Why am I getting 2 solutions for this ODE with one set of initial values?

Note: I am a beginner in math and in Matlab, any additional wisdom is gratefully received.

So, I'm getting two solutions for this singular ODE, Wolfram A gets one. My y vector in

[t,y] = ode15s(...)

is giving me a size(t) by 2 matrix. And resulting in a plot that looks like:

Here's the equation: xy''+ y' + xy = x . My code attempt to solve it:

tspan = [1e-4 50];
[x,y] = ode15s('p2ODE',tspan,[0 1]);
plot(x,y);

function:

function dxdt = p2ODE( t , x )

y = x(1);
yPrime = x(2);

dy = yPrime;
dyPrime = (-1/t)*yPrime - y + 1;

dxdt = zeros(2,1);
dxdt(1) = dy;
dxdt(2) = dyPrime;

end

But here's Wolfram A solution samples:

Answer 1

[Jan]

What is your problem? The integrator replies the position and the velocity as expected. These are not two solutions.

[Zachary Duff]

Realized that after just thinking about the code for a few minutes. Computation with vectors and matrices is still new to me. But to answer your question, my problem is exactly as stated.

Answer 2

[Torsten]

http://www.wolframalpha.com/input/?i=x*y''(x)+%2B+y'(x)+%2B+x*y(x)+-+x+%3D+0,+y(1e-4)%3D0,+y'(1e-4)%3D1

Best wishes

Torsten.