# Checked answer for separable equation but not getting right result; missing one x

#### find_the_fun

##### Active member
$$\displaystyle \frac{dy}{dx}+2xy=0$$
$$\displaystyle \frac{dy}{dx}=-2xy$$
$$\displaystyle dy=-2xy dx$$
$$\displaystyle \frac{1}{y} dy=-2x dx$$
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

##### Well-known member
MHB Math Helper
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...