\textbf{Riccati differential equation equation}

,,\qquad \boxed{y' = Py^2 + Qy + R}

\textbf{General solution}
If \[P\equiv 0 \] reduces to a linear differential equation and if \[Q\equiv 0\] it reduces to Bernoulli's equation.

Solving Riccati equation using simple integration is not possible unless one at least one particular solution \[y_1\] is known, in which case a general solution is obtained by substituting \[y = y_1 + \frac{1}{z}\]
,,\qquad y_1' - \frac{z'}{z_2} = P\left(y_1^2 + \frac{2y_1}{z} + \frac{1}{z^2}\right) + Q\left(y_1 + \frac{1}{z}\right) + R

Because \[y_1\] is known to be a solution, \[y_1' = Py_1^2 + Qy_1 + R\] will cancel resulting in:

,,\qquad z' + (2Py_1 + Q)z + P = 0

Solving this linear equation results in a solution of the form \[z=z(x,C)\], so the general solution is:

,,\qquad y = y_1 + \frac{1}{z(x, C)}

A particular solution \[y_1\] can be obtained from the general solution by letting \[C\rightarrow\pm\infty\].

\textbf{Example}

,,\qquad y'  = y^2 -xy + 1

Assuming one particular solution is in the form of  \[y_1 = ax + b\] .

\textbf{Solution}
Substituting \[y = ax+ b\]

,,\qquad a = a^2x + 2abx + b^2 - ax^2 - bx + 1
,,\qquad a^2 - a = 0,\quad 2ab - b=0,\quad a-1 =0
,,\qquad a=1, b= 0

Thus \[y_1 = x\] is a particular solution.
Now substituting \[y = x + \frac{1}z{}\]

,,\qquad 1 - \frac{z'}{z^2} = x^2 + \frac{2x}{z} + \frac{1}{z^2} - x^2 -\frac{x}{z} + 1,\quad z'+ xz = -1

Solving \[z'+xz = -1\] with the formula \[z = e^{-\int x\mathrm{d}x}\left(C - \int e^{-\int x\mathrm{d}x}\mathrm{d}x\right)\]
,,\qquad z = e^{-\frac{x^2}{2}}\left(C - \int e^\frac{x^2}{2}\mathrm{d}x\right)

Thus the general solution to the differential equation is:

,,\qquad y = x + \frac{e^{\frac{x^2}{2}}}{C - \int e^\frac{x^2}{2}\mathrm{d}x}

The integral \[\int e^{x^2/2}\mathrm{d}x\] is the Gaussian integral which of course is unsolvable.