Method of conversion to first order Riccati

ode internal name This is currently not implemented.

Given linear second order ode \(A\left ( x\right ) y^{\prime \prime }+B\left ( x\right ) y^{\prime }+C\left ( x\right ) y=0\) then using the transformation \(v\left ( x\right ) =-\frac {y^{\prime }}{y}\) converts the second order ode to a first order Riccati

\begin{align*} v^{\prime } & =\frac {-yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}}{y^{2}}\\ & =\frac {-y\left ( -\frac {B}{A}y^{\prime }-\frac {C}{A}y\right ) +\left ( y^{\prime }\right ) ^{2}}{y^{2}}\\ & =\frac {\frac {B}{A}yy^{\prime }+\frac {C}{A}y^{2}+\left ( y^{\prime }\right ) ^{2}}{y^{2}}\\ & =\frac {B}{A}\frac {y^{\prime }}{y}+\frac {C}{A}+\frac {\left ( y^{\prime }\right ) ^{2}}{y^{2}}\\ & =\frac {C}{A}+\frac {B}{A}v+v^{2}\end{align*}

Which is Riccati of the form \(v^{\prime }=f_{0}\left ( x\right ) +f_{1}\left ( x\right ) v+f_{2}v^{2}\). where \(f_{0}=\frac {C}{A},f_{1}=\frac {B}{A},f_{2}=1\). Lets say we can now find the solution to this Riccati \(v\left ( x\right ) \) (see section earlier on Riccati for algorithm). Then the solution to the second order ode is found from \(y^{\prime }=-yv\) by solving this first order ode. The solution is

\[ y=e^{-\int v\left ( x\right ) dx}+c_{2}\]

Notice there is also a second constant of integration inside \(v\left ( x\right ) \). This method of course works only if we can solve the generated Riccati ode which does not have a general method for solving and only for specific cases it can be solved. So this will be tried as last resort.

We want to look for reduced Riccati generated from the above, which is \(v^{\prime }=f_{0}+f_{2}v^{2}\). Which means \(f_{1}=0\) or \(B=0\) in the hope of solving the Riccati. This means ode of the form \(A\left ( x\right ) y^{\prime \prime }+C\left ( x\right ) y=0\) will have hope of solving using this Riccati conversion method. See Riccati section why that is.