How to find the boundary conditions of the differential equation? I have a very complicated differential equation that can not be solved analytically and I show a on the simple example:
Example 1:
$$y''(x)-y(x)=0\tag{1}$$
with boundary conditions:
  $$y(1)=1,y'(2)=1\tag{2}$$ 
solution is:
$$y(x)=\frac{e^{-x-1} \left(e^{2 x}+e^{2 x+1}-e^3+e^4\right)}{1+e^2}\tag{3}$$
then  using the substitution of 
    $$v(x)=\frac{y(x)}{x}$$
I have a NEW equation:
$$x v''(x)+2 v'(x)-x v(x)=0\tag{4}$$
How to find the boundary conditions of the NEW  differential equation?
$$v(?)=?,v'(?)=?$$
EDITED!
$$v(x)= \frac{y(x)}{x}\tag{5}$$
$$v(1)=\frac{y(1)}{1}=\frac{1}{1}=1$$
$$v'(x)= \frac{y'(x)x-y(x)}{x^2}=\frac{y'(2)*2-y(x)}{2^2}= \frac{1*2-y(x)}{2^2}$$
I don't have a $y(2)=?$
Assume $y(x)$ is $y(1)=1$ then:
$$v'(2)= \frac{1*2-1}{2^2}=1/4$$
The new boundary conditions are:
$$v(1)=1,v'(2)=1/4\tag{6}$$Solution with New equation$(4)$ and new boundary conditions $(6)$ is:
$$v(x)=\frac{e^{-x-1} \left(3 e^{2 x}+e^{2 x+1}-e^3+e^4\right)}{\left(3+e^2\right) x}$$
 then we substitute to $(5)$  and check solutions is equal:
$$\frac{e^{-x-1} \left(3 e^{2 x}+e^{2 x+1}-e^3+e^4\right)}{\left(3+e^2\right)}\neq\frac{e^{-x-1} \left(e^{2 x}+e^{2 x+1}-e^3+e^4\right)}{1+e^2}$$
Is NOT !!.
Example 2:
$$y''(x)-y(x)=0$$
with boundary conditions:
  $$y'(1)=1,y'(2)=1$$
How to find the boundary conditions of the NEW differential equation? 
 A: Hint:
we have :
$$
v(x)=\frac{y(x)}{x}
$$
than we find :
$$
v'(x)= \frac{d}{dx}v(x)=\frac{d}{dx}\frac{y(x)}{x}= \frac{y'(x)x-y(x)}{x^2}
$$
now put $x=1$ and $x=2$ and use the given initial conditions for $y(1)$ and $y'(2) $.

The  starting equation $y''(x)-y(x)=0$ has solution: $y=ae^{-x}+be^x$.
From the boundary condition $y(1)=1$ we can find:
$$
\frac{a}{e}+eb=1 \Rightarrow a=e(1-eb)
$$ 
For the derivative $y'=-ae^{-x}+be^x $ the other condition $y'(2)=1$ gives:
$$
\frac{-a}{e^2}+e^2b=1 
$$ 
and substituting the value of $a$ we have:
$$
-1+eb+e^3b=e \Rightarrow b=\frac{e+1}{e(e^2+1)}
$$
and
$$
a=\frac{e^2(e-1)}{e^2+1}
$$
and this solve the problem.  But you want solve the same problem with the substitution $v(x)=\frac{y(x)}{x}$. So the equation becomes:
$$
xv''(x)+2v'(x)-xv(x)=0
$$
that has obviously the solution:
$$
v(x)=\frac{ae^{-x}}{x}+\frac{be^{x}}{x}
$$
Now we use the new conditions to find the constants $a,b$.
From $v(1)=\frac{y(1)}{1}=1$ we find:
$$
a=e(1-eb)
$$
so:
$$
(1)\qquad \quad v(x)=\frac{e(1-eb)e^{-x}}{x}+\frac{be^x}{x}
$$
and 
$$
v'(x)= \frac{-e(1-eb)xe^{-x}-e(1-eb)e^{-x}}{x^2}+\frac{bxe^x-be^x}{x^2}
$$
so:
$$
(2) \qquad \quad v'(2)=\frac{3(eb-1)}{4e}
$$
Now use the condition for $v'(2)$ noting that $y(x)=xv(x)$ 
$$
v'(2)=\frac{2\cdot y'(2)-2\cdot v(2)}{4}=\frac{1-v(2)}{2}
$$
Now we can substitute $v(2)$ from the equation $(1)$ and, with a bit of algebra, we find:
$$
v(2)=\frac{2e-1+eb-e^3b}{4e}
$$
Equating with $(2)$ we finally find the value of $b$:
$$
b=\frac{e+1}{e(e^2+1)}
$$
so you  see that the constants are the same in the two functions.
(I hope that i've no typos !)
