If $k$ is a given non zero constant show that the functions $y = c\operatorname{exp}(kx)$ are the only solutions of the differential equation $\dfrac{dy}{dx} = ky$. Hint: assume that $f(x)$ is a solution of the equation and show that $\dfrac{f(x)}{\operatorname{exp}(kx)}$ is a constant. I tried to put $f(x)$ into the equation but I do not know how to proceed. Please help.

  • $\begingroup$ It's convenient to write $\frac{f(x)}{\exp(kx)}$ as $f(x)e^{-kx}$. Now evaluate $\frac{\mathrm{d}}{\mathrm{d}x}(f(x)e^{-kx})$. $\endgroup$ – Abel Jun 2 '13 at 10:25
  • $\begingroup$ See the proof in this answer. $\endgroup$ – Key Ideas Jun 2 '13 at 14:56

Differentiate $\frac{y}{e^{k x}}$ implicitly using the fact that $y'=k y$ and show that it is zero. Hence $\frac{y}{e^{k x}}$ must be some constant $A$ but this implies that $y=A e^{k x}$

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