Quote:
Originally Posted by ThePerfectHacker  With the x^3 look for a solution of the form Ax^3+Bx^2+Cx+D. |
So I used Bx^3+Cx^2+Dx+E since I used A for the first part.
y' = 3Bx^2+2Cx+D
y'' = 6Bx + 2C
(6Bx+2C) + (Bx^3+Cx^2+Dx+E) = x^3
isn't it just x^3(B) = x^3 , B = 1?