Find an interval I for which the given IVP is guaranteed to have a unique solution.
(x-1) y''- [y'/(x-4)] +(x-3)y=0, y(2)=0, y' (2)=1
Differential Equation :
y(2)=0, y'(2)=1
f1(x) and f2(x) must be continuous in the largest interval for given IVP to have unique solution.
is discontinuous at x = 1 and 4
and
is discontinuous at x = 1.
Therefore largest interval containing initial value x =2 having unique solution is [1 , 4]
1 and 4 are in close interval, it also contain x =2 and there is no discontinuity between x = 1 to 4.
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