I am solving the following system of equations with Mathematica. $u'=-k_{1}ux+k_{2}x^2, x'=k_{1}ux-k_{2}x^2$ and I obtain the solution for x: $x(t)=\frac{e^{k_{1}C_{1}(t+C_{2})}k_{1}C_{1}}{-1+e^{k_{1}C_{1}(t+C_{2})}(k_{1}+k_{2})}$. When I solve the system when $k_{1}=0$ I obtain the solution for x: $x(t)=\frac{x_{0}}{k_{2}tx_{0}+1}$ but if I substute $k_{1}=0$ in the first solution, I obtain $x=0$.

I could not understand why. Could you please help?

  • $\begingroup$ That is really a mathematics question, but the simple answer is that by setting $k_1=0$, you are changing the form of the differential equation by eliminating a term containing the independent variable, which also changes the form of the solution. $\endgroup$ – Bill Watts Oct 21 at 17:50
  • $\begingroup$ When posting a question you should include the code that you used so that others can reproduce your results. Your solutions include an x0 which is not included in your equations. $\endgroup$ – Bob Hanlon Oct 21 at 18:05
  • $\begingroup$ Well, x0=x(0) simply. $\endgroup$ – Ксения Цочева Oct 21 at 18:08

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