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differeq
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Homework Statement
determine the largest interval on which a unique solution exists:
y' + (t/((t^2)+1)) y = sin t with initial conditions: y(-2) = 1
Homework Equations
arent there discontinuities at t = 0 and t = -1?
(textbook says no, based on answer..i am confused why not?) i think it may have something to do with not understanding how the initial conditions, eg, y(-2) = 1 relate to the interval.
The Attempt at a Solution
the equation is already in the form y' + p(t)y = g(t)
p(t) = (( t/(t^2 ))+1))
if there are discont @ t = 0 and t= -1, then the intervals would be:
(- infin, -1)
(-1, 0)
(0, infin)
therefore largest interval for t = -2 is: - infin < t < -1
the answer in back though says largest interval is: - infin < t < infin
please help, thanks alot.
EDIT: i glossed over the square of t, so i now know that t= -1 does not make p(t) = 0.. but i am still unclear as to why t = 0 is not a discontinuity.. Also, i think i am unclear about what precisely is a discontinuity. thanks again for any help..
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