# Thread: linear homogenous equation, simple question.

1. ## linear homogenous equation, simple question.

determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution without finding the solution.

ty" + 3y = t, y(1) = 1, y'(1)=2

i set this up and got

y" + (3/t)y = 1

the interval in the book says its 0< t< infinity, but im not understanding that.
When I look at q(t) = (3/t) it looks like it would have a continuous solution everywhere except zero. why are they truncating the negative?

2. Hi, it says the longest interval right and $]-\infty,0[\cup]0,\infty[$ isn't one.

3. no its just 0 -> infinity.

im not really understanding the logic.

4. Originally Posted by p00ndawg
determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution without finding the solution.

ty" + 3y = t, y(1) = 1, y'(1)=2

i set this up and got

y" + (3/t)y = 1

the interval in the book says its 0< t< infinity, but im not understanding that.
When I look at q(t) = (3/t) it looks like it would have a continuous solution everywhere except zero. why are they truncating the negative?
your initial data is given at t = 1. thus, you want to be on the right side of zero. y(1) would make no sense otherwise. so the largest continuous interval for your solution is t > 0