Consider the differential equation below with power series solutions y(), 2 724=0 y(x) = > a; zsti j=0 (a) From the lowest power of r,
Consider the differential equation below with power series solutions y(), 2 724=0 y(x) = > a; zsti j=0 (a) From the lowest power of r, find the indicial equation. (b) Set up the recurrence relation, if it exists. (c) Verify that y = 1 + 2x + 2x2 is a solution by back into the differential equation
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