Help with advanced differential problem with an initial condition!

Question asks to solve

y' = ln(x) / (xy)

with initial condition y(1)=2

Thanks!

dy/dx = ln (x) / xy

Separate the variables

y dy = ln(x)/x dx

∫ y dy = ∫ ln(x)/x dx --------(1)

∫ ln(x)/x dx = ∫ ln(x) d[ln(x)] = ln^2(x)/2

Equation (1) becomes
y^2 /2 = ln^2(x) /2 + C
y(1)=2
2^2/2 = ln^2(1) /2 + C
2 = 0 + C
C=2

y^2 /2 = ln^2(x) /2 + 2
Multiply both sides by 2

y^2 = ln^2(x) + 1
y = sqrt ( ln^2(x) + 1)

This kind of differencial equation is know a a separable differential equation.

The idea is to utilize differentials in answering it. Take the y' and turn it into dy/dx. Next, try to move all the x related terms to one side and y related terms to the other side, acting as if the dy and dy are parts that can move around in the equation like vatiables. Specifically, you want to make 2 sides of the form of (x stuff)dx = (y stuff) dy. In this example:

dy/dx=ln(x) / (xy), dy = (ln(x) / (xy))dx, (y)dy = (ln(x) / (x))dx ---- notice the set up

Now you (indefinite) integrate both sides. The differentials tell you what to integrate with respect to:

integral (y)dy = integral (ln(x) / (x))dx -> (y^2)/2 +c1 = (ln(x))^2/2 +c2

you are suppose to consolidate the +c on both sides into a single one. Move it to the right to get just one +c : (y^2)/2=(ln(x))^2/2 +c. Now, plug in the given values you got for x=1 and y = 2 and solve for c to get a final answer. I am getting 2. Thus, the solution is :

(y^2)/2 +c1 = (ln(x))^2/2 + 2

You could solve for y if that is what is desired. You should read into simple diff eqns, specifically separable ones since they are simple to do. Also, if you ever just want a plain answer, this is a useful website:
http://www.wolframalpha.com/input/?i=solve+y%27%3Dln%28x%29%2F%28xy%29+with+y%281%29%3D2

y' = ln(x) / (xy)
dy/dx = ln(x){1/xy}
y(dy/dx) = [{ln(x)}/x]dx
integrate
y^2/2 = {ln(x)}^2/2 +k
y^2 = {ln(x)}^2 + C......................(i)
when x=1 then y= 2
4 = 0+C......................(ii)
from (i0 and(ii)
(y)^2 = {ln(x)}^2 + 4...................Ans

This is a separable equation.

y y' = ln(x) / x

Now integrate.

1/2 y^2 = 1/2 (ln(x))^2 + c
y^2 = (ln(x))^2 + c

4 = c

y^2 = (ln(x))^2 + 4