Thread: Finding 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)?

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)

$\displaystyle y''(x)+p(x)y'(x)+q(x)y(x)=g(x)$, $\displaystyle y(x_{0})=y_{0}$ and $\displaystyle y'(x_{0})=y_{1}.$

If the functions $\displaystyle p(x), q(x) $ and $\displaystyle g(x)$ are continuous on the open interval $\displaystyle I$ that contains the point $\displaystyle x_{0},$ then the IVP has exactly one solution throughout the interval $\displaystyle I$.

To start, $\displaystyle x_{0}=0$ in our example. I would suggest dividing through by $\displaystyle x-2$ 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!

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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