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I want to solve the following first- order nonlinear ordinary differntial equation:

$\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$

where a,b and c are constants. I rewrote the equation:

$\leftrightarrow 1=\frac{1}{cx^2-x(b+c)+a}\frac{dx}{dt}\\ \leftrightarrow \int 1dt=\int \frac{1}{cx^2-x(b+c)+a} dx\\ \leftrightarrow t+k= \int \frac{1}{cx^2-x(b+c)+a} dx\\ \leftrightarrow t+k= \int \frac{1}{c(x-\frac{b+c}{2c})^2+a-\frac{(b+c)^2}{4c}} dx $

for some arbitrary number k. How do I solve the last integral? Wolfram-Alpha tells me that it is

$\frac{2tan^{-1}(\frac{-c-b+2cx}{\sqrt{-c^2-b^2-2cb+4ca}})}{\sqrt{-c^2-b^2-2cb+4ca}}$

But I don't know how to calculate that on my own.

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  • $\begingroup$ Looking good so far. I would complete the square in the denominator and use a trig substitution. $\endgroup$ – Adrian Keister Nov 28 '18 at 13:48
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    $\begingroup$ What do you mean with completing the square? $\endgroup$ – Sunny Nov 28 '18 at 14:00
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    $\begingroup$ I mean you have a quadratic in the denominator of your integral, $cx^2-x(b+c)+a,$ and I'm suggesting you complete the square on it. $\endgroup$ – Adrian Keister Nov 28 '18 at 14:02
  • $\begingroup$ Indeed, remember that arctan comes from $\int\frac{1}{x^2+a^2}dx$, so try to complete the square in the denominator - a pack the linear term into the square. $\endgroup$ – orion Nov 28 '18 at 14:13
  • $\begingroup$ I have edited my question and completed the square in the denominator. $\endgroup$ – Sunny Nov 28 '18 at 14:25
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The main step is converting the fraction

$$\frac{1}{cx^2+x(b+c)+a}$$ into the form, expected from the integral tables:

$$\int\frac{1}{t^2+q^2}dt=\frac{1}{q}\arctan \frac{t}{q}$$

You take out the extra $c$, complete the square and change variables:

$$\frac{1}{c}\frac{1}{\color{red}{(x+\frac{b+c}{2c})}^2-(\frac{b+c}{2c})^2+\frac{a}{c}}$$

Now you have $t=x+\frac{b+c}{2c}$ and $q^2=\frac{a}{c}+(\frac{b+c}{2c})^2$.

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The right side has two roots $r_1, r_2$. These roots are then also constant solutions of the ODE. With these solutions the expression $$ u=\frac{x-r_1}{x-r_2} $$ has the derivative $$ u'=-\frac{(r_2-r_1)}{(x-r_2)^2}\,x'=-c(r_2-r_1)u. $$ This ODE for $u$ can now be easily solved. After that, back-substitution gives $$ x=\frac{r_2u-r_1}{u-1} $$

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After completing the square the integral has the form:

$\int \frac{1}{c(x-\frac{b+c}{2c})^2+a-\frac{(b+c)^2}{4c}} dx=\frac{1}{c}\int \frac{1}{(x-\frac{b+c}{2c})^2+\frac{a}{c}-\frac{(b+c)^2}{4}} $

By defining $y:=x-\frac{b+c}{2c}$ and $p^2:=\frac{a}{c}-\frac{(b+c)^2}{4}$ we get:

$\frac{1}{c}\int \frac{1}{(x-\frac{b+c}{2c})^2+\frac{a}{c}-\frac{(b+c)^2}{4}}=\frac{1}{c}\int\frac{1}{y^2+p^2}=\frac{1}{c}\frac{arctan(\frac{y}{p})}{p}+k_1=\frac{1}{c}\frac{arctan(\frac{x-\frac{b+c}{2c}}{\sqrt{\frac{a}{c}-\frac{(b+c)^2}{4}}})}{\sqrt{\frac{a}{c}-\frac{(b+c)^2}{4}}}+k_1$

for some arbitrary number $k_1$.

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