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I am trying to solve this differential equation analytically

$$\frac{dy}{dt}=ry(\frac{y}{\theta}-1)(1-\frac{y}{K})$$

and I have started by separating the variables to each side

$$\frac{dy}{\left(\frac{y}{\theta }-1\right)\left(1-\frac{y}{K}\right)y}=rdt$$

now however I am uncertain about how to continue, is it correct to simply split up the left side into three separate fractions?

$\frac{dy}{\left(\frac{y}{\theta }-1\right)}$ , $\frac{dy}{\left(1-\frac{y}{K}\right)}$ and $\frac{dy}{y}$

If so I know the rightmost fraction is ln y but how about the other fractions? I know a trick that is available is

$\frac{u'}{u}=ln\:u\:$ but not how/if I can apply it here.

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  • $\begingroup$ Perform a partial fraction decomposition and integrate. $\endgroup$ – Aryadeva Sep 9 '20 at 9:20
  • $\begingroup$ I guess you mean $(\ln u)'=\frac{u'}{u}$. $\endgroup$ – Miguel Sep 9 '20 at 9:35
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Hint: $$\frac{dy}{\left(\frac{y}{\theta }-1\right)\left(1-\frac{y}{K}\right)y}=rdt$$ Use fraction decomposition method and write it as: $$\dfrac 1 {\left(\frac{y}{\theta }-1\right)\left(1-\frac{y}{K}\right)y}=\frac{A}{\frac y{\theta} -1}+\dfrac B{(1-\frac{y}{K})}+\dfrac Cy$$ Find the constants $A,B,C$ then integrate.


$$1=y^2\left(-\frac{A}{K}+\frac{B}{\theta }-\frac{C}{\theta K}\right)+y\left(A-B+\frac{C}{\theta }+\frac{C}{K}\right)+\left(-C\right)$$ $$ \implies \left(-\frac{A}{K}+\frac{B}{\theta }+\frac{1}{\theta K}\right)=0$$ and also: $$\left(A-B-\frac{1}{\theta }-\frac{1}{K}\right)=0$$ It's a system of two equations. From the second equation you have ; $$A=-\left(-B-\frac{1}{\theta }-\frac{1}{K}\right)$$ Put $A$ iin the first equation and deduce $B$

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    $\begingroup$ You're welcome @P.ython then integrate with $ln$ function. $\endgroup$ – Aryadeva Sep 9 '20 at 9:27
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    $\begingroup$ collect the terms with $y^2$ since you have no y on the left side of the equation the coefficients should be equal to zero. Do the same for the coefficents of y . and collect the others terms and set them equal to 1 since you have 1 on the left side @P.ython $\endgroup$ – Aryadeva Sep 9 '20 at 9:46
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    $\begingroup$ Yes thats correct $C=-1$ Find A and B @P.ython $\endgroup$ – Aryadeva Sep 9 '20 at 9:52
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    $\begingroup$ No it cant be zero @P.ython it's the sum of coefficents of $y^2$ and of $y$ that is zero $\endgroup$ – Aryadeva Sep 9 '20 at 9:55
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    $\begingroup$ @P.ython you already know that $c=-1$ so put $-1$ instead of c in your equation thats first then the coefficents of $y^2$ is equal zero and that of $y$ too you have a system of two equations for A and B $\endgroup$ – Aryadeva Sep 9 '20 at 10:14

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