For the sulotion of the OED: \frac{dy}{dx}=1+x+y^{2}+xy.

For the sulotion of the OED: $\frac{dy}{dx}=1+x+{y}^{2}+xy$.
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yotaniwc
Changing $y=u-\frac{z}{2}$ the equation is cast to the form
$\frac{du}{dx}={u}^{2}+{\left(\frac{x}{2}-1\right)}^{2}+\frac{5}{2}$
which does not seem to possess solutions in terms of elementary functions
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limacarp4
if we know ${y}_{p}=-\frac{{x}^{2}-2x-1}{{x}^{2}-4x+3}$ is a particular solution. So we can find general solution of Riccati Equation. (Note:I did not try it .I trust GEdgar. Thanks to him. I just offer a general solution If we know a particular solution)
Use transform of $y={y}_{p}+\frac{1}{H}$
$\frac{dy}{dx}=1+x+{y}^{2}+xy$.
${y}_{p}^{{}^{\prime }}+\left(\frac{-H}{{H}^{2}}\right)=1+x+\left({y}_{p}^{2}+\frac{2{y}_{p}}{H}+\frac{1}{{H}^{2}}\right)+x{y}_{p}+\frac{x}{H}$
${y}_{p}^{{}^{\prime }}=1+x+{y}_{p}^{2}+x{y}_{p}+\frac{{H}^{\prime }}{{H}^{2}}+\frac{2{y}_{p}}{H}+\frac{1}{{H}^{2}}+\frac{x}{H}$
if ${y}_{p}$ is a solution of the equation then it must satisfy
${y}_{p}^{{}^{\prime }}=1+x+{y}_{p}^{2}+x{y}_{p}$
Thus $0=\frac{{H}^{\prime }}{{H}^{2}}+\frac{2{y}_{p}}{H}+\frac{1}{{H}^{2}}+\frac{x}{H}$