Solving Numerically a second order ODE I have a second order ordinary differential equation to solve:
The equation is as follows:
$a(x)P'(x)+b(x)P''(x)=0$, where $x \in [0,\infty)$ and $P$ is a probability distribution. 
When $0 \leq x \leq 10$, $a(x)= 3-2x$ and $b(x) = \dfrac{3-2x}{\ln(1.5x)}$. 
When $x\geq 10$, the coefficients are constants, $a(x)=-27$ and $b(x)=\dfrac{-27}{\ln(20)}$.
So I split up my ODE as follows:
$(3-2x)P_{1}'(x)+\dfrac{2-3x}{\ln(1.5x)}P_{1}''(x)=0$ for $0\leq x\leq 10$,
and
$-27P_{2}'(x)-\dfrac{27}{\ln(20)}P_{1}''(x)=0$ for $10\leq x< \infty$.
I have the following conditions:
$P_1(10)=P_2(10)$. 
$P_1'(10)=P_2'(10)$
$$\int_0^{10}P_1(x)\,dx + \int_{10}^{\infty}P_2(x)\,dx=1$$
and 
$\displaystyle\lim_{x\to\infty }P_2(x)=0$. 
$P$ is a probability distribution and in this case I can replace $\infty$ by large number, say, $1000,$ and that should be fine. I.e., $P_2(1000)=0.$
I don't know how to initialize my numerical scheme since I do not have an explicit expression for the boundary condition at $x=0$.
 A: Start with the solution in the region $10 \leq x < \infty$, in that region the solution is fairly simple to get
$$
P_2(x) = \alpha + \beta ~ 20^{-x} \tag{1}
$$
And you can get the value of one of the constants by forcing the solution to go to zero at infinity
$$
\lim_{x\to \infty} P(x) = \lim_{x\to \infty} P_2(x) = 0 = \alpha \tag{2}
$$
So the solution so far for this region is
$$
P_2(x;\beta) = \beta ~ 20^{-x}
$$
So the idea is to find a solution for $P_1$ that smoothly connects with $P_2$ at $x=10$. A possible way is to define a function 
\begin{eqnarray}
f(\beta) &=& \int_{0}^{+\infty}P(x) {\rm d}x - 1 = \int_{0}^{10}P_1(x) {\rm d}x + \int_{10}^{+\infty}P_2(x) {\rm d}x- 1 \\
&=& 
\int_{0}^{10}P_1(x) {\rm d}x + \frac{\beta}{20^{10}\ln 20} - 1
\end{eqnarray}
and find the solution of the problem $f(\beta) = 0$.
For each value of $\beta$ you find the value of $P_2(x)$ and use that as boundary conditions for the numerical integration of $P_1$. Of all possible values of $\beta$ you choose the one that normalizes the probability
