(8, 2) , y' = 2y/(3x) Find an equation of the graph that passes through the point and has the given slope

The given slope equation: y' =2y/(3x) is in form of first order ordinary differential equation. In order to evaluate this, we let y'  as (dy)/(dx) .
(dy)/(dx)=2y/(3x)
Then, express as a variable separable differential equation: N(y) dy= M(x) dx .
To accomplish this, we cross-multiply dx to the other side.
dy=2(ydx)/(3x)
Then, divide both sides by y:
(dy)/y=2(ydx)/(3xy)
(dy)/y=2(dx)/(3x)
To be able to solve for the equation of the graph, we solve for the indefinite integral on both sides.
The problem becomes: int(dy)/y= int 2(dx)/(3x)
For the left side,we integrate int(dy)/y using basic integration formula for logarithm: int (du)/u = ln|u|+C
int (dy)/y = ln|y|
For the right side, we may apply basic integration property: int c*f(x)dx =c intf(x)dx .
int 2(dx)/(3x)=(2/3)int (dx)/(x)
The integral part resembles the basic integration formula for logarithm: int (du)/u = ln|u|+C
(2/3)int (dx)/(x)=(2/3)ln|x|+C.
 
Note: Just include the constant of integration "C" on one side as the arbitrary constant of a differential equation.
Combining the results from both sides, we get the general solution of the differential equation as:
ln|y|=(2/3)ln|x|+C
or y = e^((2/3ln|x|+C))
To solve for the equation of the graph that passes to a particular point (8,2) , we plug-in x=8 and y =2 on the general solution: ln|y|=(2/3)ln|x|+C .
ln|2|=(2/3)ln|3|+C
Isolate C:
C =ln|2|-(2/3)ln|3|
Apply natural logarithm property: n*ln|x|= ln|x^n| and ln|x|-ln|y| = ln|x/y|
C =ln|2|-ln|3^(2/3)|
C=ln|2/3^(2/3)| orln|2/root(3)(9)|
Plug-in C=ln|2/root(3)(9)| on the general solution: y = e^((2/3ln|x|+C)) , we get the equation of the graph that passes through (8,2) as:
y = e^((2/3)ln|x|+ln|2/root(3)(9)|)
 Which simplifies to,
y = e^((2/3)ln(x))*e^(ln(2/root(3)(9)))
 y = 2/root(3)(9)x^(2/3) as the final answer