y''+xy'+y=0, y(0)=1, y'(0)=-1 I have used laplace transform to get
$Y'(s)-sY'(s)=-1+\frac{1}{s}$
$Y(s)=-e^\frac{s^2}{2}\int e^\frac{-s^2}{2}ds + e^\frac{s^2}{2}\int \frac{ e^\frac{-s^2}{2}}{s}ds +Ce^\frac{s^2}{2}$
what should be done? should I have used a different way. if so what?
 A: Hint:
your D.E is equivalent to ( by taking the integratition for both sides)
$$ y'+yx=C_1$$
by using the boundary condition, we will get
$$C_1=-1$$
A: EDIT: (weeks later) I have found an even better approach. The equation becomes much more trivial when fourier transform is used. It is possible to identify, immediately, that the fourier transform of a gaussian is also a gaussian, which gives one of the two solutions, which in return makes it very easy to find the second one.
I wasn't satisfied with y''+xy'+y = (y'+xy )' so I tried really hard to figure out a way.
$$Y(s)=-e^\frac{s^2}{2}\int e^\frac{-s^2}{2}ds + e^\frac{s^2}{2}\int \frac{ e^\frac{-s^2}{2}}{s}ds +Ce^\frac{s^2}{2}$$
I decided I would need this transform to vanish at infinity even though every term blows up. So I decided to pick C such that at s->inf: 
$$-e^\frac{s^2}{2}\int e^\frac{-s^2}{2}ds+Ce^\frac{s^2}{2}=0$$
which could be rewritten as (omitting ) constants
$$e^{\frac{s^2}{2}}Erfc(\frac {s} {\sqrt 2})$$
and its inverse laplace transform gives me
$$e^{t-\frac{t^2}{2}}$$
Which is one of the solutions. Knowing this solution, makes it very easy to get the second solution.
