Separable differentiable equations Which of the following is a solution to the separable differentiable equation:
$$\frac{dy}{dx}=\frac{xy}{\ln y }$$
$A.\ \displaystyle e^{|x|}$
$B.\ \displaystyle e^{\sqrt{\frac{x^2}2}}$
$C.\ \displaystyle \frac12$$e^{\sqrt{x^2+1}}$
$D.\ \displaystyle e^{-x}$
What I did was separate the functions and get:
$$\frac{\ln y\ dy}{y}=x\ dx$$
Then integrating I get:
$$\frac{\ln^2 y}{2} = \frac{x^2}{2} + C$$
But that doesn't match any of my answers. Help will be appreciated.
 A: $$\ln^2(y)=x^2+C\to \ln(y)=\pm\sqrt{x^2+C}\to y=\exp(\pm\sqrt{x^2+C})$$
A: You were right. We have:
$$\frac{\ln^2 y}{2}=\frac{x^2}{2}+C$$
Now multiply both sides by $2$:
$$\ln^2 y = x^2+C$$
NOTE: $C$ is any arbitrary constant, so we'll just make $2C=C$ because it doesn't really matter (still some constant).
Now you want to isolate for $y$, that's how you'll get the solution. Hence, first take the square root of both sides so you'll be 1 step closer:
$$\ln y = \sqrt{x^2+C}$$
And remember the definition of the natural logarithm. This can be turned into:
$$y=e^\sqrt{x^2+C}$$
Remember that $C$ can be any number, so you can figure it out now.
A: Here this might help:
Okay, so the first thing you do is:
Separate variables on either sides of the equation:
$$\frac{\ln y dy}{y} = x dx$$
If we integrate notice that if we make $u = \ln y$
$$udu = xdx$$
And it becomes an easy integral:
$$\frac{\ln^2 y}{2} = \frac{x^2}{2} + C$$
Multiply both sides by 2:
$$\ln^2 y = x^2 + 2C$$
It doesn't matter what C is, so $C = 2C$ in this instance.
Continuing:
$$\ln y = \sqrt{x^2+C}$$
$$y = e^{\sqrt{x^2+C}}$$
EDIT
The answer can be either A or D, depending on how you look at it.
