Solving a Separable Equation: What Went Wrong?

[find_the_fun]

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

[math]\frac{dy}{dx}+2xy=0[/math]
[math]\frac{dy}{dx}=-2xy[/math]
[math]dy=-2xy dx[/math]
[math]\frac{1}{y} dy=-2x dx[/math]
integrate both sides
[math]\ln{|y|}=-2x+c[/math]

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