Solving a Separable Equation: What Went Wrong?

[find_the_fun]

\(\displaystyle \frac{dy}{dx}+2xy=0\)
\(\displaystyle \frac{dy}{dx}=-2xy\)
\(\displaystyle dy=-2xy dx\)
\(\displaystyle \frac{1}{y} dy=-2x dx\)
integrate both sides
\(\displaystyle \ln{|y|}=-2x+c\)
\(\displaystyle y=e^{-2x+c}=e^{-2x}e^C=e^{-2x}k=ke^{-2x}\)
Let's check using the original equation. First calculate the derivative
\(\displaystyle \frac{dy}{dx}=k(-2e^{-2x}=-2ke^{-2x}\)
so from the original equation\(\displaystyle -2ke^{-2x}+2xke^{-2x}=0\) is false.

It looks like I'm missing an x somewhere but I'm not sure where it went. What did I do wrong?

 

[Prove It]

Re: Checked answer for seperable equation but not getting right result; missing one x

\(\displaystyle \frac{dy}{dx}+2xy=0\)
\(\displaystyle \frac{dy}{dx}=-2xy\)
\(\displaystyle dy=-2xy dx\)
\(\displaystyle \frac{1}{y} dy=-2x dx\)
integrate both sides
\(\displaystyle \ln{|y|}=-2x+c\)

The integral of -2x is NOT -2x...