Quadratic differential equation with different conditions Consider the equation $$x'=-\frac{1}{2}x^{2}+c$$ with the following cases for $c$:


*

*$c=0$,

*$c<0$,

*$c>0$.


For the first case, I found the solution 
$$x=\frac{2}{t-c_1}$$ 
(where $c_1$ is arbitrary, not necessarily the same as $c$), but for the other two solutions, I am unsure of how to proceed.
I appreciate any assistance. Thanks, Ciwan.
 A: As with general Riccati equations, you can also transform it using the parametrization of solutions $x=\frac{2u'}u$ leading to
$$
x'=\frac{2u''}{u}-\frac{2u'^2}{u^2}=-\frac12\frac{4u'^2}{u^2}+c\implies u''-\frac c2 u=0.
$$
Now you can apply the knowledge about linear DE with constant coefficients.
A: You can use Separation of variables. First, your equation is
$$\frac{dx}{dt} = -\frac{1}{2}x^2 + c \; \implies \; \frac{dx}{-\frac{1}{2}x^2 + c} = dt \tag{1}\label{eq1}$$
Integrating both sides gives
$$\int \frac{dx}{-\frac{1}{2}x^2 + c} = \int dt \tag{2}\label{eq2}$$
As you've done the case where $c = 0$, consider the non-zero cases. Let
$$y = \frac{1}{\sqrt{2|c|}}x \tag{3}\label{eq3}$$
so $-\frac{1}{2}x^2 = -|c|y^2$ and $dy = \frac{1}{\sqrt{2|c|}}dx$, i.e., $dx = \left(\sqrt{2|c|}\right)dy$. With this, \eqref{eq2} becomes
$$-\sqrt{\frac{2}{|c|}}\int \frac{dy}{y^2 - \text{sgn}(c)} = \int dt \tag{4}\label{eq4}$$
If $c \lt 0$, then $- \text{sgn}(c)$ becomes $+ 1$ so \eqref{eq4} can be solved to get
$$-\sqrt{\frac{2}{|c|}}\arctan(y) = t + c_1 \tag{5}\label{eq5}$$
Using \eqref{eq3} in \eqref{eq5} gives
$$-\sqrt{\frac{2}{|c|}}\arctan\left(\frac{x}{\sqrt{2|c|}}\right) = t + c_1 \tag{6}\label{eq6}$$
You can get $x$ by itself by dividing by the left coefficient, taking the tangent of both sides and then dividing by the coefficient in the arctan. I'll leave this to you to finish.
For the case where $c \gt 0$, then $- \text{sgn}(c)$ becomes $- 1$ and the solution is fundamentally different. In this case, since $y^2 - 1 = (y - 1)(y + 1)$, you should use Partial fraction decomposition to get
$$\frac{1}{y^2 - 1} = \frac{1}{2(y-1)} - \frac{1}{2(y+1)} \tag{7}\label{eq7}$$
With this, \eqref{eq4} can now be solved to get
$$-\frac{1}{\sqrt{2|c|}}\log\left(\frac{y-1}{y+1}\right) = t + c_1 \tag{8}\label{eq8}$$
I'll leave the substitution and other manipulations to get $x$ by itself in terms of $t$ for you to finish.
