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

1 次查看(过去 30 天)
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:

回答(2 个)

Jan
Jan 2016-5-2
What is your problem? The integrator replies the position and the velocity as expected. These are not two solutions.
  1 个评论
Zachary Duff
Zachary Duff 2016-5-2
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.

Torsten
Torsten 2016-5-3

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