# Differential equations system's solutions differ when one of the constant is 0

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 $$x=0$$.

• 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. – Bill Watts Oct 21 at 17:50
• 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. – Bob Hanlon Oct 21 at 18:05