1st order linear diff equation clarification

[differeq]

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 a lot.

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..

 

[Mark44]

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 a lot.

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..

Your function p(t) is defined and continuous for all real numbers. Why would you think there is a problem at t = 0?