Solving this linear second order Cauchy problem Here is my problem, I know that a function is a solution of this linear Cauchy problem
$$ 
\left\{
\begin{array}{rcl}
y'' &=& \frac{x^2+6}{4}y,\\
y(0)&=&0,\\
y'(0)&=& \frac{\sqrt{\pi}}{2}.
\end{array}
\right.
$$
and I want to find the solution of this problem with a simple method (I know the solution it is $x\mapsto \frac{\sqrt{\pi}}{2} e^{\frac{x^2}{4}}x$). 
In fact this function is also a solution of
$$ 
\left\{
\begin{array}{rcl}
y'' -\frac{x}{2}y'-y&=&0,\\
y(0)&=&0,\\
y'(0)&=& \frac{\sqrt{\pi}}{2}.
\end{array}
\right.
$$
So here is my question what is the simplest method (and eventually what is the good differential equation) to get the solution. The general solution of this two differential equations involves the error function so I think that we need a method for this specific Cauchy problem. My idea was : since the solution is quite simple and has a power series, we may assume at first that the solution has a power series and derive from the equation this power series and then recognize the solution. But this problem comes from an difficult exercise and solving the differential equation is the last question and I think it is supposed to be easy (to find a solution of the ODE) in order to quickly finish the exercise (in fact the first question is "here is an integral with a parameter, show that it defines a $\mathcal{C}^2$ function", second question, "find an linear second order equation such that the previous integral with a parameter is a solution of the ODE", and last question "find an simple expression for the integral with a parameter").

Apply the method in http://eqworld.ipmnet.ru/en/solutions/ode/ode0205.pdf:
Let $y=e^\frac{x^2}{4}~u$ ,
Then $y'=e^\frac{x^2}{4}~u'+\dfrac{xe^\frac{x^2}{4}}{2}u$
$y''=e^\frac{x^2}{4}~u''+\dfrac{xe^\frac{x^2}{4}}{2}u'+\dfrac{xe^\frac{x^2}{4}}{2}u'+\biggl(\dfrac{x^2e^\frac{x^2}{4}}{4}+\dfrac{e^\frac{x^2}{4}}{2}\biggr)u=e^\frac{x^2}{4}~u''+xe^\frac{x^2}{4}~u'+\dfrac{(x^2+2)e^\frac{x^2}{4}}{4}u$
$\therefore e^\frac{x^2}{4}~u''+xe^\frac{x^2}{4}~u'+\dfrac{(x^2+2)e^\frac{x^2}{4}}{4}u=\dfrac{x^2+6}{4}e^\frac{x^2}{4}~u$
$e^\frac{x^2}{4}~u''+xe^\frac{x^2}{4}~u'-e^\frac{x^2}{4}~u=0$
$u''+xu'-u=0$
$u=x$ is a particular solution
$\therefore$ Let $u=xv$ ,
Then $u'=xv'+v$
$u''=xv''+v'+v'=xv''+2v'$
$\therefore xv''+2v'+x(xv'+v)-xv=0$
$xv''+(x^2+2)v'=0$
$xv''=-(x^2+2)v'$
$\dfrac{v''}{v'}=-x-\dfrac{2}{x}$
$\int\dfrac{v''}{v'}dx=\int\left(-x-\dfrac{2}{x}\right)dx$
$\ln v'=-\dfrac{x^2}{2}-2\ln x+c$
$v'=\dfrac{c_2e^{-\frac{x^2}{2}}}{x^2}$
$v=c_2\int\dfrac{e^{-\frac{x^2}{2}}}{x^2}dx$
$v=-c_2\int e^{-\frac{x^2}{2}}~d\left(\dfrac{1}{x}\right)$
$v=-c_2\biggl(\dfrac{e^{-\frac{x^2}{2}}}{x}-\int\dfrac{1}{x}d\biggl(e^{-\frac{x^2}{2}}\biggr)\biggr)$
$v=C_2\biggl(\dfrac{e^{-\frac{x^2}{2}}}{x}+\int e^{-\frac{x^2}{2}}~dx\biggr)$
$\dfrac{e^{-\frac{x^2}{4}}y}{x}=C_1+C_2\biggl(\dfrac{e^{-\frac{x^2}{2}}}{x}+\int_0^xe^{-\frac{x^2}{2}}~dx\biggr)$
$y=C_1xe^\frac{x^2}{4}+C_2\biggl(e^{-\frac{x^2}{4}}+xe^\frac{x^2}{4}\int_0^xe^{-\frac{x^2}{2}}~dx\biggr)$
$y(0)=0$ :
$C_2=0$
$\therefore y=C_1xe^\frac{x^2}{4}$
$y'=\dfrac{C_1(x^2+2)e^\frac{x^2}{4}}{2}$
$y'(0)=\dfrac{\sqrt\pi}{2}$ :
$C_1=\dfrac{\sqrt\pi}{2}$
$\therefore y=\dfrac{\sqrt\pi xe^\frac{x^2}{4}}{2}$
