $x$ and $u$ is the function of $x$, if $u$ is the function of $x$ then i suppose that $v$ should be the function of $x$? I think that both of these are correct because if $v$ is function of $u$ and $u$ is function of $x$ then $v$ is function of $x$ too. If this is correct, why is it noted that $v$ is the function of $u$?

Now, we have to implement this substitution, and we get:

$v'=-\frac{2u+3v}{3u+4v}$

and now we have another standard form of DE which can be solved by substitution $z=\frac{u}{v} , v=zu, v'=z+uz'$. Now, a short question. Can i use substitution $z=\frac{v}{u}$ instead and why?

Now, when we implement this substitution we have:

$z+uz'=-\frac{2u+3uz}{3u+4uz}$ and then in next step we just got rid of $u$ on the right side, but this couldn't be done if u was zero. My question here is: Can $u$ be zero? If it can be zero, it means that i have to separate this solving process in two ways, one one u=0 and another when it isn't, what would happen if u=0 here(in case it is possible)? I'd like to get a proper explanation on this one even in general(which means that i would like the explanation of what would happen if $u$ is zero, even in case that it can't be zero here and other way around if the situation is opposite) because this is something that confuses me a lot.

Now, after this division we have:

$z+uz'=-\frac{2+3z}{3+4z} \Rightarrow \frac{dz}{du}u=-\frac{2(1+3z+2z^2)}{3+4z}$

and after this, we have the following expression:

$-2\frac{du}{u}=\frac{3+4z}{2z^2+3z+1}dz$

which means that we did few multiplication of both sides, for example, we multiplied both sides by $3+4z$ and there's this same question. Can $3+4z$ be zero, and if it can, how can i check what happens it case it is?

Now, after this, there are just two simple integrals to be solved in order to solve this DE, so i have these few questions i asked above and every answer that covers these questions is welcome.

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  • $\begingroup$ it is an equation of Abel type google on it $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '16 at 16:10
  • $\begingroup$ @Dr.SonnhardGraubner I don't know if you noticed, but i have a solution to it, i just want answers to the specific questions i asked about the process of solving. $\endgroup$ – cdummie Apr 13 '16 at 17:33
  • $\begingroup$ yes i noticed it see further here $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '16 at 17:35
  • $\begingroup$ arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '16 at 17:35
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    $\begingroup$ @Dr.SonnhardGraubner "The Abel equation of second kind is any nonlinear ordinary differential equation that is quadratic" that's what this document says about Abel DE's, but there's nothing quadratic in my equation, i don't know how this is related to equation i posted. $\endgroup$ – cdummie Apr 14 '16 at 14:33

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