Interval of Existance - Diff. Eq.

[theuniverse]

Homework Statement

Without solving the DE, determine an interval in which a solution to the given initial value problem is certain to exist. Is it certain to be a unique solution?
[(pi^2/16) - t^2]y' + y^(1/2)tan(t) = 0
Initial Value: y(3pi/8) = 1

Homework Equations

dy/dt + p(x)y = q(x)

The Attempt at a Solution

- It says not to solve it so I'm thinking of analyzing the functions and their intervals. I know that sqrt(y) is t>0 and y>0 and I also know that tan(t) is -pi/2<t<pi/2.
- I am not sure how the expression of the derivative is to be used.
- How is it all connecting eventually to the initial value I am given.

That's all I can think of, so it would be very helpful if you can provide me some guidelines how to continue from here.
Thanks so much!

 

[Mark44]

If you put your equation in standard form, it becomes
y&#039; = \frac{-tan(t)\cdot y^{1/2}}{\frac{\pi ^2}{16} - t^2}

That denominator is 0 when t = +/- pi/4.

 

[theuniverse]

so -pi/4 < t < pi/4 as well as anything small than -pi/4 or bigger than pi/4 is the interval in which all my solutions will have to be correct? and both -pi/4 and pi/4 will be the vertical asymptotes. Now how do I relate the initial value: y(3pi/8) = 1 into it?

 

[Mark44]

I think that where this problem is going is that your interval is -pi/4 < t < pi/4. There's also the problem of tan(t) being undefined at odd multiples of pi/2.

In the DE text I have at hand, the existence and uniqueness theorem has this to say:
"Let f(x, y) and \partial f /\partial y be continuous functions of x and y in some rectangle R of the xy plane defined by a < x < b, c < y < d. If (x₀, y₀) is any point of R, then there exists a unique solution of the initial-value problem
y'(x) = f(x, y(x)), y(x₀) = y₀.

This solution is a differentiable function and satisfies the differential equation in some interval x₀ - h < x < x₀ + h which is contained in the interval a < x < b."
Linear Algebra and Differential Equations, 2nd Ed. Charles G. Cullen.