# Homework Help: 2nd order DE largest interval confusion

1. Feb 9, 2012

### fufufu

2nd order DE..."largest interval" confusion

1. The problem statement, all variables and given/known data
Determine the largest t-interval on which therem 3.1 guarantees the existence of a unique solution:

y'' + 3t^2y' + 2y = sin(t) .....y(1) = 1 ....y'(1) = -1

2. Relevant equations
theroem 3.1 is the one that states if p(t) and q(t) and g(t) are continuous functions on the interval (a,b) then the IVP y'' + p(t)y' + q(t)y = g(t) has unique solution defined on the entire interval , (a,b) .

3. The attempt at a solution

if i therefore locate any discontinuities on each function in the equation, then that will determine what the intervals are.
p(t) = 3t^2 and g(t) sin(t) and q(t) = 2(*) are all continuous, so the largest interval is -infin<t<infin, which is the answer in back of book.
But what was the point of the initial values then?

(my geuss) --do i look at the tvalues and consider which is the largest interval that that t-value falls in? (in this case i only have one interval, so it makes it easy, but am i explaining it correctly?)

(*) my other question is: in this problem, is q(t) = 2? if there is no t in the term, then how can i ask if there are any discontinuities there?

thanks for any help in understanding this..
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 9, 2012

### LCKurtz

Re: 2nd order DE..."largest interval" confusion

The point of the initial conditions is to cause the solution to be unique. If you had arbitrary constants in there, you would have lots of solutions to the equation. And yes, q(t)= 2 is a perfectly good constant function. You get 2 for every t.