Question

use the method of variation of parameters to find a particular solution of the given equation x^2y''-4xy'+6y=x^7 where y1 = x^2 and y2 = x^3 use the method of variation of parameters to find a particular solution of the given equation x^2y''-4xy'+6y=x^7 where y1 = x^2 and y2 = x^3 @Mathematics

Answer 1

i get y = x^6/12 + c1x^3 - c2x^2

Answer 2

x^6/12 isn't right.

Answer 3

Let the particular solution be yp = Ax^7. Then yp ' = 7Ax^6, yp'' = 42Ax^5
thus
x^2y''-4xy'+6y = (42A - 28A + 6A) x^7 = x^7
and therefore
A = 1/20