Diff. eq determening IVT interval

[dlevanchuk]

Homework Statement

Consider the initial value problem

t*y' + 2y = e^(2t) ; y(1) = 0:

Determine the largest interval on which it is guaranteed to have a unique solution.

Homework Equations

The Attempt at a Solution

I just need somebody to tell me if i have a logic on this one :-S

I understand that t cannon equal zero. But also, the interval, that there is a unique solution, would be from 0 to positive infinity, because the initial condition sets y' to be positive..
If the interval would be going from negative infinity to 0, then y' would be negative, which violates the initial condition...

Am i right?

 

[vela]

How did you get y'>0 from y(1)=0?

 

[dlevanchuk]

How did you get y'>0 from y(1)=0?

i divided both sides by t and got

y' + (2/t)*y = e^(2t)/t

then i plugged in 1 for t and 0 for y.

As a result I got y' = e^2

 

[vela]

OK, so you meant y'(1)>0, not y'(t)>0. It's still not clear to me how you went from that to concluding that t>0. Why are you associating y'>0 with t>0 and y'<0 with t<0?

That's really beside the point, however. You don't need to look at the derivative. You just have to look at the initial condition itself. It's specified for t=1, so...