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By generalizing doraemonpaul's answer to Particular solution to a Riccati equation $y' = 1 + 2y + xy^2$ I have the following result.

Let $q_0(x)$, $q_1(x)$ and $q_2(x)$ by smooth functions (meaning the functions have a second derivative which is continuous). Now define the following quantities: \begin{eqnarray} P_1(x)&:=&\exp\left(\frac{1}{2} \int q_1(x) dx \right)\\ Q_1(x)&:=& q_0(x) q_2(x) - \frac{1}{4} \left( -\frac{q_0(x)^{'}}{q_0(x)} + q_1(x)\right)^2 - \frac{1}{2} \left( -\frac{q_0(x)^{'}}{q_0(x)} + q_1(x)\right)^{'}\\ Q_2(x)&:=& q_0(x) q_2(x) - \frac{1}{4} \left( -\frac{q_2(x)^{'}}{q_2(x)} - q_1(x)\right)^2 - \frac{1}{2} \left( -\frac{q_2(x)^{'}}{q_2(x)} - q_1(x)\right)^{'} \end{eqnarray} Clearly $Q_2(x)$ is produced out of $Q_1(x)$ under the transformation $(q_0(x),q_1(x),q_2(x)) \rightarrow (-q_2(x),-q_1(x),-q_0(x))$.

Consider a 2nd order ODE: \begin{equation} v^{''}(x) + Q_2(x) \cdot v(x)=0 \end{equation} whose solution $v(x)$ is known.

Now define a new function $V(x)$ as follows: \begin{eqnarray} V(x)&:=& \frac{1}{\sqrt{q_0(x)}} P_1(x) \cdot \exp\left(-\int q_0(x) q_2(x) \frac{\sqrt{q_2(x)} P_1(x) v(x)}{(\sqrt{q_2(x)} P_1(x) v(x))^{'}}dx \right)\\ v(x)&:=& \frac{1}{\sqrt{q_2(x)}} \frac{1}{P_1(x)} \cdot \exp\left(-\int q_0(x) q_2(x) \frac{\sqrt{q_0(x)} \frac{1}{P_1(x)} V(x)}{(\sqrt{q_0(x)} \frac{1}{P_1(x)} V(x))^{'}}dx \right) \end{eqnarray} Then the function $V(x)$ satisfies the follwoing 2nd order ODE: \begin{equation} V^{''}(x) + Q_1(x) \cdot V(x)=0 \end{equation}

As usual we provide a Mathematica code snippet for that:

In[1310]:= Clear[v]; Clear[V]; Clear[Q1]; Clear[Q2]; Clear[P1]; \
Clear[q0]; Clear[q1]; Clear[q2];
Q1[x_] = (q0[x] q2[x] - 1/4 (-q0'[x]/q0[x] + q1[x])^2 - 
    1/2 D[(-q0'[x]/q0[x] + q1[x]), x]);
Q2[x_] = (q0[x] q2[x] - 1/4 (-q2'[x]/q2[x] - q1[x])^2 - 
    1/2 D[(-q2'[x]/q2[x] - q1[x]), x]);
P1[x_] = Exp[1/2 Integrate[q1[x], x]];
myeqn = V''[x] + Q1[x] V[x];
V[x_] = 1/Sqrt[q0[x] ] P1[
    x] Exp[-Integrate[
      q0[x] q2[x]  (Sqrt[q2[x]] P1[x] v[x])/
       D[Sqrt[q2[x]] P1[x] v[x], x], x]];
FullSimplify[
 Simplify[myeqn/V[x]] /. Derivative[2][v][x] :> -Q2[x] v[x]]

Out[1316]= 0

The result follows from comparing two 2nd order linear ODEs have being produced firstly from a Riccati equation and secondly from the same Ricatti equation where the dependent variable has been replaced by its inverse.

Note that this result can be quite useful as one can see in the following examples.

Example 1:

Take : \begin{eqnarray} q_0(x)&:=&1\\ q_1(x)&:=&2\\ q_2(x)&=&x \end{eqnarray} then $Q_1(x)=(x-1)$ and $Q_2(x)=(x-1)-1/x-3/4 \cdot 1/x^2$ . Clearly the ODE related to $Q_1$ is solved in terms of Airy functions whereas the other one is not solved in terms of any known special function (i.e. those that are available in Mathematica for example) and one has to use Heun function to solve it.

Example 2:

Take: \begin{eqnarray} q_0(x)&:=&a x^2+b x + c\\ q_1(x)&:=&b_1 x+ c_1\\ q_2(x)&=&c_2 \end{eqnarray} Then \begin{eqnarray} Q_1(x) &=& x^2 \left(a c_2-\frac{b_1^2}{4}\right)+x \left(b c_2-\frac{b_1 c_1}{2}\right)-\frac{b_1}{2}+c c_2-\frac{c_1^2}{4}\\ Q_2(x) &=& Q_1(x) + \frac{{\mathcal M}(x)}{4 \left(a x^2+b x+c\right)^2} \end{eqnarray} where \begin{eqnarray} {\mathcal M}(x):=x^2 \left(-8 a^2-6 a b c_1+4 a b_1 c+2 b^2 b_1\right)-2 x \left(4 a b+2 a c c_1+b^2 c_1-3 b b_1 c\right)+2 a x^3 (b b_1-2 a c_1)+4 c (a+b_1 c)-3 b^2-2 b c c_1 \end{eqnarray}

Again, clearly the ODE related to $Q_1$ is solved in terms of Gaussian functions and Hermite polynomials whereas the other ODE (related to $Q_2$) has unknown solutions (at least Computer Algebraic systems wouldn't be able to cope with it).

Now my question would be what other transformations of ODEs can one come up with in order to express solutions of 2nd order linear ODEs with complicated coefficients through those with simpler coeffcients?

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Hint:

A general Riccati ODE $y'=q_0(x)+q_1(x)y+q_2(x)y^2$ itself can generate another classes of other Riccati ODEs.

Let $y=a(x)u$ ,

Then $y'=a(x)u'+a'(x)u$

$\therefore a(x)u'+a'(x)u=q_0(x)+q_1(x)a(x)u+q_2(x)a^2(x)u^2$

$a(x)u'=q_0(x)+(q_1(x)a(x)-a'(x))u+q_2(x)a^2(x)u^2$

And repeat your process in above can generate a network of second-order linear ODEs.

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  • $\begingroup$ @ doraemonpaul Unfortunately this idea does not bring anything new. The mapping $\left(q_0(x), q_1(x), q_2(x) \right) \rightarrow \left(q_0/a(x), q_1(x) - a^{'}(x)/a(x), q_2(x) a(x)\right)$ leaves both $Q_1(x)$ and $Q_2(x)$ unaltered. $\endgroup$ – Przemo Oct 22 '18 at 10:47

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