Differential Equations involving exponentials

Question

It's been sometime since I've had to solve a differential equation involving an exponential. The DE is a separable

$$\begin{align} & 2\cdot\frac{\mathrm{d}y(x)}{\mathrm{d}x} + e^{y(x)} = 0\\ \implies& \frac{\mathrm{d}y(x)}{\mathrm{d}x} = -\frac{1}{2}e^{y(x)}\\ \implies& \frac{1}{e^{y(x)}}\cdot\frac{\mathrm{d}y(x)}{\mathrm{d}x} = -\frac{1}{2}\\ \implies& \int e^{-y(x)}\ \mathrm{d}y(x) = -\frac{1}{2}\int \mathrm{d}x\\ \implies& - e^{y(x)}= -\frac{1}{2}x+C\\ \implies& y(x) = \ln\left(\frac{1}{2}x+C\right)\\ \end{align}$$

I think this is the answer, but for some reason Wolfram Alpha gives $y(x) = -\ln\left(\frac{1}{2}x + C\right)$. Where did the negative come from?

Answer 1

Your second last line should be $$-e^{-y}=\frac{-1}2x+c$$ since $$\int e^{-y}dy=-e^{-y}$$

Answer 2

You have made an error in evaluating $$\int e^{-y(x)}dy(x)$$

The negative sign does not go away.