Q:

What is wrong with the steps of this very basic differential equation solution?

$x\frac {dy}{dx}=2y$
Now of course we can rewrite as $\frac{1}{2y}dy=\frac1xdx$
and end on LHS after integrating with $\ln(2y)$.
But couldn't we think of LHS as $\frac12 \times \frac 1y$
Then when we integrate we just get $\frac12 \int \frac1y dy$ which is just $\ln (y^{\frac12})$
Thanks for any explanation!
Rob

A:

The derivative of $\ln{2y}$ is $\displaystyle 2\cdot \ln'{2y}=2 \cdot \frac{1}{2y}=\frac{1}{y}$. An other way to see it is $\ln{2y}=\ln{2} + \ln{y}$ so it has the same derivative than $\ln $ and the primitive of $\displaystyle \frac{1}{2y}$ is not $\ln{2y}$.
Hence your first integration is wrong, you need to add a factor to integrate and this is exactly what you're doing in the second method, which is correct.

