Question

solve the equation y''+xy'+y=0

Answer 1

This isn't so pretty. By general theory of ODEs, we know there should be two linearly indep solutions.

One of them by inspection is y1 = e^{-x^2/2}, as y1' = -x y1 and y'' = - y1 + x^2 y1
so y1'' + xy1' + y1 = -y1 + x^2 y1 -x^2 y1 + y1 = 0.

So far, so good. Now for y2 ... now I'm stuck and will need to think a minute.

Answer 2

To clean up the notation for minute, let f(x) = y1(x), and let g(x) = primitive of f(x) = int_0^x exp(-t^2/2) dt

Notice that g is not a solution as g' = f, g'' = f' = -x f and hence
g'' + xg' + g = -x f + x f + g which is not zero.

But how about gf, call this h. Well,
h' = f g' + f' g = f^2 - x f g
h'' = 2 f f' - f g - x f' g - x f g' = 2x f^2 - h + x^2 f g - x f^2 = -x f^2 - h + x^2 f g

So h'' + x h' + h = -x f^2 - h - x^2 f g + x f^2 + x^2 f g + h = 0.

So y2 = fg is also a solution and the general solution is

y = A.y1 + B.y2

Answer 3

But there's something very unsatisfactory about this. What is a method we could have applied to find f1 and f2?! vs. a big of hard looking for y1 and for y2 a bit of luck.

Ideas?

Answer 4

i just got confused--- the first thing i did was to check if the function was exact- since it is then i could solve for it but i mean i dont think we need to do all that bc we are bearly studying wronskian