I have the following differential equation:
$y''(x) + \frac{1}{x}y'(x)+\frac{\lambda x}{D}y(x) = 0$
From what I understand, DE's of this kind have a general solution of:
$y(x)=Ay_1(x)+By_2(x)$
but at least $y_1$ or $y_2$ must be known to solve completely (I don't know either) - but that sometimes it is possible to make an initial guess for $y_1$ or $y_2$.
Are there any other options for solving this? Perhaps Laplace transforms?
EDIT:
Boundary conditions are
$y(0)=A$
$y(X)=B$
$$y''(x) + \frac{1}{x}y'(x)+Cxy(x) = 0 \qquad C=\frac{\lambda}{D}$$ Obviously this is an ODE of the Bessel kind (generalized). A convenient change of variable $X=\beta x^\gamma$ reduces it to the standard Bessel form.
Lasy people can use ready made formula : Eqs.$(6-7)$ in https://mathworld.wolfram.com/BesselDifferentialEquation.html giving $\alpha=0$ , $\beta=\frac23\sqrt{C}$ , $\gamma=\frac32$ , $n=0$ with the symbols in this paper. $$y(x)=c_1 J_0\left(\frac23\sqrt{C}x^{3/2} \right)+c_2 Y_0\left(\frac23\sqrt{C}x^{3/2} \right)$$ $J_0$ and $Y_0$ are the Bessel functions of first and second kind.
UPDATE after the addition of conditions in the question :
Condition $y(0)=A$
Since $Y_0(0)$ is infinite this implies $c_2=0 \quad\implies\quad y(x)=c_1 J_0\left(\frac23\sqrt{C}x^{3/2} \right)$
And $J_0(0)=1\quad\implies\quad c_1=A$
$$y(x)=A\;J_0\left(\frac23\sqrt{C}x^{3/2} \right)$$
With condition $y(X)=B$ : $$\begin{cases} \text{if } B\neq A\;J_0\left(\frac23\sqrt{C}X^{3/2} \right) \text{ there is no solution.}\\ \text{if } B= A\;J_0\left(\frac23\sqrt{C}X^{3/2} \right)\text{ the solution is }y(x)=A\;J_0\left(\frac23\sqrt{C}x^{3/2} \right) \end{cases}$$ Thus in the general case (any $A$ and $B$ except the particular value of $B$ defined above) there is no solution which satisfies both the ODE and the two conditions.