How to solve differential-functional equation?

Question

$$x (x+1) y'(x)+\left(\frac{3 x}{2}+1\right) y(x)-\frac{n^2 y(x+2)}{\pi }=0$$ and $$y(1)=\frac{1}{\sqrt{2}}$$

where: $n>0$ and $n\in \mathbb{Z}$

for $n = 0$ is simple differential-equation:

$$x (x+1) y'(x)+\left(\frac{3 x}{2}+1\right) y(x)=0$$

$$y(x)=\frac{c_1}{x \sqrt{1+x}}$$

I'm tried with Fourier Transform (see below),but can't find Inverse Transform.

$$-is{\frac {\partial ^{2}}{\partial {s}^{2}}}{\it fourier} \left( y \left( x \right) ,x,s \right) - \left( i/2+s \right) {\frac { \partial }{\partial s}}{\it fourier} \left( y \left( x \right) ,x,s \right) -{\frac {{n}^{2}{{\rm e}^{2\,is}}{\it fourier} \left( y \left( x \right) ,x,s \right) }{\pi}}=0 $$

Substituting: ${\it fourier} \left( y \left( x \right) ,x,s \right) =y \left( s \right) $

$$-is{\frac {{\rm d}^{2}}{{\rm d}{s}^{2}}}y \left( s \right) - \left( i/ 2+s \right) {\frac {\rm d}{{\rm d}s}}y \left( s \right) -{\frac {{n}^{ 2}{{\rm e}^{2\,is}}y \left( s \right) }{\pi}}=0 $$

$$y(s)=\text{C1} \cos \left(n* \text{erf}\left((-1)^{3/4} \sqrt{s}\right)\right)-\text{C2} \sin \left(n* \text{erf}\left((-1)^{3/4} \sqrt{s}\right)\right) $$

$$y(x)=\mathcal{F}_s^{-1}\left[\text{C1} \cos \left(n* \text{erf}\left((-1)^{3/4} \sqrt{s}\right)\right)-\text{C2} \sin \left(n* \text{erf}\left((-1)^{3/4} \sqrt{s}\right)\right)\right](x) $$ Maybe there is another way to solve this equation to closed-form?

EDITED: 20 October 2017.

Solution to this differential-functional equation i general formula finding Laplace Transform of:

$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^{1+n}\right](x)=?$

for n=0.

$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^{1+n}\right](x)=\frac{1}{x \sqrt{1+x}}$

for n=1.

$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^2\right](x)=\frac{4 \tan ^{-1}\left(\frac{1}{\sqrt{1+x}}\right)}{\pi x \sqrt{1+x}}$

for n=2.

$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^3\right](x)=\frac{6 \tan ^{-1}\left(\frac{1}{\sqrt{1+x} \sqrt{3+x}}\right)}{\pi x \sqrt{1+x}}$

for n=3.

$\mathcal{L}_t\left[\text{erf}\left(\sqrt{t}\right)^4\right](x)=\frac{4 \sqrt{\frac{1}{1+x}}}{x}-\frac{96 \int_0^{\sqrt{x+1}} \frac{\cot ^{-1}\left(\sqrt{2+a^2}\right)}{\left(1+a^2\right) \sqrt{2+a^2}} \, da}{\pi ^2 x \sqrt{1+x}}$

In the last example I can not find close-form integral.

Answer 1

Because $y(x)$ drops of quite quickly as $x$ increases I reckon that you will get a pretty good approximation by treating $y(x+2)$ as a source term then iterate the solution. I.e. first solve for no source as you have already done to get $y_0$, then use this solution for $y(x+2)$ as a new source. It is possible to do the integration to get a new solution $y_1$ in closed form. For small $n$ at least it looks like $y_0$ is close to $y_1$.

If I got it right $y_1(x)= \frac{1}{x \sqrt{1+x}}(1+\frac{2 n^2}{\pi}(tan^{-1}(\sqrt{\frac{1+x}{3+x}})-tan^{-1}(\frac{1}{\sqrt{2}})))$

Answer 2

I think if you will use a series representation you will have more insight. For example:

$ y(x) = \sum_{i=0}^\infty a_ix^i $ and you will be able to find some relationship between different $a_i$ with $a_0$ being a free variable (which will be related to the condition $y(1)$).