Differential Equations

show that the general solutions of dy/dx=y/2x

I did:
dy/y=dx/2x
integrate both sides
ln y= ln 2x +a
ln y =ln 2x +ln a
ln y= ln 2ax
y = 2ax
so y^2 = 4a^2x^2
And if I use this method:
ln y = 1/2 ln 2x + ln a
2ln y = ln 2x+ 2ln a
2ln y - 2 ln a= ln 2x
2ln (y/a) = ln 2x
ln (y/a)^2 = ln 2x
y^2/a^2 = 2x
y^2 = 2xa^2

Neither method is giving me the correct answer.

Are you sure about these bits?

Yes.

Then why did you differentiate dx/2x differently for each method?

Any why does 'a' become 'ln a' seemingly at will?

show that the general solutions of dy/dx=y/2x

I did:
dy/y=dx/2x
integrate both sides
ln y= ln 2x +a
ln y =ln 2x +ln a
ln y= ln 2ax
y = 2ax
so y^2 = 4a^2x^2
And if I use this method:
ln y = 1/2 ln 2x + ln a
2ln y = ln 2x+ 2ln a
2ln y - 2 ln a= ln 2x
2ln (y/a) = ln 2x
ln (y/a)^2 = ln 2x
y^2/a^2 = 2x
y^2 = 2xa^2

Neither method is giving me the correct answer.

I thought that the integration of dx/2x in both methods are equivalent, but that didn't happen.
The took the constant to be ln a. Because ln a gives you a constant, so I took that as the constant of integration. I didn't mean to suggest that I turned the constant a, into ln a.

Perhaps post what the solution you've been given is and I can check what they've done from there?

show that the general solutions of dy/dx=y/2x

I did:
dy/y=dx/2x
integrate both sides
ln y= ln 2x +a
ln y =ln 2x +ln a
ln y= ln 2ax
y = 2ax
so y^2 = 4a^2x^2
And if I use this method:
ln y = 1/2 ln 2x + ln a
2ln y = ln 2x+ 2ln a
2ln y - 2 ln a= ln 2x
2ln (y/a) = ln 2x
ln (y/a)^2 = ln 2x
y^2/a^2 = 2x
y^2 = 2xa^2

Neither method is giving me the correct answer.

The book doesn't give answers to the prove that questions. :/

I believe your mistake comes at logging the costant of integration randomly

The integral of $\displaystyle \frac{1}{2x}$ is not equal to $\displaystyle \ln(2x)$!!!!!!!

I'm not doing the integral of 1/2x. I am doing the integral of dx/2x
The diffentiation of ln f(x) = dx/f(x). so the integration of dx/2x should be ln (2x).

I believe your mistake comes at logging the costant of integration randomly

I'm not doing the integral of 1/2x. I am doing the integral of dx/2x
The diffentiation of ln f(x) = dx/f(x). so the integration of dx/2x should be ln (2x).

What do you get if you differentiate $\ln{2x}$ ?

Let me ask another way - you say that those methods are not giving you the correct answer? Then what is the correct answer?

No, the derivative of $\displaystyle \ln(f(x))$ is $\displaystyle \frac{f'(x)}{f(x)}$

It asks you to prove that the general solution is y^2= 4ax. But I am not getting y^2= 4ax.