Solution to $y''-\frac{2y'}x+cy=0$ by order reduction $$y''-\frac{2y'}x+cy=0$$
I have tried many ways to reduce this form to a first-order equation but they seem to be useless. Can anyone give me a hint?
Is there any method that is better than changing to first-order?
 A: *

*Multiply both sides by $x$:
$$y''(x)-\frac{2\cdot y'(x)}{x}+\text{c}\cdot y(x)=0\Longleftrightarrow x\cdot y''(x)-2\cdot y'(x)+\text{c}\cdot x\cdot y(x)=0$$

*Apply the Laplace transformation, $\mathcal{L}_x\left[f(x)\right]_{(s)}=\int_0^\infty f(x)e^{-sx}\space\text{d}x$:
$$\mathcal{L}_x\left[x\cdot y''(x)-2\cdot y'(x)+\text{c}\cdot x\cdot y(x)\right]_{(s)}=\mathcal{L}_x\left[0\right]_{(s)}$$

*Find the Laplace transformation term by term and factor out constants:
$$\mathcal{L}_x\left[x\cdot y''(x)\right]_{(s)}-2\cdot\mathcal{L}_x\left[y'(x)\right]_{(s)}+\text{c}\cdot\mathcal{L}_x\left[x\cdot y(x)\right]_{(s)}=\mathcal{L}_x\left[0\right]_{(s)}$$

*$$\mathcal{L}_x\left[x\cdot y''(x)\right]_{(s)}=-\frac{\text{d}}{\text{d}s}\left(\mathcal{L}_x\left[y''(x)\right]_{(s)}\right)$$

*$$\mathcal{L}_x\left[y''(x)\right]_{(s)}=s^2\cdot\text{Y}(s)-s\cdot y(0)-y'(0)$$

*$$\mathcal{L}_x\left[y'(x)\right]_{(s)}=s\cdot\text{Y}(s)-y(0)$$

*$$\mathcal{L}_x\left[x\cdot y(x)\right]_{(s)}=-\frac{\text{d}}{\text{d}s}\left(\mathcal{L}_x\left[y(x)\right]_{(s)}\right)$$

*$$\mathcal{L}_x\left[y(x)\right]_{(s)}=\text{Y}(s)$$

*$$\mathcal{L}_x\left[0\right]_{(s)}=0$$

*Rewrite the equation:
$$\text{Y}'(s)+\frac{4\cdot s\cdot\text{Y}(s)}{\text{c}+s^2}=\frac{3\cdot y(0)}{\text{c}+s^2}$$

*Now, let
 $v(s)=\exp\left[\int\frac{4s}{\text{c}+s^2}\space\text{d}s\right]=\left(\text{c}+s^2\right)^2$
 Multiply both sides by $v(s)$:
$$\left(\text{c}+s^2\right)^2\cdot\text{Y}'(s)+\left(4\cdot s\cdot\left(\text{c}+s^2\right)\right)\cdot\text{Y}(s)=3\cdot y(0)\cdot\left(\text{c}+s^2\right)$$

*Substitute $4\cdot s\cdot\left(\text{c}+s^2\right)=\frac{\text{d}}{\text{d}s}\left(\left(\text{c}+s^2\right)^2\right)$:
$$\left(\text{c}+s^2\right)^2\cdot\text{Y}'(s)+\frac{\text{d}}{\text{d}s}\left(\left(\text{c}+s^2\right)^2\right)\cdot\text{Y}(s)=3\cdot y(0)\cdot\left(\text{c}+s^2\right)$$

*Apply the reverse product rule:
$$\frac{\text{d}}{\text{d}s}\left(\left(\text{c}+s^2\right)^2\cdot\text{Y}(s)\right)=3\cdot y(0)\cdot\left(\text{c}+s^2\right)$$

*Integrate both sides with respect to $s$:
$$\int\frac{\text{d}}{\text{d}s}\left(\left(\text{c}+s^2\right)^2\cdot\text{Y}(s)\right)\space\text{d}s=\int3\cdot y(0)\cdot\left(\text{c}+s^2\right)\space\text{d}s$$

*Evaluate the integrals:
$$\left(\text{c}+s^2\right)^2\cdot\text{Y}(s)=3\cdot y(0)\cdot\left(\frac{s^3}{3}+\text{c}\cdot s\right)+\text{K}$$

*Solving for $\text{Y}(s)$:
$$\text{Y}(s)=\frac{y(0)\cdot s\cdot\left(3\cdot\text{c}+s^2\right)+\text{K}}{\left(\text{c}+s^2\right)^2}$$

*With inverse Laplace transform, we find:
$$y(x)=\frac{\sqrt{\text{c}}\cdot\cos\left(x\cdot\sqrt{\text{c}}\right)\cdot\left(2\cdot\text{c}\cdot y(0)-\text{K}\cdot x\right)+\sin\left(x\cdot\sqrt{\text{c}}\right)\cdot\left(2\cdot\text{c}^2\cdot x\cdot y(0)+\text{K}\cdot x\right)}{2\cdot\text{c}^{\frac{3}{2}}}$$

