Answer
See below.
Work Step by Step
Take the derivatives of the function. $$y(x)=c_1e^{2x}+c_2e^{-3x}$$ $$y'(x)=2c_1e^{2x}-3c_2e^{-3x}$$ $$y''(x)=4c_1e^{2x}+9c^2e^{-3x}$$ Substituting these functions into the differential equation yields $$y''+y'-6y=0$$ $$4c_1e^{2x}+9c^2e^{-3x}+2c_1e^{2x}-3c_2e^{-3x}-6(c_1e^{2x}+c_2e^{-3x})=0$$ $$0=0$$ This statement is true, therefore, the equation must be a valid solution to the differential equation.
