How can I guess particular solution of a Riccati differential equation? I am given the two following differential equations.
$$y'=6+5y+y^2$$
$$y'=9+6y+y^2$$
I have learned how to solve one with a given particular solution but cannot seem to get my head around this. I need help with guessing the particular solution of both of these.
 A: $$y'=6+5y+y^2=\frac{dy}{dx}$$
There is no need for particular solution. Solving is straightforward since it is an autonomous and separable EDO.
$$dx=\frac{dy}{6+5y+y^2}$$
$$x=\int \frac{dy}{6+5y+y^2} = \ln\Big|\frac{y+2}{y+3}\Big|+\text{constant}$$
$$\frac{y+2}{y+3}=c\:e^x$$
$$y(x)=\frac{2-3c\:e^x}{c\:e^x-1}$$
By the way, a particular solution is $y(x)=-2$ which corresponds to the particular case $c=0$. Another particular solution is $y(x)=-3$ which corresponds to the particular case $c=\infty$.
Proceed on the same manner to solve the second ODE, which is even simpler to integrate.
NOTE:
If you really insist to solve the equations as Riccati ODE, you don't necessarily need guessing a particular solution. There is an alternative method without requiring to find a particular solution first. See :
General Solution to Ricatti Differential Equations
A: This might seem a difficult one at first sight  but believe me this is almost the same as one with a given particular solution. You can easily find a particular solution by putting whatever is on the right hand side equal to zero and by a simple factorization you'll get two solutions. Either of them can be your particular solution i.e your "$y_1$". For the above problem $y'=6+5y+y^2$ which by coincidence  I recently did a minute ago for my assignment is as follows
$$6+5y+y^2 = 0 $$
$$ 6+3y+2y+y^2 = 0 $$
$$ (y+3)(y+2) = 0 $$
Either $y=-3$ or $y=-2$, this is your particular solution i.e $y_1=-2$ or $y_1=-3$. Hope this helps. 
