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#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

\(\displaystyle \frac{dy}{dx}=-2xy\)

\(\displaystyle dy=-2xy dx\)

\(\displaystyle \frac{1}{y} dy=-2x dx\)

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?