Function that transforms a differential equation What function $u$ can transform the equation $$fy''-4f'y'+gy=0$$ to an equation of the form $$v''+kv=0$$. Here, $f,g,y,k,u,v$ are all functions of $x$ and $y=uv$. 
Is there any theory for such type of transformations? By simple calculation, it is seen that $v''=fy''-4f'y'$. But how to proceed further? Any hints? Thanks beforehand.
 A: As I know there is no any closed form method to find the general solution of  $y''+h(x)y'+p(x)y=0$ without knowing a particular solution. You will need to write infinite series to express the solution.  If you know a particular solution $y_p$, it is easy to find the general solution. I would like to offer  my approach to find a particular solution then how to find general solution . 
More generally, we can write that second order homogeneous linear differential equation: 
$$y''+h(x)y'+p(x)y=0$$
$$y=z.w$$
$$z''.w+2z'w'+zw''+h(x)z'w+h(x)zw'+p(x)zw=0$$
$$z''.w+z'(2w'+h(x)w)+z(w''+h(x)w'+p(x)w)=0$$
Let's 
$$2w'+h(x)w=0$$
$$w=c.e^{-\frac{1}{2}\int{h(x)}dx}$$  where c is a constant.
We will get 
$$z''+s(x)z=0$$
$$s(x)=\frac{w''+h(x)w'+p(x)w}{w}$$
$$s(x)=-\frac{h'(x)}{2}-\frac{h^2(x)}{4}+p(x)$$
Let's transform into  via using;
$$u=\frac{z'}{z}$$
Then If we derivate both sides 
$$u'=\frac{z''}{z}-(\frac{z'}{z})^2$$
$$u'=\frac{-s(x)z}{z}-(u)^2$$
$$u'+u^2=-s(x)$$
$$u'+u^2=\frac{h'(x)}{2}+\frac{h^2(x)}{4}-p(x)$$
$$u'+u^2=f(x)=\frac{h'(x)}{2}+\frac{h^2(x)}{4}-p(x)$$
$$u=\frac{f(x)}{\frac{f'(x)}{2f(x)}+u_1} $$ on Riccati Differential Equation
We get same type Riccati Differantial Equation after the transform.
$$u_1 '+u_1^{2}=f_1(x)=f(x)+(\frac{-f'(x)}{2f(x)})^{2}+(\frac{-f'(x)}{2f(x)})'$$
Then, if  we apply the transform  $$u_1=\frac{f_1(x)}{\frac{f_1'(x)}{2f_1(x)}+u_2} $$  on $u_1 '+u_1^{2}=f_1(x)$ , we get the same type equation  again $$u_2 '+u_2^{2}=f_2(x)=f_1(x)+(\frac{-f_1'(x)}{2f_1(x)})^{2}+(\frac{-f_1'(x)}{2f_1(x)})'$$ 
It never ends , we always get the same type equation ($y'+y^{2}=f(x)$) , Thus I named it as endless transform.
Finally, the particular solution can be written in infinite terms as
$$u_p(x)=\frac{f(x)}{\frac{f'(x)}{2f(x)}+\frac{f_1(x)}{\frac{f_1'(x)}{2f_1(x)}+\frac{f_2(x)}{\frac{f_2'(x)}{2f_2(x)}+.....}} } $$
Where $$f_{n+1}(x)=f_n(x)+(\frac{-f_n'(x)}{2f_n(x)})^{2}+(\frac{-f_n'(x)}{2f_n(x)})'$$
and $$f_0(x)=f(x)$$
$u_0(x)=u_p(x)  $  is our particular solution
$u=u_p+\frac{1}{H} $
$u_p'+(\frac{-H'}{H^{2}})+u_p^{2}+\frac{2u_p}{H}+\frac{1}{H^{2}}=f(x) $
$\frac{-H'}{H^{2}}+\frac{2u_p}{H}+\frac{1}{H^{2}}=0 $
$H'-2u_p.H=1 $
$H(x)=e^{2\int{u_p}dx}\int{e^{-2\int{u_p}dx}}dx $
$u(x)=u_p(x)+\frac{e^{-2\int{u_p(x)}dx}}{\int{e^{-2\int{u_p(x)}dx}}dx} $
(This is general solution)
I do not know the name of this transform in literature, So I named it as endless transform. Maybe someone else can help to find what the name of the transform is in literature.
I posted it in other questions before here .I hope it can be helpful for you. 
