Answer
See below.
Work Step by Step
Find the derivatives of the function. Use chain rule. $$y(x)=c_1e^x+c_2e^{-2x}$$ $$y'(x)=c_1e^x-2c_2e^{-2x}$$ $$y''(x)=c_1e^x+4c_2e^{-2x}$$ Substitute these equations into the differential equation to get $$y''+y'-2y=0$$ $$(c_1e^x+4c_2e^{-2x})+(c_1e^x-2c_2e^{-2x})-2(c_1e^{x}+c_2e^{-2x})=0$$ $$0=0$$ This satisfies the differential equation, therefore, it must be a solution. This solution is valid for all values.
