y''+xy'+y=0, y(0)=1, y'(0)=-1

Question

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?

Answer 1

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.

Answer 2

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 $$C_1=-1$$