- #26

Mark44

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Aren't you skipping a step? You have y' = f(t, y) = (tok, i am very grateful for your help. i have my exam in 3 days.

can you please kindly tell me how do i answer part e. wether there is a unique solution through the point (1,2).

i know that to prove uniqueness, we must have y'=f(t,y), and differentiate w.r.t to y. i.e find (f(t,y))'

and then show that (f(t,y))' is continuous.

to show whether or not it has a unique solution at (1,2), should i sub t=1 and y=2 into (f(t,y))', and if i get a unique solution. does that mean there is a unique solution through the point (1,2)??

^{2}- y - 2)(1 - y

^{2}).

For there to be a unique solution at (1, 2), both f and f

_{y}have to be defined and continous on some rectangle around (1, 2). The part that you seem to have skipped is finding

[tex]f_y = \frac{\partial }{\partial y}f(t, y)[/tex]