Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 13

For the given problem: yln(x)-xy'=0 , we can evaluate this by applying variable separable differential equation in which we express it in a form of f(y) dy = f(x)dx .
to able to apply direct integration: int f(y) dy = int f(x)dx .
Rearranging the problem:
yln(x)-xy'=0
yln(x)=xy' or xy' = y ln(x)
(xy')/(yx) = (y ln(x))/(yx)
(y') /y = ln(x)/x
Applying direct integration, we denote y' = (dy)/(dx) :
int (y') /y = int ln(x)/x
int 1 /y (dy)/(dx) = int ln(x)/x
int 1 /y (dy)= int ln(x)/x dx

For the left side, we apply the basic integration formula for logarithm: int (du)/u = ln|u|+C
int 1 /y (dy) = ln|y|
For the right side, we apply u-substitution by letting u= ln(x) then du = 1/x dx .
int ln(x)/x dx=int udu
Applying the Power Rule for integration : int x^n= x^(n+1)/(n+1)+C .
int udu=u^(1+1)/(1+1)+C
=u^2/2+C
Plug-in u = ln(x) in u^2/2+C , we get:
int ln(x)/x dx =(ln(x))^2/2+C
Combining the results, we get the general solution for differential equation (yln(x)-xy'=0) as:
ln|y|=(ln|x|)^2/2+C

The general solution: ln|y|=(ln|x|)^2/2+C can be expressed as:
y = C_1e^((ln|x|)^2/2)+C .

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