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.
A: The integral $\displaystyle{\int \frac{1}{2y}~\mathrm{d}y}$ is not $\ln 2y$. You are right to say that
$$\int \frac{1}{2y}~\mathrm{d}y = \frac{1}{2}\int \frac{1}{y}~\mathrm{d}y = \frac{1}{2}\ln|y| + C$$
For the integral to be an exact log, you need the numerator to be the derivative of the denominator:
$$\int \frac{\mathrm{f}'(x)}{\mathrm{f}(x)}~\mathrm{d}x = \ln|\mathrm{f}(x)| + C$$
To continue with your solution: from $\displaystyle{\frac{1}{2y}~\mathrm{d}y = \frac{1}{x}~\mathrm{d}x}$ we get:
$$\int \frac{1}{2y}~\mathrm{d}y = \int \frac{1}{x}~\mathrm{d}x$$
$$\frac{1}{2}\ln|y| = \ln|x|+C$$
$$\ln|y| = 2\ln|x| + D$$
$$\ln |y| = \ln x^2 + D$$
$$|y| = \mathrm{e}^{\ln x^2 + D}=\mathrm{e}^{\ln x^2}\cdot\mathrm{e}^D$$
$$|y| = Ex^2 \ \ \text{where} \ \ E=\mathrm{e}^D >0$$
$$y = Fx^2 \ \ \text{where} \ \ F \in \mathbb{R}$$
