What is the particular $Y_p$ for the general solution $y''+y=6\sin x$?

Question

In this case, We set $Y_p=Cx(A\cos x+B\sin x)$ beacuse of $D= \pm i$ then, $$Y_p''=2C(-A\sin x+B\cos x)+Cx(-A\cos x-B\sin x)$$ $$Y_p''+Y_p=2C(-A\sin x+B\cos x)=6\sin x$$ So $AC=-3$, BC=0, if C=1, then we have $Y_p=--3xsin x$ .However... According to the reference, $$Y_p=-3x\cos x$$ but I don't understand why.

Answer 1

y = A xsin x + Bxcos x

y' = Asin x + Bcos x - B xsin x + Axcos x

y'' = -2Bsin x + 2Acos x - Axsin x + Bxcos x y''+y = -2Bsinx + 2Acosx = -6sin x

$B = -\frac {1}{2}*6=-3 , A = 0$

$y = c_1sint + c_2cosx -3xcosx$

$y(0) = 6sin0=C_2 = 0$

$y'(0) = C_1 = 0$

$y = -3xcosx$