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An alternative approach to solving the linear differential equation [math]\dydx + ky = 0[/math] is to write it as [math]\dydx = -ky[/math]. The latter equation is similar to [math]\dydx = y[/math], which has [math]e^x[/math] for a solution. With this similarity in mind, it is not hard to guess, and then verify, that [math]y = e^{-kx}[/math] is a solution to the original equation. The problem is now to show that every solution is a constant multiple of [math]e^{-kx}[/math]. Prove this fact by assuming that [math]y = f(x)[/math] is an arbitrary solution of [math]\dydx + ky = 0[/math] and then showing that the derivative of the quotient [math]\frac{f(x)}{e^{-kx}}[/math] is zero. (See Problem \ref{ex5.3.8}.)