(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

determine the largest interval on which a unique solution exists:

y' + (t/((t^2)+1)) y = sin t with initial conditions: y(-2) = 1

2. Relevant 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.

3. 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..

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: 1st order linear diff equation clarification

**Physics Forums | Science Articles, Homework Help, Discussion**