Answer
See below
Work Step by Step
The general solution for the given differential equation is: $y(x)=c_1e^x+c_2 e^{-2x}+A_0\sin x+B_0\cos x$
The trial solution for $y_p=A_0\sin x+B_0\cos x$ can be computed as by plugging back into the given differential equation.
So, we have: $y''+y'-2y=-10\sin x\\(D^2+D-2)y_p(x)=-10\sin x\\ (-3A_0-B_0)\sin x +(A_0-3B_0)\cos x=-10\sin x$
On comparing coefficients, we get: $A_0=3,B_0=1$
Therefore, the general solution for the given differential equation is: $y(x)=c_1e^x+c_2e^{-2x}+3\sin x+\cos x$
Given $y(0)=2\\y'(0)=1$
Substitute $c_1+c_2=1\\
c_1-2c_2=-2$
Then we have $c_1=0\\
c_2=1$
Hence, $y=e^{-2x}+3\sin x+\cos x$
