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Okay so I have a simple diffrential equation where I get two different answers depending on the method I use. Additionally, Wolfram has yet a other solution. The differential equation to be solved is:

$y'+ ry = $$-\frac{ry^2}{b}$

So using the Bernoulli's equation we have the following substitution:

$y = z^{-1}$ -> $z'-rz = \frac{r}{b}$

Solving the equation gives:

$\frac{1}{e^{rt}-\frac{1}{b}}$

On the other hand if separated equation is used, the equation becomes:

$\frac{y'}{-ry-\frac{ry^2}{b}} = 1$

Integrating we have:

$\frac{ln\frac{y+b}{y}}{r}=t$

Solving the equation:

$y=\frac{-b}{1-e^{rt}}$

Any ideas where I have gone wrong? Wolfram alpha gives the following answer:

$-\frac{be^b}{e^b-e^{rt}}$

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  • $\begingroup$ What phappened to $a$? It does not appear in the solution. $\endgroup$ – Julián Aguirre Dec 11 '15 at 10:37
  • $\begingroup$ It looks like the $a$ should be $r$ and there is an initial condition you didn't write. $\endgroup$ – KittyL Dec 11 '15 at 10:40
  • $\begingroup$ @JuliánAguirre Apologies it's r and not a. Good catch. $\endgroup$ – Dole Dec 11 '15 at 10:40
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    $\begingroup$ What initial conditions are you using in the different solutions? $\endgroup$ – Mankind Dec 11 '15 at 11:15
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    $\begingroup$ And you have also forgotten the constant of integration in both your solutions (which I believe is the root of all your troubles). $\endgroup$ – Hans Lundmark Dec 11 '15 at 11:31
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As pointed out by the comments, the answer is that the above solutions are the same, except they are shifted left or right relative to each other. This can be easily seen by substituting (x+a) in place of X and choosing a value that shifts the functions on top of each other. Knowing this, the answer is that the constants(which I have not written down) in the equations must be different for a given problem.

Wolfram Alpha's answer is the answer to the separable equation, it just has the constant added.

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