Thread: Solutions to ODE y'(t) = t^2 +y(t)^2

Hi all,

The problem assigned to me was to show that the equation y'(t)==t^2+y(t)^2 has a solution for initial conditions y(0) = 0 over the interval 0<= t <= min(a, b/(a^2+b^2)), where we are considering a rectangle R, 0<= t <= a, -b<=y<=b.

What puzzles me is that the uniqueness and existence theorem says that if the DE is of the form y'(t) = f(t,y), where f and df/dy (partial derivative) are continuous over some rectangle containing the initial conditions (t0,y0) then the solution exists and is unique in that rectangle. In this case, f(t,y) is simply t^2 + y^2, and df/dy = 2 y. Both are clearly continuous for all values of t and y, so what is up with this question?

Also, trying to solve both numerically and analytically in Mathematica state that y(0)=0 is a funky point (the error statest that singularity or stiff system suspected).

Thanks in advance,
Julian

Originally Posted by aznmaven

Hi all,

The problem assigned to me was to show that the equation y'(t)==t^2+y(t)^2 has a solution for initial conditions y(0) = 0 over the interval 0<= t <= min(a, b/(a^2+b^2)), where we are considering a rectangle R, 0<= t <= a, -b<=y<=b.

What puzzles me is that the uniqueness and existence theorem says that if the DE is of the form y'(t) = f(t,y), where f and df/dy (partial derivative) are continuous over some rectangle containing the initial conditions (t0,y0) then the solution exists and is unique in that rectangle. In this case, f(t,y) is simply t^2 + y^2, and df/dy = 2 y. Both are clearly continuous for all values of t and y, so what is up with this question?

I think that the clause in blue is the key to this. The question is asking for a solution y(t) that is defined in the interval 0 ≤ t ≤ a and satisfies the condition |y(t)| ≤ b in that interval. In order to prevent the solution from getting too large, it may be necessary to restrict the interval so that instead of going up to t=a it has to stop short at t = b/(a^2+b^2).

There is a very clear statement of the existence and uniqueness theorem here.