2.6.5.4 Solved using first_order_ode_riccati_by_guessing_particular_solution

Entering first order ode riccati guess solver

\[\begin {aligned} y^{\prime }&=x \left (a y^{2}+b \right ) \end {aligned}\]

This is a Riccati ODE. Comparing the above ODE to solve with the Riccati standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that

\begin{align*} f_0(x) & =b x\\ f_1(x) & =0\\ f_2(x) &=a x \end{align*}

Using trial and error, the following particular solution was found

\begin{align*} y_p &= \frac {\sqrt {-a b}}{a} \end{align*}

Since a particular solution is known, then the general solution is given by

\begin{align*} y &= y_p + \frac { \phi (x) }{ c_1 - \int { \phi (x) f_2 \,dx} } \end{align*}

Where

\begin{align*} \phi (x) &= e^{ \int 2 f_2 y_p + f_1 \,dx } \end{align*}

Evaluating the above gives the general solution as

\begin{align*} y = \frac {b \left ({\mathrm e}^{\sqrt {-a b}\, x^{2}} a +2 \sqrt {-a b}\, c_1 \right )}{a \left (\sqrt {-a b}\, {\mathrm e}^{\sqrt {-a b}\, x^{2}}+2 c_1 b \right )} \end{align*}

Summary of solutions found

\[ y = \frac {b \left ({\mathrm e}^{\sqrt {-a b}\, x^{2}} a +2 \sqrt {-a b}\, c_1 \right )}{a \left (\sqrt {-a b}\, {\mathrm e}^{\sqrt {-a b}\, x^{2}}+2 c_1 b \right )} \]