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#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

\(\displaystyle x \frac{dy}{dx} = 4y\)

becomes \(\displaystyle \frac{1}{4y} dy = \frac{1}{x} dx\) Integrate to get

\(\displaystyle \frac{1}{4} \ln{|y|} = \ln{|x|}+C\)

I'm stuck here because I want to raise e to the power of both sides of the expression like

\(\displaystyle e^{ \frac{1}{4} \ln{|y|}} = e^{\ln{|x|}+C}\) but I'm not sure what affect that would have on \(\displaystyle \frac{1}{4}\)?