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

find_the_fun

Active member
Feb 1, 2012
166
\(\displaystyle \frac{dy}{dx}+2xy=0\)
\(\displaystyle \frac{dy}{dx}=-2xy\)
\(\displaystyle dy=-2xy dx\)
\(\displaystyle \frac{1}{y} dy=-2x dx\)
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
Jan 26, 2012
1,403
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...