"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)?
There is an existence theorem for second order linear differential equations that goes as follows:
Consider the Initial Value Problem (IVP)
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.
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.
chaussures jordan 6
timberland bottes timberland bottes pas cher