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Trying to work backward to the solution for this nonlinear ODE in a control book I have:

$\frac{\mathrm{dx} }{\mathrm{d} t}=-x+x^2$

Book lists the solution to the ODE as:

$x(t)=\frac{x_0e^-t}{1-x_0+x_0e^-t}$

I used the ODE45 function in MATLAB to generate a solution from a family of initial conditions and I get the same plot as in the book.

I run into issues when I try to analytically solve this equation through separation of variables:

$\int \frac{dx}{-x+x^2}=\int dt$

$ln\left|\frac{x+1}{x} \right|=t+C$

Add my initial conditions

$ln\left|\frac{x+1}{x} \right|=t+ln\left|\frac{x_0+1}{x_0}\right|$

Multiply both sides by e

$\frac{x+1}{x}=e^t\left ( \frac{x_0+1}{x_0} \right )$

Rearrange

$x=\frac{1}{e^t\left ( \frac{x_0+1}{x_0} \right )-1}$

I use dsolve on Matlab to try come up with a general solution and I get almost the same as I worked by hand:

enter image description here

However, it does not match the plot from the textbook.

Unsure where I am going wrong. Can someone review my working.

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    $\begingroup$ From $\int \frac{dx}{x(-1+x)}$ you can not get a factor $(x+1)$ in the integral. $\endgroup$ Commented Apr 20, 2023 at 10:13
  • $\begingroup$ Thank you @LutzLehmann and Goncalo $\endgroup$
    – SS1
    Commented Apr 20, 2023 at 10:19
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    $\begingroup$ It is in the end simpler to treat this as Bernoulli equation, that is, divide by $-x^2$ and get a linear DE. $\endgroup$ Commented Apr 20, 2023 at 10:32

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