-2.4.2 interval of initial value problem

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SUMMARY

The discussion focuses on determining the interval for the existence of solutions to the initial value problem defined by the differential equation $t(t-4)y' + y = 0$ with the initial condition $y(2) = 2$ and the constraint $0 < t < 4$. The key conclusion is that the solution exists within the interval $0 < t < 4$, as the derivative $y'$ is undefined at the boundaries $t = 0$ and $t = 4$. The presence of the initial condition at $t = 2$ confirms that the solution is valid within this specified interval.

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karush
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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
 
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Well, the problem specifically says "0< t< 4" and y' does not exist at t= 0 and t= 4.
 
how does y(2)=2 fit into this
doesn't that give us specific y interval
 
karush said:
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
 
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".
 

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