Reduced Riccati with \(n\neq -2\)

For all other cases, there is direct solution to the reduced Riccati given by Eqworld ode0106 and Dr Dobrushkin web page as

\begin{align} w & =\sqrt {x}\left \{ \begin {array} [c]{cc}c_{1}\operatorname {BesselJ}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) & ab>0\\ c_{1}\operatorname {BesselI}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-ab}x^{k}\right ) +c_{2}\operatorname {BesselK}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-ab}x^{k}\right ) & ab<0 \end {array} \right . \tag {2}\\ y & =-\frac {1}{b}\frac {w^{\prime }}{w}\nonumber \\ k & =1+\frac {n}{2}\nonumber \end{align}

If \(n\) satisfies constraint that

\[ \frac {n}{2n+4}\]

Is an integer, then the solution \(y\left ( x\right ) \) will come out using algebraic, exponential and logarithmic functions (including circular functions, such as sin and cosine). If however, \(n\) does not satisfy the above constraint, then (2) can still be used but the solution will come out using Bessel function (also called cylindrical functions).

Hence (2) can be used for any \(n\) to solve the special or reduced Riccati ode.

The constraint that \(\frac {n}{2n+4}\) is an integer, can also be given by saying that \(n=\frac {4k}{1-2k}\) where \(k=\pm 1,\pm 2,\cdots \).

When \(n\) satisfies this, then as mentioned above Eq (2) gives the solution in algebraic, exponential and logarithmic functions. For all other values, Liouville proved no solution exist in terms of elementary functions.

These \(n\) values come out to be \(n=\left \{ \cdots ,-\frac {40}{21},\cdots ,-\frac {8}{5},\frac {-4}{3},-4,-\frac {8}{3},-\frac {12}{5},\cdots ,-\frac {40}{19}\right \} \). We notice that the limit on both ends goes to \(n=-2\) which is the first special case above. Below are two examples to illustrate this. First example will use \(n\) that meets this constraint, and the second example will use \(n\) that does not meet the constraint.

Example 1

\[ y^{\prime }=x^{-4}+y^{2}\]

Comparing this to \(y^{\prime }=ax^{n}+by^{2}\) shows that \(a=1,b=1,n=-4\). We see that \(n\) satisfies that \(\frac {n}{2n+4}=1\) which is integer. Hence we expect that applying (2) will give solution in elementary functions. Since \(ab>0\) then applying

\begin{align*} w & =\sqrt {x}c_{1}\operatorname {BesselJ}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) \\ k & =1+\frac {-4}{2}=1-2=-1 \end{align*}

Hence

\[ w=\sqrt {x}c_{1}\operatorname {BesselJ}\left ( \frac {-1}{2},-x^{-1}\right ) +c_{2}\operatorname {BesselY}\left ( -\frac {1}{2},-x^{-1}\right ) \]

Hence

\[ y=-\frac {w^{\prime }}{w}\]

Simplifying the above gives

\[ y=\frac {1}{x^{2}}\left ( \tan \left ( -\frac {1}{x}+c_{1}\right ) -x\right ) \]

Example 2

\[ y^{\prime }=x^{3}+y^{2}\]

Comparing this to \(y^{\prime }=ax^{n}+by^{2}\) shows that \(a=1,b=1,n=3\). We see that \(n\) do not satisfy that \(\frac {n}{2n+4}=\frac {3}{6+4}=\frac {3}{10}\) being an integer. Hence we expect that applying (2) will give solution in cylinderical functions and not elemetrary functions. Since \(ab>0\) then applying

\begin{align*} w & =\sqrt {x}c_{1}\operatorname {BesselJ}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) \\ k & =1+\frac {3}{2}=\frac {5}{2}\end{align*}

Hence

\[ w=\sqrt {x}c_{1}\operatorname {BesselJ}\left ( \frac {1}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) \]

Hence

\[ y=-\frac {w^{\prime }}{w}\]

Simplifying the above gives

\[ y=\frac {x^{\frac {3}{2}}\left ( -c_{1}\operatorname {BesselJ}\left ( \frac {-4}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) -\operatorname {BesselY}\left ( \frac {-4}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) \right ) }{c_{1}\operatorname {BesselJ}\left ( \frac {1}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) +\operatorname {BesselY}\left ( \frac {1}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) }\]

We see that the solution is in terms of cylinderical functions. Because \(n\) did not satisfy that \(\frac {n}{2n+4}\) is integer. But the main point is that (2) can still be used to solve the special Riccati ode.