Answer
Given below.
Work Step by Step
Take the derivatives of the function.
$y(x)=c_1x^2+c_2x^3-x^2sinx$
$y'(x)=2xc_1+3x^2c_2-2xsinx - x^2cosx$
$y''(x)=2c_1+6xc_2-2sinx-4xcosx+x^2sinx$
Substituting these functions into the differential equation yields
$x^2y'' -4xy' +6y= x^4sinx$
$x^2(2c_1+6xc_2-2sinx-4xcosx+x^2sinx)-4x(2xc_1+3x^2c_2-2xsinx - x^2cosx)+6(c_1x^2+c_2x^3-x^2sinx)=x^4sinx$
$2c_1x^2+6c_2x^3-2x^2sinx-4x^3cosx+x^4sinx-8c_1x^2-12c_2x^3+8x^2sinx+4x^3cosx+6c_1x^2+6c_2x^3-6x^2sinx-x^4sinx=0$
$0=0$
This means that the solution is valid for all real values. The interval linked to this solution is $(−∞,∞)$.
