Calculus 10th Edition

$y=e^{-2x}$ is a solution
First take the derivative of y to find y' $y= e^{-2x}$ $y'= -2e^{-2x}$ Then substitute y and y' into the differential equation $3(-2e^{-2x} ) + 5(e^{-2x} )=-e^{-2x}$ $-6e^{-2x} +5e^{-2x} = -e^{-2x}$ $-e^{-2x} = -e^{-2x}$ The solution checks with the differential equation so $y=e^{-2x}$ is a solution.