Answer
$$A=-\frac{3}{10}$$ and $$B=-\frac{1}{10}$$
Work Step by Step
$$y=A\sin x+B\cos x$$
1) Find $y'$ and $y''$ $$y'= A\cos x-B\sin x$$ and $$y''=-A\sin x-B\cos x$$
2) Now consider the equation $$y''+y'-2y=\sin x$$ $$-A\sin x-B\cos x+A\cos x-B\sin x-2(A\sin x+B\cos x)=\sin x$$ $$-A\sin x-B\cos x+A\cos x-B\sin x-2A\sin x-2B\cos x=\sin x$$ $$(-A-B-2A)\sin x+(A-B-2B)\cos x=\sin x$$ $$(-3A-B)\sin x+(A-3B)\cos x=1\sin x+0\cos x$$
Comparing both sides of the equation, we can see that $$-3A-B=1$$ and $$A-3B=0$$
Consider $A-3B=0$ then $A=3B$
Combine $A=3B$ to the first equation, we have $$-3\times3B-B=1$$ $$-10B=1$$ $$B=-\frac{1}{10}$$
