MHB -2.4.2 interval of initial value problem

[karush]

Determine an interval in which the solution of the given initial value problem is certain to exist

$t(t-4)y'+y=0 \quad y(2)=2\quad 0<t<4$

ok my first step was isolate y'
s
$y'=-\dfrac{y}{t(t-4)}$

not sure what direction to go since we are concerned about an interval

 

[HOI]

Well, the problem specifically says "0< t< 4" and y' does not exist at t= 0 and t= 4.

 

[karush]

how does y(2)=2 fit into this
doesn't that give us specific y interval

 

[topsquark]

how does y(2)=2 fit into this
doesn't that give us specific y interval

You could always do "brute force" if you can't figure out a work around. Boundary conditions get rid of integration constants. Solve the differential equation. What is y(t)? What does that tell you about the solution interval(s)?

-Dan

 

[HOI]

IF the "0< t< 4" were not there you would still be able to observe that if x= 0 or x= 4, the denominator would be 0 so y' would not exist. That would tell you that a solution cannot be extended across x= 0 or x= 4. But you could have solutions for x< 0 or x> 4. The fact that you are given y at x= 2, which is between 0 and 2 tells you that the correct interval is 0< x< 4, not x< 0 or x> 4. But, as I said, that is redundant since the problem itself says "0< x< 4".