Calculus: Early Transcendentals, Chapter 9, 9.3, Section 9.3, Problem 12

This is differential equation with separable variables. We can separate the variables by putting y on the left side and dx to right side to get
int y dy=int (ln x)/x dx (1)
This looks like we multiplied the whole equation by y dx but that is not really the case. If you want to learn more about theory behind this I would suggest you read
W. E. Boyce, R. C. DiPrima, Elementary Differential Equations and Boundary Value Problems
or some other book on ordinary differential equations.
Now back to solving the equation by integrating. Let us first integrate the right side because it is harder.
int ln x/x dx=
We make substitution t=ln x=>dx/x=dt
int t dt=t^2/2=(ln^2x)/2+C
Now we return to (1).
y^2/2=(ln^2x)/2+C
y^2=ln^2x+C
y=pm sqrt(ln^2x+C)
Now we use initial condition y(1)=2.
2=pm sqrt(ln^2 1+C)
2=pm sqrt C
Obviously there is no solution for minus sign.
C=4
Therefore, the solution is y=sqrt(ln^2x+4)