"Find an interval around x = 0 for which the given initial-value problem has a unique solution."

(x-2)y" + 3y = x, y(0) = 0, y'(0) = 1

How do I do this without solving the DE, which I haven't learned yet (but will soon)?

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- August 17th 2012, 05:58 PMphys251Finding unique solutions
"Find an interval around x = 0 for which the given initial-value problem has a unique solution."

(x-2)y" + 3y = x, y(0) = 0, y'(0) = 1

How do I do this without solving the DE, which I haven't learned yet (but will soon)? - August 17th 2012, 06:49 PMGJARe: Finding unique solutions
Hi, phys251.

There is an existence theorem for second order linear differential equations that goes as follows:

__Main Theorem__

Consider the Initial Value Problem (IVP)

, and

If the functions and are continuous on the open interval that contains the point then the IVP has exactly one solution throughout the interval .

To start, in our example. I would suggest dividing through by and seeing if you can use the Main Theorem from there. Think about it a little more, and if you're still stuck I'll detail a little further what I had in mind for this exercise.

Good luck! - November 9th 2012, 05:48 AMhappyroseRe: Finding unique solutions
some crazy partial fractions, but I think this definitely can't be right as I'm supposed to end up with another piecewise function as my answer. Could anyone point me in the right direction? I have never done a piecewise function with functions of t as the values.

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