Show that #y_1=x^2# is a solution to the differential equation #x^2y'' -(x^2+4x)y' +(2x+6)y=0 #?
1 Answer
May 10, 2018
We seek to show that
#x^2y'' -(x^2+4x)y' +(2x+6)y=0 #
Differentiating the given function we have:
# y_1' \ \= 2x #
# y_1'' = 2 #
And substituting into the given ODE, we have:
#x^2y'' -(x^2+4x)y' +(2x+6)y #
# \ \ \ \ \ \ \ \ \ \ = x^2(2) -(x^2+4x)(2x) +(2x+6)(x^2) #
# \ \ \ \ \ \ \ \ \ \ = 2x^2 -2x^3-8x^2 +2x^3+6x^2 #
# \ \ \ \ \ \ \ \ \ \ = 0 \ \ \ \ # QED
