Vtyjkyuk

How to find the general solution of $y''+2xy'-2ny=0$?

I was solving a problem in this thread. The posted solution involves the error function which I am not aware of. I found this link.

How to find solution of $y''+2xy'-2ny=0$? The solution of this differential equation is given in the link as $$y=A,texterfc_n(x)+B,texterfc_n(-x),.$$ Even I am not aware of this notation. Kindly explain. If not, please suggest some online sources. My textbook does not have this topic.

I can give you a polynomial solution. For each integer $ngeq 0$, let $p_n:mathbbRtomathbbR$ denote the polynomial function
$$p_n(x)=sum_r=0^leftlfloorfracn2rightrfloor,frac12^2r,r!,(n-2r)!,x^n-2rtext for all xinmathbbR,.$$
Then, $y:=p_n$ is a solution to the differential equation
$$y''(x)+2x,y'(x)-2n,y(x)=0$$
for each integer $ngeq 0$. You can obtain the general solution by the Reduction-of-Order Method, but I am not sure how complicated this route of attempt will be. However, you can at least show that all solutions $y$ look like
$$y(x)=A,p_n(x)+B,p_n(x),int_x^infty,fracexp(-t^2)big(p_n(t)big)^2,textdt$$
for some constants $A$ and $B$.

For $n=0$ with $p_0(x)=1$, we obtain the general solution
$$y(x)=A+frac2sqrtpi,B,texterfc(x)=a,texterfc_0(+x)+b,texterfc_0(-x),,$$
where $a:=dfracA2+dfrac2sqrtpi,B$ and $b:=dfracA2$, as $texterfc_0=texterfc$ and $texterfc(+x)+texterfc(-x)=2$.
In general, we note that
$$(-1)^n,texterfc_n(+x)+texterfc_n(-x)=frac2sqrtpi,int_-infty^+infty,frac(x-t)^nn!,expleft(-t^2right),textdt=2,p_n(x),.$$
Hence, it remains to show that the function $q_n:mathbbRtomathbbR$ defined by
$$q_n(x):=frac2sqrtpi,left(fracp_n(x)2^n,n!right),int_x^infty,fracexp(-t^2)big(p_n(t)big)^2,textdttext for all xinmathbbR$$
is a linear combination of $texterfc_n(+x)$ and $texterfc_n(-x)$, noting that
$$texterfc_n(z)=frac2sqrtpi,int_z^infty,frac(t-z)^nn!,exp(-t^2),textdt,.$$
Perhaps, it helps to know that $p_n'(x)=p_n-1(x)$ and $texterfc'_n(x)=-texterfc_n-1(x)$ for every $n=1,2,3,ldots$. As far as I know,
$$q_0(x)=texterfc(x)=texterfc_0(x),,,,q_1(x)=texterfc_1(x),,text and q_2(x)=texterfc_2(x),.$$
I conjecture that
$$q_n(x)=texterfc_n(x)text for all n=0,1,2,3,ldots,.$$
It will help tremendously if I can show that
$$p_n(x),texterfc_n-1(x)+p_n-1(x),texterfc_n(x)=frac12^n,n!,left(frac2sqrtpi,expleft(-x^2right)right),,tag*$$
or equivalently,
$$texterfc_n(-x),texterfc_n-1(+x)+texterfc_n(+x),texterfc_n-1(-x)=frac22^n,n!,left(frac2sqrtpi,expleft(-x^2right)right),.tag#$$

On the other hand, it is quite easy to show that $y(x):=texterfc_n(x)$ satisfies the homogeneous differential equation $y''(x)+2x,y'(x)-2n,y(x)=0$. For $n<2$, this can be easily checked by hand. For $ngeq 2$, we first note that, applying integration by parts, w have
$$texterfc_n-2(x)=frac2sqrtpi,int_x^infty,frac2t(t-x)^n-1(n-1)!,exp(-t^2),textdt,.$$
That is,
$$texterfc_n''(x)=texterfc_n-2(x)=frac2sqrtpi,int_x^infty,fracbig(2x+2(t-x)big)(t-x)^n-1(n-1)!,exp(-t^2),textdt,.$$
Expanding the integral above yields
$$texterfc_n''(x)=small 2x,left(frac2sqrtpi,int_x^infty,frac(t-x)^n-1(n-1)!,exp(-t^2),textdtright)+2n,left(frac2sqrtpi,int_x^infty,frac(t-x)^nn!,exp(-t^2),textdtright),.$$
That is,
$$texterfc_n''(x)=2x,texterfc_n-1(x)+2n,texterfc_n(x),.$$
Because $texterfc_n'(x)=-texterfc_n-1(x)$, we deduce that
$$texterfc_n''(x)+2x,texterfc'_n(x)-2n,texterfc(x)=0,,$$ establishing our claim.

Now, we must have
$$texterfc_n(x)=lambda_n,p_n(x)+mu_n,q_n(x)$$
for some constants $lambda_n,mu_n$. For $n=0$, it is obvious that $lambda_0=0$ and $mu_0=1$. For $n>0$, we note that $texterfc_n(x)$ and $q_n(x)$ tend to $0$ as $xtoinfty$, whereas $p_n(x)toinfty$ when $xtoinfty$. This means $lambda_n=0$. It can be easily checked that, for an even integer $n>0$,
$$left.fractextdtextdxright|_x=0,left(fractexterfc_n(x)p_n(x)right)=frac2^nbinomnfracn2,left(frac2sqrtpiright)=left.fractextdtextdxright|_x=0,left(fracq_n(x)p_n(x)right),,$$
whence $mu_n=1$. For an odd integer $n>0$, we note that
$$lim_xto 0,Biggl(x,left(fractexterfc_n(x)p_n(x)right)Biggr)=frac2^n-1n,binomn-1fracn-12,left(frac2sqrtpiright)=lim_xto 0,Biggl(x,left(fracq_n(x)p_n(x)right)Biggr),,$$
so $mu_n=1$, as well. This shows that $q_n(x)=texterfc_n(x)$ for all $ninmathbbZ_geq0$. As a consequence, both (*) and (#) hold.