Question

I'm given two solutions to a differential equation and I have to say where these solutions are valid. (I will write them below :)

Answer 1

\[y _{1}(t)=1-t\] and \[y _{2}(t)=-t^2/4\] are solutions of the initial value problem \[dy/dt = 1/2(-t +\sqrt{t^2 +4y} )\], \[y(2) = -1\]

Answer 2

we have to differentaite both the equations??

Answer 3

if you differentiate both solutions and sub them in they both appear to be correct, but I need to state where each solution is valid. That's the part that confuses me :(

Answer 4

differentiation of y1=-1

Answer 5

differentiation of y2=-t/2

Answer 6

one thing that comes to mind is, a solution to a diffyQ along an interval has to be continuous along the interval, and something about uniqueness as well

Answer 7

that dy/dt has a sqrt in it, and is that whole set up in the denominator as well?

if so, then we would want to know what intervals we can have such that we do not get a negatvie value in the sqrt, or a zero in the denominator

Answer 8

I think you're right, as the next question is: 'Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of the Existence and Uniqueness Theorem.'

And this theorem is given to me as: If f(x,y) and the partial derivative with respect to x are continous, for a<x<b and for c<t<d, then for any k belonging to (a,b) and s belonging to (c,d), the initial value problem has a unique solution on some open interval containing s.

Answer 9

Oh no it isn't all in the denominator, sorry!

Answer 10

id try to take that further, but the memory of all this is a bit fuzzy :)

Answer 11

thanks :) I tried considering when the square root bit would be less than zero, but when I sub in the two given solutions for y, it always seems to be greater than or equal to zero?