## College Algebra (10th Edition)

y= C$e^{4x}$ y' = 4C$e^{4x}$ True
By taking the derivative of y, we have y=C$e^{4x}$ and its derivative y'=4C$e^{4x}$. By plugging these two equations into the differential equation results in: 4C$e^{4x}$=4$\times$C$e^{4x}$. Which is true, so y = C$e^{4x}$ is a solution.